Peak Index in a Mountain Array
10 Prerequisite LeetCode Problems
“852. Peak Index in a Mountain Array” involves finding the peak of a mountainlike array (an array that strictly increases then strictly decreases). Here are 10 problems to prepare for this problem:
Find Minimum in Rotated Sorted Array (LeetCode 153): Although this problem is slightly more complex, it is similar in the sense that you’re looking for a pivot in the array, albeit in a slightly different context.
Search Insert Position (LeetCode 35): This problem involves determining the correct location to insert a new element in a sorted array.
First Bad Version (LeetCode 278): This problem is about finding a transition point in a sorted array.
Valid Mountain Array (LeetCode 941): This problem asks you to identify if an array is a mountain array.
Missing Number (LeetCode 268): This problem requires you to find a missing number in an array, reinforcing your ability to work with indices and values.
Single Number (LeetCode 136): This problem will help you with the ability to manipulate and compare array elements.
Best Time to Buy and Sell Stock (LeetCode 121): While a bit different, this problem also involves finding a kind of peak: the biggest difference between a later and earlier number.
Find All Numbers Disappeared in an Array (LeetCode 448): This problem helps you understand how to handle values and their indices.
Squares of a Sorted Array (LeetCode 977): In this problem, you learn how to deal with arrays in a way that modifies the elements and might alter their order.
Move Zeroes (LeetCode 283): This problem helps with understanding of how to manipulate an array inplace.
These cover how to work with arrays, how to use array indices and values, and how to find a certain condition in an array (like a peak), are valuable for solving problem 852.
Problem Analysis and Key Insights
Upon analyzing the problem statement, we can gather several key insights:
Mountain Array: The problem involves a specific pattern of data: a mountain array. This means the numbers in the array first increase to a certain point (the peak) and then decrease. Recognizing this pattern is crucial to solving the problem.
Peak Index: The task is to find the index of the peak of the mountain, not the peak value itself. This means the solution will focus on the position of the maximum value, not the value.
Logarithmic Time Complexity: The problem explicitly states that the solution should have a time complexity of O(log n). This implies that a simple linear search through the array will not be efficient enough. Instead, we should look to use a more efficient search algorithm, such as binary search.
Array Constraints: The constraints of the problem specify that the array will always have at least three elements and will always form a mountain. This ensures that there will always be a peak in the array to find.
From these insights, we can infer that the problem will likely involve implementing a binary search algorithm tailored to find the peak in a mountain array.
Problem Boundary
The scope of this problem is:
Input: The problem receives an array of integers as input. The array will always form a mountain, meaning it will first strictly increase and then strictly decrease. The length of the array will be between 3 and 10^5, and individual elements will be between 0 and 10^6.
Output: The output of the problem is an integer that represents the index of the peak of the mountain.
Process: The problem needs to be solved in logarithmic time complexity, O(log n), which suggests that a binary search algorithm or similar is needed. The solution needs to correctly identify the index of the peak in the mountain array.
Constraints: The problem explicitly specifies that the array is guaranteed to be a mountain array and the solution should run in logarithmic time. No other constraints or edge cases are mentioned in the problem statement, implying that we do not need to handle other scenarios such as empty arrays, arrays without a peak, or arrays with multiple peaks.
So, the problem is about implementing an efficient algorithm to find the peak index in a given mountain array within the specified constraints.
To establish the boundary of this problem, we should consider the following aspects:
Input Size: The size of the input array is a significant boundary condition. The array length can be between 3 and 10^5. The problem specifies a limit on the size of the array, so we don’t need to consider arrays smaller than 3 elements or larger than 10^5 elements.
Input Values: The array elements are integers ranging from 0 to 10^6. The elements are guaranteed to form a mountain, with values strictly increasing to a peak and then strictly decreasing.
Output: The output is the index of the peak of the mountain. This index will always be within the range of the array indices, which is between 0 and the length of the array minus 1.
Time Complexity: The problem asks for a solution with a time complexity of O(log n), which suggests that a binary search or a similar algorithm should be used. This imposes a boundary on the type of algorithms we can consider for solving this problem.
