Predict the Winner

This can be solved using dynamic programming. The key insight here is to realize that in order to maximize the score, each player will choose a number that maximizes their future score.

When a player makes a choice, there are two possibilities - either the left end or the right end of the remaining array. After this choice, the remaining array and the turn to choose again will be left for the opponent. Since the opponent also plays optimally, he will leave the minimum score for the player. Therefore, the player will choose the one that maximizes his score.

We can use a 2D dp array to represent the maximum score the current player can get from the i-th number to the j-th number. For each range [i, j], there are two options:

  • If the current player takes nums[i], the opponent will face the range [i+1, j]. Since the opponent also plays optimally, he will leave the minimum score for the current player, which is dp[i+1][j]. So the current player can get nums[i] + dp[i+1][j] in total in this case.
  • If the current player takes nums[j], similarly, he can get nums[j] + dp[i][j-1] in total.

Finally, the player wants to maximize his score, so dp[i][j] will be the max of the two options.

Python solution:

class Solution:
    def PredictTheWinner(self, nums: List[int]) -> bool:
        length = len(nums)
        dp = [[0]*length for _ in range(length)]

        for i, value in enumerate(nums):
            dp[i][i] = value

        for i in range(length-2, -1, -1):
            for j in range(i+1, length):
                dp[i][j] = max(nums[i] - dp[i+1][j], nums[j] - dp[i][j-1])

        return dp[0][length-1] >= 0

In the end, if dp[0][length-1] is non-negative, it means the first player’s score is not less than the second player’s score, so the first player can win or tie.

After Translating the Problem into Maximal Sum Difference b/w [A, B]. This problem becomes exactly similar to this one:

It’s insightful to draw a connection between these two problems.

The Stone Game VII problem is also about two players alternately taking stones from either end of a row of stones, and the score is the sum of the values of the stones left in the row. This game also involves strategic thinking about which stone to take to minimize the future potential score of the opponent.

However, the scoring rules in these two games are different. In the Predict the Winner game, the score is the sum of the values of the stones taken by the player, while in the Stone Game VII, the score is the sum of the values of the stones left in the row, not the ones taken.

Despite these differences, the common theme of strategic choice and prediction of the opponent’s optimal moves, makes the dynamic programming approach applicable in both cases. The optimal substructure of these problems allows us to use the results of smaller sub-problems to solve larger ones.

The state transition equation will be different due to the different scoring rules, but the underlying principle of maximizing one’s score while minimizing the opponent’s future score remains the same.

The dynamic programming approach helps to avoid re-computing the same sub-problems multiple times, thereby increasing the efficiency of the solution. However, the implementation details will be different for the two problems due to their different rules and conditions.

Can Bitmask DP be used to attack this problem?

Yes, Bitmask Dynamic Programming (DP) can indeed be used to solve this problem.

The basic idea is that we maintain a state for each possible subset of the numbers in the game (represented as a binary mask), and for each state, we keep track of the maximum possible score difference between the two players if they play optimally.

To start, we initialize a DP array of size 2^n (where n is the length of the input array), and set all elements to a large negative value. We then set DP[0] to 0, because when there are no elements left in the game, the score difference is 0.

Then, for each subset (represented as a binary mask), we iterate through all its elements (represented as bits in the binary mask). For each element, we consider two scenarios: the current player picks it, or they don’t.

If the current player picks it, the score difference changes by the value of the element (if it’s player 1’s turn) or by the negative of the value of the element (if it’s player 2’s turn). We then recursively call our DP function on the remaining elements, and update our DP array with the maximum score difference.

In the end, we return whether DP[2^n - 1] is non-negative. This represents the maximum score difference when all elements have been picked, and if it’s non-negative, player 1 can win or tie.

Please note that the Bitmask DP approach to this problem has a time complexity of O(n*2^n), which can be quite slow for large n. Therefore, it’s more suitable for cases where the length of the input array is small. If the length of the input array is large, a more efficient approach might be needed.

it looks similar to 877(stone game)

Indeed, this problem bears similarities with the problem in LeetCode 877, the Stone Game. Both problems involve two players picking items from an array, starting from either end, with the aim to maximize their total sum.

However, there are some key differences as well. In LeetCode 877, it’s guaranteed that the length of the array is even, and the total sum of the array is also guaranteed to be odd. This means that it’s not possible for the two players to end up with the same total sum, and it’s always possible for the first player to win if they play optimally.

In this problem, on the other hand, it’s not guaranteed that the first player can always win. The players can end up with the same total sum, and the first player wins in the case of a tie. Therefore, while both problems can be solved using a similar dynamic programming approach, the conditions for victory are different.

Can anyone tell the error in this,, thanks in advance

class Solution {

    int dp[501][501];
  int maxStone(int i, int n, int AliceTurn, vector < int > & arr) {
    if (i > n) return 0;
    if (dp[i][n] != -1) return dp[i][n];

    int first = 0, last = 0;
    if (AliceTurn) {
      return dp[i][n] = arr[i] + max(maxStone(i + 1, n, 0, arr), maxStone(i, n - 1, 0, arr));
    } else {
      return dp[i][n] = max(maxStone(i + 1, n, 0, arr), maxStone(i, n - 1, 1, arr));
  bool PredictTheWinner(vector < int > & nums) {
    int n = piles.size();
    memset(dp, -1, sizeof(dp));

    // TURN ==1 -> MEAN ALICE
    int sumAlice = maxStone(0, n - 1, 1, piles);
    int sum = 0;
    for (int i = 0; i < n; i++) sum += piles[i];
    int sumBon = sum - sumAlice;
    cout << sumAlice << " " << sumBon;
    return (sumAlice >= sumBon);

Your function is looking quite right, but there are a few minor issues:

  1. It looks like you have mixed variable names. In your PredictTheWinner function, you’re using the variable piles which hasn’t been defined anywhere. You probably meant to use nums, which is the input to your function.

  2. In your maxStone function, you’re recursively calling it with 0 for AliceTurn in both cases within your if (AliceTurn) block. In the else condition, you should call maxStone(i+1, n, 1, arr) and maxStone(i, n-1, 1, arr), otherwise you’re effectively making it always Alice’s turn and Bob will never play.

Here is the corrected code:

class Solution {

    int dp[501][501];
    int maxStone(int i, int n, int AliceTurn, vector<int>&arr){
        if(i > n)return 0;
        if(dp[i][n] != -1)return dp[i][n];

            return dp[i][n] = arr[i] + max(maxStone(i+1, n, 0, arr), maxStone(i, n-1, 0, arr));
            return dp[i][n] = max(maxStone(i+1, n, 1, arr), maxStone(i, n-1, 1, arr));
    bool PredictTheWinner(vector<int>& nums) {
        int n = nums.size();
        memset(dp, -1, sizeof(dp));
        int sumAlice = maxStone(0,n-1, 1, nums);
        int sum =0;
        for(int i=0; i <n; i++) sum += nums[i];
        int sumBon = sum - sumAlice;
        cout<<sumAlice<<" "<<sumBon;
        return ( sumAlice >= sumBon);

Remember, it’s important to maintain consistency in your variable names to avoid these types of errors. Also, make sure you’re setting the right parameters in your recursive calls.

Test Case # 60 answer is wrong?!

nums = [3606449,6,5,9,452429,7,9580316,9857582,8514433,9,6,6614512,753594,5474165,4,2697293,8,7,1]

A selection: [3606449, 5, 7, 452429, 2697293, 5474165, 6614512, 9, 9857582, 7] B selection: [6, 1, 9, 8, 4, 753594, 6, 8514433, 9580316]

A must be winner.

Answer is True, but test case is False.

The provided test case is a demonstration of the fact that both players are playing optimally. They’re not just picking the largest value from either end, they’re thinking ahead to ensure the highest possible score for themselves.

To make it clearer, let’s run through the first few steps of this game with both players playing optimally:

  1. Player 1 (P1) can choose from 3606449 and 1. P1 picks 3606449, because it’s larger.
  2. Player 2 (P2) can choose from 6 and 1. P2 picks 6, because it’s larger.
  3. P1 can choose from 5 and 1. P1 picks 5, because it’s larger.
  4. P2 can choose from 9 and 1. P2 picks 9, because it’s larger.
  5. P1 can choose from 452429 and 1. P1 picks 452429, because it’s larger.
  6. P2 can choose from 7 and 1. P2 picks 7, because it’s larger.

