Jump Game II
You can use a greedy algorithm to calculate the minimum number of jumps needed to reach the end of the array. Here’s a Python function to do that:


Explanation:
 Initialize variables for the maximum reachable position in one step and the maximum reachable position in the next step.
 Iterate through the array, and at each step, update the maximum reachable position in the next step.
 If you reach a position greater than the current maximum reachable position, increase the jump counter, and update the current maximum reachable position with the maximum reachable position in the next step.
 When the loop finishes, the
jumps
variable will hold the minimum number of jumps needed to reach the last index.
In the provided examples, the output will be 2 for both cases, as explained in the problem statement.
Identifying Problem Isomorphism
“Jump Game II” is isomorphic to the problem “Word Ladder”.
In “Jump Game II”, the objective is to reach the end of an array (destination) starting from the first element (source), with the constraint that from a given position, you can jump to a number of next positions specified by the value at that position. The problem asks for the minimum number of jumps to reach the end.
In “Word Ladder”, the aim is to reach a target word starting from a source word, where you can change one character at a time, and the intermediate words must be in the given word list. The problem seeks the minimum number of word transformations required to reach the target word from the source word.
The isomorphism is in the need to find the minimum number of steps (jumps or transformations) to reach a destination from a source. The array in “Jump Game II” is similar to the word list in “Word Ladder”. The number of positions one can jump to in “Jump Game II” is analogous to the transformations of a word in “Word Ladder”.
“Jump Game II” is simpler as it deals with numbers and straightforward jumps, while “Word Ladder” involves character manipulations and more complex validity checks.
10 Prerequisite LeetCode Problems
“Jump Game II” involves understanding and implementation of concepts such as dynamic programming and greedy algorithms. Here are 10 problems to prepare for this problem:
Climbing Stairs (LeetCode 70): This problem is a simple dynamic programming problem that introduces the concept of finding different ways to reach a target.
Jump Game (LeetCode 55): This problem is a simpler version of “Jump Game II”, where you only need to decide whether it is possible to reach the end of the array, not the minimum number of jumps.
Best Time to Buy and Sell Stock (LeetCode 121): This problem introduces the concept of finding maximum profit with only one transaction which is an introduction to greedy algorithms.
Maximum Subarray (LeetCode 53): This problem helps to understand the concept of finding contiguous subarray with the largest sum.
House Robber (LeetCode 198): This problem introduces you to dynamic programming in the context of optimizing a decision (whether to rob a house) based on previous decisions.
Coin Change (LeetCode 322): This problem introduces the concept of finding minimum number of coins that make a given value.
Minimum Path Sum (LeetCode 64): This problem is about finding a path from top left to bottom right which minimizes the sum of all numbers along its path.
Can Place Flowers (LeetCode 605): This is a relatively simple greedy algorithm problem that can help you get a feel for this kind of problem.
Partition Equal Subset Sum (LeetCode 416): This is a dynamic programming problem that focuses on partitioning an array into two subsets with equal sum.
Nonoverlapping Intervals (LeetCode 435): This is a problem where you need to find the minimum number of intervals you need to remove to make the rest of the intervals nonoverlapping.
These problems cover key techniques such as dynamic programming and greedy algorithms, which are vital for solving the problem “Jump Game II”.


Problem Classification
You are given a 0indexed array of integers nums of length n. You are initially positioned at nums[0].
Each element nums[i] represents the maximum length of a forward jump from index i. In other words, if you are at nums[i], you can jump to any nums[i + j] where:
0 <= j <= nums[i] and i + j < n Return the minimum number of jumps to reach nums[n  1]. The test cases are generated such that you can reach nums[n  1].
Example 1:
Input: nums = [2,3,1,1,4] Output: 2 Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.
Example 2:
Input: nums = [2,3,0,1,4] Output: 2
Constraints:
1 <= nums.length <= 10^4 0 <= nums[i] <= 1000 It’s guaranteed that you can reach nums[n  1].
Language Agnostic Coding Drills
Variables: Understanding how to declare and use variables. This concept is used to hold the value for the final answer (
ans
), the current end (end
), and the farthest index that can be reached (farthest
).Lists (Arrays): Understanding how to declare and manipulate lists (arrays). This concept is used for the input
nums
.Accessing Array Elements: Understanding how to access elements in an array using indices. This concept is used to get the value at each index (
i
) innums
.Looping: Understanding how to use loops to iterate over an array. The
for
loop is used to go through each index (i
) innums
.Arithmetic Operations: Understanding basic arithmetic operations such as addition (
+
) and subtraction (
). These operations are used to calculatefarthest
and incrementans
.Relational Operators: Understanding how to use relational operators like
==
and>=
. These operators are used in theif
statements.Conditional Statements (If Else): Understanding how to use conditional statements to make decisions in code. These are used to check if
farthest
is greater than or equal to the last index innums
and if the current index is equal toend
.Functions: Understanding how to define and use functions. The
jump
function is defined and used here.ObjectOriented Programming: Understanding how to define classes and methods in OOP. The
Solution
class is defined here.Finding the Maximum: Understanding how to find the maximum between two numbers. This concept is used to calculate the maximum
farthest
index.Algorithmic Thinking (Greedy Algorithms): Understanding how to approach problems using algorithms, specifically greedy algorithms in this case. This is the overall algorithm used in the problem to find the minimum number of jumps to reach the end of the array.
This is the overall structure of the problem and each concept builds upon the previous one, gradually increasing in complexity.
Targeted Drills in Python
Variables
1 2 3 4
x = 5 y = 10 sum = x + y print(sum)
Lists (Arrays)
1 2
nums = [1, 2, 3, 4, 5] print(nums)
Accessing Array Elements
1 2 3
nums = [1, 2, 3, 4, 5] print(nums[0]) # Outputs: 1 print(nums[3]) # Outputs: 4
Looping
1 2 3
nums = [1, 2, 3, 4, 5] for i in nums: print(i)
Arithmetic Operations
1 2 3 4
x = 5 y = 10 print(x + y) # Outputs: 15 print(y  x) # Outputs: 5
Relational Operators
1 2 3 4
x = 5 y = 10 print(x == y) # Outputs: False print(x < y) # Outputs: True
Conditional Statements (If Else)
1 2 3 4 5
x = 5 if x > 10: print("x is greater than 10") else: print("x is not greater than 10")
Functions
1 2 3 4
def add_numbers(x, y): return x + y print(add_numbers(5, 10)) # Outputs: 15
ObjectOriented Programming
1 2 3 4 5 6
class Solution: def add_numbers(self, x, y): return x + y s = Solution() print(s.add_numbers(5, 10)) # Outputs: 15
Finding the Maximum
1 2 3
x = 5 y = 10 print(max(x, y)) # Outputs: 10
Algorithmic Thinking (Greedy Algorithms)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
class Solution: def canCompleteCircuit(self, gas: List[int], cost: List[int]) > int: if sum(gas) < sum(cost): return 1 start = 0 surplus = 0 for i in range(len(gas)): if gas[i] + surplus < cost[i]: start = i + 1 surplus = 0 else: surplus += gas[i]  cost[i] return start
In the last example, a greedy algorithm is used to solve the gas station problem (LeetCode 134). It’s a problem where you have to find the starting gas station’s index if you can travel around the circuit once in the clockwise direction.