We can see that both players are taking the higher value available to them, and are not considering future moves. However, optimal play isn’t about maximizing the immediate gain. It’s about thinking ahead and choosing values that will lead to a higher final score.

In the given case, the returned answer is false because with optimal play, Player 1 does not win. Even though it seems like Player 1 can win with the selected values, keep in mind that the order of the choices and the strategy used to make these choices matter significantly, which is why your test case returns False.

Let’s illustrate the decision process with a small example, [1, 5, 233, 7], where we already know that Player 1 can win. The process can be represented in the following table:

TurnPlayerAvailable ChoicesChoicePlayer 1 ScorePlayer 2 Score
111, 7110
225, 7717
315, 2332332347

Player 1 is making their choices to ensure that Player 2 always has to choose between two numbers where one is significantly smaller than the other. In the second round, Player 1 forces Player 2 to choose between 5 and 7 instead of, say, 7 and 233, thus ensuring that they can grab the large number in the third round.

This is a simplified example, and the real game has more rounds and requires more complex strategic thinking. But it illustrates the basic idea: Playing optimally isn’t just about grabbing the largest available number, it’s also about controlling the choices available to the other player.

If you want to create a similar table for a larger array, the table will have more rows and the strategic decisions will be more complex, but the basic principle is the same.

it looks similar to 877(stone game)

Yes, you’re right. This problem does have similarities with LeetCode problem 877 “Stone Game”. In both problems, two players are playing a game where they can choose an element from either end of the array, and the goal is to maximize their score.

The key difference between the two problems lies in the win condition:

  • In the Stone Game problem, Alex (the first player) always wins if he plays optimally. The proof for this is outside the scope of this explanation, but it has to do with the fact that the number of piles is even and the total number of stones is odd. Therefore, there is no need to calculate the actual scores for each player.

  • In the “Predict the Winner” problem, whether player 1 wins can vary depending on the specific values in the array. Therefore, you do need to calculate the scores for each player.

Because of this difference, the “Predict the Winner” problem is slightly more complex than the Stone Game problem. It requires a dynamic programming solution to calculate the maximum score for each player, whereas the Stone Game problem does not.


Player 1 will win ,the answer must be TRUE for this test case,but the expected answer is FALSE ,any idea ?

Let’s analyze the case you provided:

Here, the array is [2, 4, 55, 6, 8].

The game would progress as follows:

  1. Player 1 can choose between 2 and 8. If player 1 chooses 8 (which is the optimal choice because 8 is bigger than 2), the array becomes [2, 4, 55, 6].

  2. Player 2 can choose between 2 and 6. If player 2 chooses 6 (the optimal choice), the array becomes [2, 4, 55].

  3. Player 1 can choose between 2 and 55. The optimal choice is 55, so the array becomes [2, 4].

  4. Player 2 can choose between 2 and 4. The optimal choice is 4, leaving [2].

  5. Player 1 can only choose 2.

So, the final scores would be:

  • Player 1: 8 + 55 + 2 = 65
  • Player 2: 6 + 4 = 10

Player 1 wins in this case. Hence, for this particular test case, the output should be true.

It seems like there might be an error in the expected answer provided. If the game is being played optimally by both players, player 1 should be able to win with this array.

Always remember, the players are playing optimally, they will always choose the move that maximizes their final score.

bool PredictTheWinner(vector & nums) {
  int s1 = 0;
  int s2 = 0;
  for (int i = 0; i < nums.size(); i++) {
    if (i % 2 == 0) {
      s1 += nums[i];
    } else {
      s2 += nums[i];
  if (s1 >= s2) {
    return true;
  return false;

for this code , {1,2} , it should return false, but compiler showing true , can anyone tell the problem , please?

The code you posted doesn’t seem to follow the rules of the game as described in the problem statement. You’re simply taking alternate elements starting from the beginning of the array, assuming player 1 always takes the elements at even indices and player 2 always takes the elements at odd indices. This doesn’t follow the rule of the game where players can pick from either end of the array.

Let’s consider the array [1,2]. Player 1 can take 2 (the optimal choice) from the end, then Player 2 takes 1. So Player 1 has a total of 2 and Player 2 has a total of 1. Therefore, Player 1 wins, and the output should be true.

But if you run your code on this input, it will output false, because it just adds up the elements at even indices (1 in this case) for Player 1 and the elements at odd indices (2 in this case) for Player 2, incorrectly determining that Player 2 wins.

Your code does not account for the fact that players are playing optimally, meaning they will always choose the move that maximizes their final score, and can choose from either end of the array.

For the correct solution, consider using a dynamic programming approach where you store the maximum score difference a player can achieve over their opponent when it’s their turn, given a subarray of the original array.

The problem description is vague. What if we go down the decision tree deep and both players are winning randomly when we hit the base condition. What should be returned?

The problem assumes that both players are playing optimally. That means they will always make the decision that leads to the maximum score for themselves. So in this game, randomness doesn’t play a part; every decision is made with the intention of maximizing the player’s own score or minimizing the opponent’s score.

When we reach the base condition, it means there are no more elements to pick from the array. At that point, if Player 1’s score is equal to or higher than Player 2’s, Player 1 is considered the winner.

As for the decision tree, you don’t necessarily have to explore all the possible moves. Instead, you can apply a dynamic programming approach to remember the score difference for each possible range of the array, reducing the time complexity of the problem.

I do not understand example 1 nums = [1,5,2] plyer 1 takes [1] player 2 takes [2] player 1 takes [5]

player 1 can still win and it returns True..?

Your misunderstanding comes from the assumption that players can choose from any position in the array, but according to the problem statement, a player can only choose from either end of the array. In other words, they can only pick the first or the last element in the array.

So for the array [1,5,2], here is how the game could play out optimally:

  • Player 1 can choose between 1 and 2, let’s say Player 1 chooses 1.
  • Now the array is [5,2]. Player 2 can choose between 5 and 2, and will choose 5 (the higher number) because both players are playing optimally.
  • Finally, the array is [2], and Player 1 has no choice but to pick 2.

So the final score would be:

  • Player 1: 1 + 2 = 3
  • Player 2: 5

Player 2 ends up with a higher score, so Player 1 does not win. The function would return False for the input [1,5,2].

Array - [0,0,7,6,5,6,1] Index - [0,1,2,3,4,5,6]

Can someone please explain why P1 won’t win in this case. P1 selects 6th Index First Now P2 turns, Array contains Even Element, and P1 will get 12 and P2 will get 12. So P1 sum will become 13 and P2 will be 12.

In the given array [0,0,7,6,5,6,1], we assume that both players play optimally, meaning they will always make the decision that leads to their highest possible final score.

Let’s say Player 1 (P1) indeed starts by choosing 6th index, which is 1.

Now the array is [0,0,7,6,5,6] and it’s Player 2’s (P2) turn. P2 can choose either from the start or end of the array, and being optimal P2 will choose 6, the largest number.

The array is now [0,0,7,6,5] and it’s P1’s turn. P1 will choose 5.

The array becomes [0,0,7,6] and P2 chooses 6.

The array now is [0,0,7] and P1 can only choose 7, as it is the larger number.

Finally, the array is [0,0] and P2 can choose either number as they are both 0.

So, the final score would be:

  • Player 1: 1 + 5 + 7 = 13
  • Player 2: 6 + 6 = 12

P1 does win in this case if they both play optimally. The result should be true.

If you’re getting a different result from a function or code, it might not be implementing the game’s rules correctly or the strategy for optimal play might not be implemented correctly.

[2,4,55,6,8] I’m stuck with this test case and I do not understand why it’s expected FALSE

The game involves two players, Player 1 (P1) and Player 2 (P2), who alternately take one number from either end of the array, and add that number to their respective scores. The goal is to maximize one’s own score.

Let’s examine the given array: [2,4,55,6,8]

First, observe that both players are supposed to play optimally. This means they will each take the step that maximizes their potential score. Let’s walk through the game:

  1. P1 has the first move and will choose between 2 and 8. P1 chooses 8 (the maximum value) because P1 is playing optimally. The array becomes: [2,4,55,6]

  2. Now, it’s P2’s turn. P2 will choose between 2 and 6. P2 chooses 6 (the maximum value). The array becomes: [2,4,55]

  3. P1 now chooses between 2 and 55. P1 chooses 55 (the maximum value). The array becomes: [2,4]

  4. P2 now chooses between 2 and 4. P2 chooses 4 (the maximum value). The array becomes: [2]

  5. Finally, P1 picks the last remaining number, 2.

Summing up the scores:

  • P1: 8 + 55 + 2 = 65
  • P2: 6 + 4 = 10

Clearly, P1 wins.

If you’re getting a result that indicates that P2 wins (i.e., FALSE), then there might be an issue with your algorithm implementation. It’s crucial to ensure that your approach correctly implements the rules of the game and the optimal play strategy.

10 Prerequisite LeetCode Problems

This problem is a classic example of a game theory problem, involving dynamic programming and recursive strategies. It requires you to consider two players picking numbers from an array and predict who will win based on the sum of the numbers they have picked.

Here are some problems to prepare for it:

  1. 70. Climbing Stairs: This problem introduces you to basic dynamic programming concepts, which are vital for the “Predict the Winner” problem.

  2. 53. Maximum Subarray: Understanding how to find the maximum subarray sum can be beneficial when trying to maximize the sum in a game.

  3. 198. House Robber: This problem involves picking numbers from an array in a way that maximizes the total, a concept that’s very useful for “Predict the Winner”.

  4. 322. Coin Change: This problem requires a dynamic programming approach to find the fewest number of coins that you need to make up a given amount.

  5. 300. Longest Increasing Subsequence: This problem requires finding a subsequence in a way that maximizes a certain criterion, which is a common theme in game theory problems.

  6. 416. Partition Equal Subset Sum: This problem requires determining if a set can be partitioned into two subsets of equal sum, similar to how “Predict the Winner” requires partitioning the numbers between two players.

  7. 1143. Longest Common Subsequence: This problem asks for a dynamic programming solution to find the length of the longest common subsequence, a useful exercise in managing two separate sequences.

  8. 486. Predict the Winner: This is the actual problem, but it’s worth attempting multiple times and reviewing your solution to improve your understanding of the problem.

  9. 877. Stone Game: This problem is very similar to “Predict the Winner”, but with the added complexity of having piles of different sizes.

  10. DP Strategy Game Problems: More generally, any problems involving game strategy and dynamic programming will be helpful, as they’ll get you thinking in the right direction for “Predict the Winner”.

The key to these problems is understanding how to optimize your strategy based on the current state of the game and the decisions you can make. Understanding dynamic programming and recursion is key to these types of problems.

Problem Analysis and Key Insights

What are the key insights from analyzing the problem statement?

Problem Boundary

What is the scope of this problem?

How to establish the boundary of this problem?

Problem Classification

Domain Categorization

The problem is in the domain of algorithms, specifically in the category of game theory and dynamic programming.

What Components

  1. Integer Array (nums) - The game board.
  2. Two Players - Player 1 and Player 2.
  3. Score - Initially 0 for both players.
  4. Turn-Based Gameplay - Players take turns to select numbers.
  5. Selection Rule - Players can choose a number from either end of the array.
  6. Win Condition - Player 1 wins if their score is equal to or greater than Player 2’s score.

Further Classification

  1. Input/Output: Given an array, the function returns a boolean value indicating if Player 1 can win or not.
  2. Constraints: Array size and element constraints are provided.
  3. Optimization Problem: Both players are assumed to be playing optimally, implying we need to find the best strategy.
  4. Game Theory: Decision-making is crucial here, as choices directly impact the end result.
  5. Dynamic Programming: Due to the nature of the problem (optimal substructure, overlapping subproblems), dynamic programming could be a useful approach.


The problem can be understood as a game theory problem with two players where each player tries to maximize their score under certain rules. It’s also an optimization problem because we assume that both players are playing optimally. Dynamic programming is often used for solving such kinds of optimization problems.

Distilling the Problem to Its Core Elements

Fundamental Concept

The fundamental concept this problem is based upon is “Minimax,” which is a decision rule used in game theory and decision theory for minimizing the possible loss or maximizing the potential gain. In this case, each player tries to maximize their score while minimizing the opponent’s score. Dynamic programming can be used to solve such Minimax problems efficiently.

Simplest Description

Two players take turns picking numbers from either end of a list. They add these numbers to their score. The first player wins if their score is equal to or higher than the second player’s score at the end. Can the first player always win?

Core Problem

The core problem is to find out if the first player has a strategy to win the game, assuming both players are making the best possible moves.

Key Components

  1. Array of Integers: Represents the game board.
  2. Two Players: Player 1 and Player 2.
  3. Turn-Based Gameplay: Alternating turns between the two players.
  4. Selection from Ends: Players can only choose numbers from either end of the array.
  5. Scoring: Accumulation of selected numbers as scores.
  6. Win Condition: First player wins if their final score is equal to or greater than the second player’s.

Minimal Set of Operations

  1. Initialize Scores: Zero for both players.
  2. Turn Taking: Determine which player’s turn it is.
  3. Selection: Choose the optimal number from either end of the array.
  4. Update Array: Remove the chosen number from the array.
  5. Update Score: Add the chosen number to the player’s score.
  6. Check Win Condition: Evaluate if Player 1 wins based on the final scores.

By focusing on these key components and minimal operations, you can create a structured approach to solve this problem.

Visual Model of the Problem

Visualizing the Problem Statement

To visualize the problem, you can use a few different approaches:

1. Number Line or Array Representation:

Imagine the given array as a physical line of numbers. Place markers at both ends to represent where players can make their selections.

Array: [1, 5, 2]

1 - 5 - 2
^       ^

2. Decision Tree:

Each decision leads to a new state of the array, creating branches in the tree. The nodes of the tree represent the current array state, whose turn it is, and the scores.

    [1,5,2] (P1: 0, P2: 0)
    /                   \
[5,2] (P1: 1, P2: 0)   [1,5] (P1: 2, P2: 0)
    |                    |
...                    ...

3. Scoreboard:

Maintain a virtual scoreboard to keep track of each player’s score as you traverse through decisions.

Player 1: 0
Player 2: 0

4. Move History:

You can visualize past moves to understand how each player’s strategy unfolds. For instance, for array [1,5,2]:

Move 1: Player 1 picks 2
Move 2: Player 2 picks 5
Move 3: Player 1 picks 1

These visualization techniques can be used individually or in combination to gain a clearer understanding of the problem. They allow you to better grasp the decision-making process, the game state after each move, and the scoring mechanics.

Problem Restatement

Two players are playing a turn-based game using an array of integers. Player 1 starts, and each player can pick a number from either end of the array on their turn. They add this number to their score. The game ends when the array is empty. The goal is to determine if Player 1 can win or tie the game when both players are making the best choices to maximize their scores. Player 1 wins if their score is the same or higher than Player 2’s.

Essential Elements:

  1. Array of Integers: The ‘board’ for the game.
  2. Two Players: Each with a score, starting at zero.
  3. Turn-based: Player 1 goes first.
  4. Number Selection: From either end of the array.
  5. Score Accumulation: Adding chosen numbers to players’ scores.
  6. End Condition: The array becomes empty.
  7. Winning Criteria: Player 1 wins if their score >= Player 2’s score.

Requirements and Constraints:

  • Array length is between 1 and 20.
  • Each array element is between 0 and 10^7.
  • Players play optimally to maximize their score.

By paraphrasing the problem like this, we clarify what exactly we need to focus on. Now we can better target our approach for solving it.

Abstract Representation of the Problem

In an abstract setting, consider a sequence ( S ) of ( n ) elements. Two agents, ( A ) and ( B ), alternate in selecting an element from either the start or end of ( S ). Each element has a value ( v ), and the agents accumulate the value of their selected elements in a sum ( \Sigma ).

Agent ( A ) goes first and aims to maximize ( \Sigma_A ), while ( B ) aims to maximize ( \Sigma_B ). The sequence ( S ) becomes empty after ( n ) turns.

The question: Is there a strategy for ( A ) such that ( \Sigma_A \geq \Sigma_B ), assuming both agents make choices to maximize their respective sums?

Key Elements:

  1. Sequence ( S ): A finite, ordered set of ( n ) elements.
  2. Agents ( A ) and ( B ): Two participants making selections.
  3. Turns: ( A ) and ( B ) alternate, starting with ( A ).
  4. Selection: An element ( v ) from either end of ( S ).
  5. Accumulated Sum ( \Sigma ): The total value each agent accumulates.
  6. Optimal Strategy: Agents choose elements to maximize ( \Sigma ).

This abstract representation captures the essence of the problem, stripping it of specific real-world details and emphasizing the structure and key elements.


  1. Turn-Based Game: A type of gameplay where players alternate turns. Understanding the turn-based nature is key for simulating the game.

  2. Minimax Algorithm: A decision rule for minimizing the worst-case scenario or maximizing the best-case scenario. This is essential for modeling optimal play by both players.

  3. Dynamic Programming: A method for efficiently solving a complex problem by breaking it down into simpler subproblems. This is often used to solve problems like this one, where past decisions affect future ones.

  4. Game State: The current situation in the game, which in this case includes the remaining elements in the array and the scores of both players. Keeping track of game states is crucial for implementing the solution.

  5. Game Theory: The study of mathematical models of strategic interaction. Understanding the basics can help you realize why each player aims to maximize their score.

  6. Optimal Strategy: The best possible way to play to achieve the objective, often used in the context of game theory. Both players in this problem are assumed to use optimal strategies.

  7. Memoization: Storing the results of expensive function calls and reusing them when the same inputs occur again. This concept often appears in dynamic programming solutions to improve efficiency.

  8. Base Case and Recursive Case: In dynamic programming, the base case is the simplest, smallest instance of the problem that can be solved directly. The recursive case builds upon the base case(s) to solve more complex instances of the problem.

  9. Constraints: The limits within which the problem must be solved (e.g., size of the array, range of numbers). These often influence the choice of algorithm in terms of time and space complexity.

Each of these terms plays a role in defining, understanding, or solving the problem. Understanding these can provide a solid foundation for tackling the problem efficiently.

Problem Simplification and Explanation

Breaking Down into Simpler Terms

  1. Array of Numbers: Think of this as a row of coins on a table.
  2. Two Players: Two people, Player 1 and Player 2, are playing a game.
  3. Turns: Each person takes a turn, one after the other, starting with Player 1.
  4. Pick from Ends: On each turn, a player can only take a coin from one of the ends of the row.
  5. Scoring: The value of the coin goes to the player who picks it.
  6. Winning: Player 1 wins if they collect coins worth as much as or more than Player 2.

Key Concepts and Interactions

  1. Turn-Based Mechanism: One person picks a coin, then the other person picks a coin, and it keeps going like this.
  2. Strategy: Both players are smart and will try to get the most valuable coins.
  3. End of Game: The game ends when there are no more coins to pick.
  4. Win Condition: At the end, if Player 1 has a total coin value equal to or higher than Player 2, then Player 1 wins.

Metaphor or Analogy

Imagine you and a friend are in a candy store. There’s a row of jars of candy in front of you. Each jar has a label that tells you how many pieces of candy are in it. You and your friend take turns picking a jar from either end of the row. Whoever ends up with more or equal pieces of candy wins.

The catch is, you both always want to pick the jar that has the most candy to get ahead. The question is, if you go first, can you always win or at least tie the game?


Understanding the characteristics and constraints of the problem can guide us toward an efficient solution. Here’s what can be useful:

Characteristics to Exploit:

  1. Optimal Play: Both players aim to maximize their scores. This behavior can be used to predict their moves and plan accordingly.

  2. Array Size Constraint (1 <= nums.length <= 20): The maximum array size is small, making it suitable for algorithms that might not scale well for larger sizes. Dynamic programming or recursion could be useful here.

  3. Limited Number Range (0 <= nums[i] <= 10^7): The values are bounded, which can help in calculations, especially if you’re considering bitwise operations or using an array to track states.

  4. Turn-Based Nature: This allows us to consider the problem from a “single-player” perspective by alternating between maximizing and minimizing scores during recursive evaluation.

  5. Symmetry: The players can only pick from the two ends of the array. This imposes a certain symmetry on the problem. If one end is more favorable than the other at any point, the opposite will also be true in a later similar state but with roles reversed.

Patterns and Numerical Ranges:

  1. Even-Odd Lengths: The outcome may depend on whether the array length is odd or even. If it’s odd, Player 1 has an extra turn, which could be an advantage.

  2. High-Low Value Ends: If high values are clustered at one end, it may become a focal point for strategy.

  3. Contiguous Subarray Sums: Understanding which contiguous subarrays yield the highest sums might help predict optimal plays.

Understanding these specific elements enables us to consider targeted strategies and algorithms that operate efficiently within these constraints and characteristics.

The key insights drawn from analyzing the constraints are:

  1. Small Array Size: The maximum length of the array is 20. This small size opens the door for solutions that involve recursion or dynamic programming, even if they are not the most time-efficient for larger data sets.

  2. Bounded Values: The elements in the array range from 0 to (10^7). This constraint can simplify calculations and decision-making during the algorithm design.

  3. Optimal Play: Both players are assumed to make the best possible moves. This constraint implies that we can apply a minimax algorithm to model the behavior of the players.

  4. Even vs. Odd Array Length: Since Player 1 goes first, an odd-length array would give Player 1 an extra move. This could be an advantage that affects the outcome.

These insights can be harnessed to create an algorithm that is both efficient and effective for the problem at hand.

Case Analysis

Certainly. Below are different test cases that encompass various facets of the problem.

Base Case

Case: Single Element

  • Input: [5]
  • Output: true
  • Reasoning: With only one element, Player 1 picks it and wins because Player 2 has no moves left.

Small Arrays

Case: Two Elements of Equal Value

  • Input: [3, 3]
  • Output: true
  • Reasoning: Player 1 picks one of the 3s and wins since both will end up having 3.

Case: Two Elements of Different Values

  • Input: [2, 4]
  • Output: true
  • Reasoning: Player 1 picks 4 and wins as Player 2 can only pick 2.

Uneven Length Arrays

Case: Favorable Extra Turn

  • Input: [1, 2, 1]
  • Output: true
  • Reasoning: With an extra turn, Player 1 can win by picking 1, 1, while Player 2 can only pick 2.

Even Length Arrays

Case: Unfavorable Distribution

  • Input: [2, 10, 1, 2]
  • Output: false
  • Reasoning: Player 1 picks 2, then Player 2 picks 10, and the game ends with Player 1 only being able to make their score 4, while Player 2’s score is 10.

Large Elements

Case: High-Value Edges

  • Input: [10000000, 1, 1, 10000000]
  • Output: true
  • Reasoning: Player 1 picks one of the edges (10,000,000), ensuring a win, as Player 2 can only end up with a score of 1 + 10,000,000.

Edge Cases

Case: All Elements are Zero

  • Input: [0, 0, 0, 0]
  • Output: true
  • Reasoning: All values are 0. Player 1 will have at least as much as Player 2, fulfilling the win condition.

Case: Descending Order

  • Input: [5, 4, 3, 2, 1]
  • Output: true
  • Reasoning: Player 1 picks 5, Player 2 picks 4, then Player 1 picks 3, ending up with 8 vs. Player 2’s 4, guaranteeing a win.

Each test case provides a unique look into the behavior of the system, considering the key constraints and potential pitfalls, making sure that the solution handles all scenarios.

Analyzing the different cases provides several key insights:

  1. Starting Advantage: When Player 1 starts, they have the first pick, which can offer an initial advantage. In uneven-length arrays, this advantage becomes even more pronounced, as Player 1 gets an extra turn.

  2. Value Clustering: If high or low values are clustered at one end, the first player to pick from that end can secure a significant advantage.

  3. Array Length: Even-length arrays can make the game more balanced, whereas odd-length arrays give Player 1 an additional turn, which can be a decisive factor in winning.

  4. Element Distribution: If the array elements are not uniformly distributed, and especially if high values are towards the center, Player 2 may have a better chance of winning.

  5. Zero Elements: When all elements are the same (e.g., all zeros), Player 1 will always win because they can match or exceed Player 2’s score.

  6. Single-Element Array: In a single-element array, Player 1 always wins as they are the only one who gets to pick an element.

These insights help us understand the nuances of the problem, enabling us to consider all possible scenarios when devising a solution.

Identification of Applicable Theoretical Concepts

Various mathematical and algorithmic concepts can be applied to this problem to simplify it or make it more manageable:

Minimax Algorithm

The game can be framed as a zero-sum game where one player tries to maximize the score while the other tries to minimize it. The Minimax algorithm is well-suited for such scenarios.

Dynamic Programming

The optimal choice for a subarray can be derived from smaller subarrays. Thus, Dynamic Programming (DP) can help store and reuse these computations to make the algorithm more efficient.


In a recursive solution, the same subproblems can occur multiple times. Memoization can store the results of these subproblems to avoid redundant calculations.

Game Theory

Certain Game Theory principles can be applied here. For instance, finding a “Nash Equilibrium” can offer insights into what an optimal strategy might look like for both players.


The problem has an inherent recursive structure: the optimal move at any stage can be determined by considering optimal moves in subsequent stages.

Greedy Algorithm

Though not sufficient alone, greedy choices can sometimes give insights or simplify subproblems. For instance, if high-value elements are clustered at one end, a greedy approach might dictate taking from that end.

Turn-based Strategy

Understanding that the game is turn-based allows us to simplify the problem by considering it from a single-player perspective and then alternating the maximizing and minimizing roles.

By applying these mathematical and algorithmic concepts, the problem becomes more tractable, allowing us to develop an efficient and effective solution.

Simple Explanation

Can you explain this problem in simple terms or like you would explain to a non-technical person? Imagine you’re explaining this problem to someone without a background in programming. How would you describe it? If you had to explain this problem to a child or someone who doesn’t know anything about coding, how would you do it? In layman’s terms, how would you explain the concept of this problem? Could you provide a metaphor or everyday example to explain the idea of this problem?

Problem Breakdown and Solution Methodology

Given the problem statement, can you explain in detail how you would approach solving it? Please break down the process into smaller steps, illustrating how each step contributes to the overall solution. If applicable, consider using metaphors, analogies, or visual representations to make your explanation more intuitive. After explaining the process, can you also discuss how specific operations or changes in the problem’s parameters would affect the solution? Lastly, demonstrate the workings of your approach using one or more example cases.

Inference of Problem-Solving Approach from the Problem Statement

The key terms and their impact on the solving strategy are as follows:

1. Integer Array

  • Impact: The array type specifies that we are dealing with discrete values, ruling out the need for complex mathematical operations.

2. Two Players

  • Impact: Since it’s a two-player game, strategies like Minimax can be directly applicable to model both players’ optimal behavior.

3. Turn-based

  • Impact: This term guides us toward a recursive or Dynamic Programming approach as we can break the problem down into subproblems based on whose turn it is.

4. Score

  • Impact: The objective of the game is to maximize this metric, focusing the problem on optimization methods.

5. Optimal Play

  • Impact: This assumption means both players will make the best moves, making the Minimax algorithm an ideal candidate for solving the problem.

6. Array Ends

  • Impact: Players can only choose from either end of the array, which simplifies the choices available at each step and enables a more efficient algorithm.

7. Constraints (Array Length and Element Range)

  • Impact: Knowing the constraints informs us that brute-force approaches are also viable due to the small array size, though they may not be the most efficient.

8. Winner Conditions

  • Impact: Knowing that a tie also means Player 1 wins helps in defining the base case in recursion or the initial state in DP.

Each of these key terms or concepts directly informs the strategy and methods that could be employed to solve the problem effectively.

How did you infer from the problem statement that this problem can be solved using ?

Simple Explanation of the Proof

I’m having trouble understanding the proof of this algorithm. Could you explain it in a way that’s easy to understand?

Stepwise Refinement

1. Stepwise Refinement of Approach

Step 1: Problem Understanding

  • Read and understand the problem statement thoroughly, paying special attention to the rules, constraints, and conditions for winning.

Step 2: Choose Methodology

  • Decide on an algorithmic approach. The Minimax algorithm is highly suitable here, possibly enhanced with Dynamic Programming for efficiency.

Step 3: Define Data Structures

  • Choose appropriate data structures for storing scores, array state, and intermediate results (if using DP or memoization).

Step 4: Base Case and Initialization

  • Initialize the score for both players as 0. Set up the base case where the array is empty or contains a single element.

Step 5: Recursion or Iteration

  • Implement the main logic using recursion or iteration, considering both players’ turns.

Step 6: Memoization (Optional)

  • If using recursion, add memoization to store and reuse results of subproblems.

Step 7: Return Result

  • Determine the winner based on the final scores and return the result.

2. Granular, Actionable Steps

  • Step 2.1: If using Minimax, define a function that takes the current state and returns the maximum attainable score.
  • Step 2.2: Within the Minimax function, consider both choices: taking the first or last element.
  • Step 3.1: If memoizing, create a hash table to store results of subproblems.
  • Step 5.1: In the main logic, make a choice at each turn that maximizes your score while minimizing the opponent’s score.
  • Step 5.2: Recursively call the Minimax function for the remaining array and the other player’s turn.

3. Parts that Can Be Solved Independently

  • Memoization logic can be implemented independently of the main algorithm.
  • Determining the base case and initialization can also be isolated.

4. Repeatable Patterns

  • The choice of taking an element from either end of the array is a repeatable action.
  • The action of alternating turns between two players is a consistent pattern.
  • If using recursion, the pattern of solving the problem for smaller subarrays and building up the solution is repeatable.

By breaking down the problem and solution into these elements, we can systematically approach solving it.

Solution Approach and Analysis

Let’s think of this problem as two players picking coins from a line of coins. Each coin has a different value, and the goal is to end up with a sum of coin values that’s greater than or equal to the opponent’s sum. The catch is, you can only pick coins from either end of the line.

Step-by-Step Solution

  1. Understand the Rules: Know that Player 1 always goes first and that both players will play optimally.

  2. Initial Setup:

    • Initialize scores for Player 1 and Player 2 to zero.
    • Create a memoization table to store intermediate results.
  3. Start the Game:

    • For each turn, define a function, say bestScore, that will return the best score achievable from that point onward.
  4. Base Case:

    • If only one coin is left, the best score is simply its value. Return it.
    • If no coin is left, the best score is zero. Return it.
  5. Choice Making:

    • At each turn, you have two choices: pick the coin at the beginning or the end of the line.
    • For each choice, use recursion to calculate the opponent’s best score and subtract it from your chosen coin’s value to get your net score.
  6. Memoization:

    • Before recursion, check if this subproblem was already solved (look it up in the memoization table).
    • After computing the score for a subproblem, store it in the memoization table.
  7. Result Extraction:

    • At the root call, compare the scores. If Player 1’s score is >= Player 2’s score, return True. Otherwise, return False.

How Parameters Affect the Solution

  • Array Length: As the length increases, computational time will increase, although using memoization can mitigate this.

  • Element Values: If all elements are equal, Player 1 will always win or tie, making the problem trivial.

  • Player Strategies: The assumption that players play optimally is crucial. If they don’t, then the Minimax approach won’t be accurate.

Example Case: [1, 5, 2]

  1. Initial Setup: Player 1 Score = 0, Player 2 Score = 0, Memoization Table = {}

  2. Start the Game: Call bestScore([1, 5, 2])

  3. Choice Making:

    • If Player 1 picks 1, then Player 2 can pick 5. Net score = 1 - 5 = -4
    • If Player 1 picks 2, then Player 2 can pick 5. Net score = 2 - 5 = -3
  4. Memoization: Store these results in the table for quick future lookup.

  5. Result Extraction: The best score Player 1 can achieve is -3, which is less than Player 2’s score of 0. So, return False.

By following this approach, we can determine whether Player 1 can win given a certain array of integers.

Identify Invariant

The invariant in this problem is the rule that each player can only pick one number from either the beginning or the end of the array during their turn. This rule never changes and applies throughout the game, regardless of the state of the array or the scores of the players. It forms the basis for constructing the algorithm and influences how both players make their choices optimally.

Identify Loop Invariant

In the context of this problem, if you were to implement it iteratively, a loop invariant could be the remaining subarray that players can pick numbers from. Specifically, after each iteration (or recursive call in a recursive approach), the remaining subarray should always be a valid subarray from which the next player can choose a number. This means it should always be a contiguous sequence from the original array and its length should be reduced by exactly one element after each turn.

Another loop invariant could be the sum of scores for Player 1 and Player 2. At any point in the game, the sum should equal the sum of the elements that have been picked so far. This invariant helps ensure the integrity of the game logic, validating that scores are being updated correctly.

These invariants help ensure the correctness of the solution by providing conditions that must hold true before and after each iteration or recursive call.

The invariant and loop invariant are not the same for this problem.

  • Invariant: This is a rule or condition that holds true throughout the problem. In this problem, the invariant is the rule that a player can only pick from either end of the array on their turn. This holds true no matter what subarray is being considered or whose turn it is.

  • Loop Invariant: This is a condition or set of conditions that holds true before and after each iteration of a loop (or recursive call in a recursive solution). In this problem, a loop invariant could be the integrity of the remaining subarray from which players can choose numbers or the sum of the scores for both players.

The invariant serves as a consistent rule for the game, while the loop invariant helps verify the correctness of each iterative or recursive step in your algorithm.

Thought Process

Basic Thought Process

  1. Understand the Game Rules: The first step is to understand how the game works. Both players take turns picking numbers from either end of the array.

  2. Identify the Goal: The problem asks us to find out if Player 1 can win or tie the game. We have to find a way to simulate optimal play for both players to reach a conclusion.

  3. Consider Strategy: Players are playing optimally, meaning they will make the best possible move at each step. The challenge is how to calculate these moves efficiently.

  4. Recursive Insight: At any turn, a player has two choices. Each of these choices leads to a new state of the game. This hints at a recursive approach.

  5. Memoization: The problem has a small constraint size (up to 20 elements), but calculating each possibility individually could still be costly. Memoization can help avoid recalculating the same game states.

  6. Implement: Code the algorithm considering all the above insights and validate it using test cases.

Python Code

Here’s how you can implement the solution in Python3:

def PredictTheWinner(nums):
    memo = {}  # To store already computed game states

    def bestScore(start, end):
        # If this subproblem is already solved, return the answer
        if (start, end) in memo:
            return memo[(start, end)]

        # Base case: one element left
        if start == end:
            return nums[start]

        # Recursive case: Player chooses an end, subtracting the opponent's best score
        pick_start = nums[start] - bestScore(start + 1, end)
        pick_end = nums[end] - bestScore(start, end - 1)

        # Store the best score in the memo table and return it
        memo[(start, end)] = max(pick_start, pick_end)
        return memo[(start, end)]

    # If Player 1's best score is >= 0, they can win or tie
    return bestScore(0, len(nums) - 1) >= 0


Cues and Insights from Problem Statement

  • The game ends when there are no more elements in the array: This tells us we need a base case for our recursion.
  • Both players are playing optimally: This suggests we may need to use a Minimax algorithm or a similar optimal strategy.
  • You can choose from either end of the array: This provides the choices that make up the branching factor in our recursion.

Applying these insights and cues helps us design a clean and effective solution to the problem.

Establishing Preconditions and Postconditions


  1. Inputs: The input to the method is nums.
  2. Types: nums is a list of integers.
  3. Representation: nums represents the integer array in the context of the game, from which players choose numbers.


  1. State of Program: There is no specific state that the program needs to be in before this method is called.
  2. Constraints:
    • The length of the array nums must be between 1 and 20.
    • Each integer in the array nums must be between 0 and 10^7.

Method Functionality:

  1. Expected to do: The method is expected to determine if Player 1 can win or tie the game while both players are playing optimally.
  2. Interaction: It iteratively (or recursively) explores the optimal choices for both players starting from the initial state represented by nums.


  1. State of Program: The memo dictionary is populated with solutions to subproblems.
  2. Return Value: Returns a boolean value True or False. True indicates that Player 1 can win or tie; False means Player 1 will lose.
  3. Side Effects: The memo dictionary is filled, but since it’s local to the function, this doesn’t affect anything outside the function.

Error Handling:

  1. Preconditions Not Met: The method assumes that the preconditions (input constraints) are met and does not explicitly handle errors.
  2. Response: If input constraints are not met, behavior is undefined. In a production environment, you would typically add validation and potentially throw exceptions for invalid input.

Problem Decomposition

The problem asks us to find out if Player 1 can win a game where two players take turns picking numbers from either end of an array. Both start with a score of 0. Player 1 starts the game. The game is over when all numbers from the array are picked. Player 1 wins if they have an equal or higher score than Player 2 at the end.

Initial Breakdown:

  1. Initialize Memoization Store: Create a data structure to save results of subproblems.
  2. Subarray Scoring: Determine the maximum possible score for Player 1 for each subarray.
  3. Turn Handling: Account for the turns of Player 1 and Player 2 while calculating the scores.
  4. End-Game Check: Decide if Player 1 can win based on the final scores.

Subproblem Refinement:

  1. Initialize Memoization Store

    • Choose the data structure (likely a dictionary).
  2. Subarray Scoring

    • Determine the score if the leftmost element is chosen.
    • Determine the score if the rightmost element is chosen.
    • Save the best score for the current player in the memoization store.
  3. Turn Handling

    • Include a parameter to indicate whose turn it is.
    • Make recursive calls to swap turns.
  4. End-Game Check

    • Base case: Return the score when the array is empty.

Task Identification:

  1. Picking an element from the array is a repeated task.
  2. Recursively determining scores for subarrays is a repeated task.

Task Abstraction:

  1. pick_element(): Abstracts the logic for picking an element from the array’s end.
  2. calculate_score(): Recursively determines and returns the best score for the current player for a given subarray and turn.

Method Naming:

  1. initialize_memo(): Initializes memoization data structure.
  2. pick_element(): Chooses and removes an element from one end of the array.
  3. calculate_score(): Determines the best possible score for the current player for a given subarray.
  4. can_player1_win(): Checks if Player 1 can win the game, initiates the recursive solving.

Subproblem Interactions:

  1. Memoization is initialized before the game starts.
  2. can_player1_win() kicks off the recursive score calculations with calculate_score().
  3. calculate_score() uses pick_element() to evaluate choices at both ends.
  4. Each call to calculate_score() depends on the results of its previous recursive calls.
  5. Results of subproblems are stored in memoization data structure for future use.

From Brute Force to Optimal Solution

Brute Force Solution:

A brute force solution would involve trying out every possible combination of numbers that both players can pick and then determining if Player 1 can win.

In Python3:

def can_player1_win(nums):
    def dfs(left, right, is_player1):
        if left > right:
            return 0
        if is_player1:
            return max(nums[left] + dfs(left + 1, right, not is_player1),
                       nums[right] + dfs(left, right - 1, not is_player1))
            return min(dfs(left + 1, right, not is_player1),
                       dfs(left, right - 1, not is_player1))

    return dfs(0, len(nums) - 1, True) >= 0


  1. Time Complexity: The time complexity here is exponential, O(2^n). Each recursive call branches into two other calls, making it very inefficient for larger arrays.
  2. Space Complexity: Recursive calls also use stack space, and in the worst case, it could be O(2^n).


We can eliminate the repetitive work by using Dynamic Programming (DP).


We’ll use memoization to save the computed result for each subarray [left, right] for Player 1 and Player 2. This will help to avoid recomputing the score for the same subarray and turn.

Modified Python3 code:

def can_player1_win(nums):
    memo = {}

    def dfs(left, right, is_player1):
        if (left, right, is_player1) in memo:
            return memo[(left, right, is_player1)]

        if left > right:
            return 0

        if is_player1:
            score = max(nums[left] + dfs(left + 1, right, not is_player1),
                        nums[right] + dfs(left, right - 1, not is_player1))
            score = min(dfs(left + 1, right, not is_player1),
                        dfs(left, right - 1, not is_player1))
        memo[(left, right, is_player1)] = score
        return score

    return dfs(0, len(nums) - 1, True) >= 0

Impact on Time and Space Complexity:

  1. Time Complexity: With memoization, each subarray [left, right] for each player is calculated only once. This leads to a time complexity of O(n^2), where n is the length of the array.

  2. Space Complexity: The memoization table will also have at most O(n^2) entries, making the space complexity O(n^2) as well.

By using Dynamic Programming, we’ve optimized the solution from exponential time and space complexity to polynomial time and space complexity.


Code Explanation and Design Decisions

1. Initial Parameters:

  • nums: This is the input list of integers. It represents the numbers that players can choose from.
  • memo: This is a dictionary used for memoization. It caches the result for each subarray [left, right] for each player’s turn.

2. Primary Loop:

The primary loop is not explicit but happens through recursive calls in dfs(left, right, is_player1). Each recursive call considers a subarray [left, right] of nums and calculates the best score for Player 1 or Player 2 for that subarray.

3. Conditions and Branches:

  • if (left, right, is_player1) in memo: Checks if the result for this subarray is already computed. If so, returns it.
  • if left > right: Checks if we’ve reached the end of the subarray. If so, returns 0 as no more numbers can be picked.
  • if is_player1: Determines whose turn it is to pick a number. If True, it’s Player 1’s turn, otherwise, Player 2’s.

4. Updates or Modifications:

  • memo[(left, right, is_player1)] = score: Saves the computed score for a specific subarray and player’s turn. This is crucial to avoid redundant calculations and make the algorithm efficient.

5. Invariant:

The invariant here is the sum of scores for Player 1 and Player 2 in any subarray [left, right]. This sum will always be equal to the sum of all elements in that subarray. This is maintained implicitly and ensures that all elements are considered.

6. Final Output:

The final output is a Boolean value, indicating whether Player 1 can win or not. It directly answers the problem’s main question: “Can Player 1 end up with a score greater than or equal to Player 2 after both players have made their moves?”. If True, Player 1 can win; otherwise, he can’t. This fulfills the requirement set by the problem statement.

Coding Constructs

1. High-level Strategies:

  • Dynamic Programming: Used to store and retrieve already computed results for subproblems.
  • Recursion: Used to break down the problem into smaller, manageable subproblems.

2. Explanation for Non-Programmer:

The code is like a game planner for a two-player game where each player picks numbers from a list. It calculates the best strategy for the first player to end up with a higher total number than the second player, if possible.

3. Logical Elements:

  • Recursion: Solving a larger problem by solving smaller instances of the same problem.
  • Memoization: Storing solutions to subproblems for future use.
  • Conditions: Making decisions based on whose turn it is or whether we have reached the end of the list.

4. Algorithmic Approach in Plain English:

Start from the full list of numbers. On each turn, a player can pick either the first or the last number in the remaining list. Calculate the best possible outcome for the first player by trying all combinations of moves for both players. Remember solutions to smaller problems to avoid doing the same calculations multiple times.

5. Key Steps or Operations:

  • Choosing Elements: On each turn, choose either the first or last element from the remaining list.
  • Recursive Call: After a number is chosen, call the same procedure on the remaining list.
  • Store Result: If this situation has been calculated before, retrieve the stored result; otherwise, store the new result.

6. Algorithmic Patterns:

  • Dynamic Programming: For storing and retrieving solutions to subproblems.
  • Optimal Decision Making: Using recursive calls to simulate each player making the best possible move at each turn.

Language Agnostic Coding Drills

1. Distinct Concepts:

  1. Variable Declaration: Storing values for later use.
  2. Loops: Iterating through sequences.
  3. Conditional Statements: Making decisions in code.
  4. Functions: Encapsulating a block of code.
  5. Recursion: Functions calling themselves.
  6. Array Slicing: Extracting subsets from arrays.
  7. Dynamic Programming/Memoization: Storing already-computed results.

2. List by Difficulty:

  1. Variable Declaration: Easiest, as this is fundamental and usually one of the first concepts learned.

  2. Loops: A little more complex, but still elementary. Requires understanding of iteration and control flow.

  3. Conditional Statements: Adds complexity by introducing decision-making based on variables’ values.

  4. Functions: Introduces the concept of code modularity and reuse.

  5. Array Slicing: Requires understanding how data is stored and accessed in arrays.

  6. Recursion: Conceptually challenging, as it requires understanding how functions can call themselves and the implications thereof.

  7. Dynamic Programming/Memoization: Most difficult, as it needs a deep understanding of problem decomposition and optimization techniques.

3. Problem-Solving Approach:

  1. Variable Declaration: Start by identifying the variables you will need to store data like the list of numbers, players’ scores, and memoization cache.

  2. Loops: Initially, consider looping through the list to understand its structure. A loop will help you simulate different steps in the game but will eventually be replaced by recursion.

  3. Conditional Statements: Implement conditions to decide who picks next and to check whether you’ve reached the end of the list.

  4. Functions: Encapsulate your core logic inside a function. This makes the code reusable and sets the stage for recursion.

  5. Recursion: Replace the loop logic with recursive calls. Each call should represent a single move for a player, with the function calling itself to proceed to the next move.

  6. Array Slicing: To simplify recursion, slice the array to represent the remaining numbers to choose from.

  7. Dynamic Programming/Memoization: To optimize, introduce a memoization cache. Before diving into calculations, check if the current subproblem has already been solved. If it has, retrieve the stored value; if not, store the new value after calculation.

Each of these drills contributes to building an optimized solution for solving the problem in the most efficient way. You start with simple data storage and control flow, move to modularization, and then leverage more advanced concepts like recursion and dynamic programming for optimization.

Targeted Drills in Python

1. Python-based Coding Drills for General Concepts:

Variable Declaration

# Declaring a variable to store an integer
my_integer = 10


# Loop through a list of numbers
for i in range(5):

Conditional Statements

# Check if a number is even
if my_integer % 2 == 0:


# A function to add two numbers
def add(a, b):
    return a + b


# A recursive function to find factorial of a number
def factorial(n):
    if n == 0:
        return 1
    return n * factorial(n-1)

Array Slicing

# Slicing a list
my_list = [0, 1, 2, 3, 4]
sliced_list = my_list[1:4]

Dynamic Programming/Memoization

# Using memoization to store factorial results
memo = {}
def memoized_factorial(n):
    if n in memo:
        return memo[n]
    if n == 0:
        return 1
    memo[n] = n * memoized_factorial(n-1)
    return memo[n]

2. Problem-Specific Drills:

Max Score Calculator

# A function to calculate the max score from a list
def max_score(nums):
    return max(nums)

This drill is essential because finding the max score among available choices is central to the problem.

3. Integrating the Drills:

  1. Variable Declaration: Declare variables to keep track of scores and memoization cache.
  2. Functions: Define a function called calculate_score that takes in the list of scores and calls itself recursively.
  3. Loops: Inside this function, you might initially loop through possible choices (though you’ll replace this with recursion).
  4. Conditional Statements: Use conditions to determine what choice to make next.
  5. Recursion: Replace your loop with recursive calls to calculate_score.
  6. Array Slicing: Slice the array to represent the remaining elements before each recursive call.
  7. Dynamic Programming/Memoization: Store the result of each subproblem in a memoization cache. Before making a recursive call, check if the result for the current state is already in the cache.
  8. Max Score Calculator: Use this function inside your recursive function to determine the best choice at each step.

Each drill feeds into the next, allowing you to build a coherent, optimized solution from simple, understandable components.


Could anyone please help me understand the fault in below approach. Approach:- I’m trying to find the max score each player can make by recursively making a choice to pick/not pick the 1st and the last element.

`public boolean PredictTheWinner(int[] nums) { int[] winner = {0, 0}; if(nums.length>1){ LinkedList list = new LinkedList<>();

    for(int i : nums){

    simulateWinner(list, 0, 0, 1, winner);
return winner[0] >= winner[1];


void simulateWinner(LinkedList list, int score1, int score2,int player, int[] winner){

    winner[0] = Math.max(winner[0], score1);
    winner[1] = Math.max(winner[1], score2);

if(player == 1){
    int num = list.pollFirst();
    simulateWinner(list, score1+num, score2, 2, winner);

    num = list.pollLast();
    simulateWinner(list, score1+num, score2, 2, winner);
    int num = list.pollFirst();
    simulateWinner(list, score1, score2+num, 1, winner);

    num = list.pollLast();
    simulateWinner(list, score1, score2+num, 1, winner);

}` Thanks in advance.

Your approach to recursively simulate both players picking the first and last numbers is a good start. However, there is a critical flaw in this approach. In a game where both players play optimally, each player, when making a choice, must consider not just the immediate gain, but also what the opponent will do in the next turn. In your code, each player only tries to maximize their own score in the current turn without considering the opponent’s possible moves in the next turn.

In your simulateWinner function, when a player chooses a number, you recursively call simulateWinner for the other player with the remaining numbers. However, the current player doesn’t take into consideration what the score would be after the other player makes their optimal move.

To fix this, you should change the function so that when a player is about to make a move, they consider both options (picking the first number and picking the last number), simulate the remaining game for both options (i.e., call simulateWinner twice for each option), and then choose the option that results in the maximum final score for themselves.

To do this, you need to change simulateWinner to return the final score difference (player 1’s score minus player 2’s score), and when simulating the game for each option, the current player should choose the option that maximizes this score difference. This is because a positive score difference means player 1 wins, and a negative score difference means player 2 wins.

Finally, in PredictTheWinner, instead of checking if winner[0] >= winner[1], you should check if the final score difference is greater than or equal to 0, as a non-negative score difference means player 1 can win or draw the game.

For predict the winner: Confused about a test case here:

nums = [3606449,6,5,9,452429,7,9580316,9857582,8514433,9,6,6614512,753594,5474165,4,2697293,8,7,1]

{‘p1’: 3606449, ‘p2’: 0} [6, 5, 9, 452429, 7, 9580316, 9857582, 8514433, 9, 6, 6614512, 753594, 5474165, 4, 2697293, 8, 7, 1]

{‘p1’: 3606449, ‘p2’: 6} [5, 9, 452429, 7, 9580316, 9857582, 8514433, 9, 6, 6614512, 753594, 5474165, 4, 2697293, 8, 7, 1]

{‘p1’: 3606454, ‘p2’: 6} [9, 452429, 7, 9580316, 9857582, 8514433, 9, 6, 6614512, 753594, 5474165, 4, 2697293, 8, 7, 1]

{‘p1’: 3606454, ‘p2’: 7} [9, 452429, 7, 9580316, 9857582, 8514433, 9, 6, 6614512, 753594, 5474165, 4, 2697293, 8, 7]

{‘p1’: 3606461, ‘p2’: 7} [9, 452429, 7, 9580316, 9857582, 8514433, 9, 6, 6614512, 753594, 5474165, 4, 2697293, 8]

{‘p1’: 3606461, ‘p2’: 16} [452429, 7, 9580316, 9857582, 8514433, 9, 6, 6614512, 753594, 5474165, 4, 2697293, 8]

{‘p1’: 4058890, ‘p2’: 16} [7, 9580316, 9857582, 8514433, 9, 6, 6614512, 753594, 5474165, 4, 2697293, 8]

{‘p1’: 4058890, ‘p2’: 24} [7, 9580316, 9857582, 8514433, 9, 6, 6614512, 753594, 5474165, 4, 2697293]

{‘p1’: 6756183, ‘p2’: 24} [7, 9580316, 9857582, 8514433, 9, 6, 6614512, 753594, 5474165, 4]

{‘p1’: 6756183, ‘p2’: 28} [7, 9580316, 9857582, 8514433, 9, 6, 6614512, 753594, 5474165]

{‘p1’: 12230348, ‘p2’: 28} [7, 9580316, 9857582, 8514433, 9, 6, 6614512, 753594]

{‘p1’: 12230348, ‘p2’: 753622} [7, 9580316, 9857582, 8514433, 9, 6, 6614512]

{‘p1’: 18844860, ‘p2’: 753622} [7, 9580316, 9857582, 8514433, 9, 6]

{‘p1’: 18844860, ‘p2’: 753628} [7, 9580316, 9857582, 8514433, 9]

{‘p1’: 18844869, ‘p2’: 753628} [7, 9580316, 9857582, 8514433]

{‘p1’: 18844869, ‘p2’: 9268061} [7, 9580316, 9857582]

{‘p1’: 28702451, ‘p2’: 9268061} [7, 9580316]

{‘p1’: 28702451, ‘p2’: 18848377} [7]

{‘p1’: 28702458, ‘p2’: 18848377} [] So I would expect the output to be true, but it isn’t. Why can’t player 1 win here?

The game played here is based on the Minimax algorithm, which is a kind of backtracking algorithm used in decision making and game theory to find the optimal move for a player, assuming that the opponent also plays optimally.

In your specific test case, even though Player 1 seems to have the upper hand at first glance due to higher score accumulation, the decision to pick the number is not based solely on the current highest available number. Instead, the algorithm looks ahead to evaluate potential future states of the game, trying to minimize the possible loss for a worst-case (maximum loss) scenario.

The selection of numbers for both players might seem counter-intuitive, but the decision-making process assumes that both players are playing optimally. This means that each player chooses their number not just based on the current state, but also on how their choice would influence the options available for the next player and, by extension, the options that would be available to themselves on the next round.

Thus, in some cases, a player might choose a lower number over a higher one because it leads to a future state where they would ultimately have the advantage.

The scenario you’ve posted shows the step-by-step decision-making process of the players under the Minimax algorithm, and it seems that even though Player 1 has more points throughout the game, Player 2 wins in the end when playing optimally. This is why the function returns False, meaning Player 1 (the first player to start the game) cannot win assuming both players play optimally.

Yes, the explanation provided does indeed relate to the LeetCode problem 486 “Predict the Winner”. This problem involves a two-player game, where players take turns picking numbers from either the start or end of an array. The player with the highest total sum of numbers when all numbers have been picked is the winner.

The Minimax algorithm is a common strategy for determining the optimal play for such games. The algorithm assumes that both players are playing optimally, and it evaluates the game tree to determine the best possible move for the current player, taking into account the future possible moves of the opponent.

The question asked why Player 1 doesn’t win in the given test case, despite frequently having a higher score during the game. The reason is that the algorithm doesn’t just consider the current scores, but also the potential future states of the game. This means that a player may sometimes pick a lower-scoring option if it leads to a better overall outcome when the opponent’s responses are taken into account.

So even though it may seem like Player 1 should win based on the scores during the game, the algorithm determines that Player 1 can’t actually win when both players are playing optimally, which is why it returns False.

Similar Problems

  1. Two Sum - This problem is related because it also requires you to loop through an array and use a cache (hash table) to remember past elements for an efficient solution.

  2. Longest Substring Without Repeating Characters - This problem requires scanning through an array and keeping track of the visited characters in a hash table, similar to memoization techniques we discussed.

  3. Climbing Stairs - Like our original problem, it requires you to find an optimal solution by considering a limited set of choices at each step, and it can be solved with dynamic programming.

  4. Maximum Subarray - It’s related in the sense that you have to find the maximum sum of a contiguous subarray. The key insight is similar; finding the optimal solution by breaking down the problem.

  5. Coin Change - This problem is about finding the fewest number of coins that you need to make up a certain amount. It’s similar in the way it uses dynamic programming and recursion.

  6. Longest Palindromic Substring - Like our original problem, it asks you to find an optimal solution (longest palindrome) from a given string. You’ll also use a memoization table here.

  7. House Robber - This problem involves dynamic programming and has an almost identical approach to our problem, especially when considering optimal choices.

  8. Unique Paths - It requires you to find the number of unique paths from the top-left corner to the bottom-right corner of a grid, similar to how you’d look for an optimal sequence of choices.

  9. Palindrome Partitioning - This problem is similar in that it requires you to divide the problem into subproblems and can be solved using dynamic programming and recursion.

  10. Longest Increasing Subsequence - Like our original problem, this requires finding an optimal sequence within a given array. Dynamic programming is key to solving it efficiently.

These problems share techniques like dynamic programming, recursion, memoization, or array manipulation, which are relevant to the original problem we discussed.