Special Conditions: The problem explicitly states that the given array will always form a mountain, meaning that it will strictly increase to a peak and then strictly decrease. This condition helps to establish the boundary by removing the need to handle cases where the array does not form a mountain.
By considering these aspects, we can define the boundary of the problem, which helps to guide our approach to solving it and to define the range of valid inputs and outputs.
Problem Classification
This problem belongs to the domain of Search Algorithms.
The problem is about finding the peak index in a mountain array. Here’s the breakdown of the problem:
Data Structure: The problem is based on an array, which is a fundamental data structure in computer science. The characteristics of the array, such as the fact that it forms a “mountain”, are key to the problem.
Search Algorithm: The problem requires finding an index in the array, which implies a search operation. However, the problem also specifies that the solution must have a logarithmic time complexity, hinting at the use of a more efficient search algorithm, such as Binary Search.
Pattern Recognition: The problem involves recognizing a specific pattern in the data (the mountain) and finding the peak of this pattern.
Order and Sorting: The problem involves an array that is first increasing and then decreasing. This ascending and descending order is a key characteristic of the problem.
The problem can be classified as a Binary Search problem that requires understanding of array manipulation and pattern recognition in data.
Distilling the Problem to Its Core Elements
Fundamental Concept: This problem is based on the principle of binary search, which is a divideandconquer algorithm used in computer science to efficiently find a specific value in a sorted array. Here, we’re not looking for a particular value but the peak of a mountain, which is the highest value in a sequence that increases and then decreases.
Simple Description: Imagine you’re hiking on a mountain trail that goes up to a peak and then back down. Your task is to find the peak of the mountain. The catch is you want to do it in the shortest time, so you try to divide your search area into halves each time to quickly locate the peak.
Core Problem: The core problem we are trying to solve is finding the peak of a mountain in an array, with the condition that we have to do it in logarithmic time complexity. The mountain in an array is represented by numbers increasing to a peak and then decreasing.
Key Components:
 An array that forms a mountain.
 A requirement to find the peak of the mountain.
 A constraint to find it in O(log n) time complexity.
Minimal Set of Operations:
 Start from the middle of the array.
 Check if it’s the peak (the numbers before and after it are smaller).
 If it’s not the peak, check if the number after it is greater. If so, the peak is in the second half of the array. Otherwise, it’s in the first half.
 Repeat the process with the new half until the peak is found.
Visual Model of the Problem
This problem can be visualized as a mountain hike where you are trying to find the highest peak. Consider the array as a series of steps on the mountain with each value in the array representing the height at that step. The hike begins at the first index (0) and ends at the last index (n1).
In the beginning, the hike goes uphill, with each step higher than the last one (values in the array are increasing). Then, after reaching the peak, the hike goes downhill, with each step lower than the previous one (values in the array are decreasing).
The goal is to find the peak of the mountain, which is the highest step in this hike. In the context of the array, it is the maximum value that is larger than both its neighbors.
Here is a simple visualization:
/\
/ \
/ \
/ \
The slash (/) and backslash (\) represent the hike going uphill and downhill respectively, and the peak is the highest point in the middle. In an array like [1,2,3,2,1], 3 would be the peak of the mountain.
Your task is to find this peak efficiently, with the stipulation that the search time should be proportional to the logarithm of the length of the array. This makes the problem more challenging and interesting.
Problem Restatement
We have an array of integers that resembles a mountain in profile. In this mountainlike array, values first increase to a certain peak point, and then decrease. We need to identify the index of this peak point.
However, it’s not just about finding the peak. The solution needs to be efficient, specifically, it should run in logarithmic time complexity, meaning it should not scan through the entire array to find the peak. This is a hint that we can leverage some form of divideandconquer or binary search technique to solve this problem more efficiently.
In summary, the goal is to efficiently find the index of the peak element in a mountainlike array.
Abstract Representation of the Problem
We are given a sequence of elements. This sequence has the unique characteristic of being unimodal  it strictly increases until it reaches a single peak, and then strictly decreases afterwards. Our task is to find the position of this peak in the sequence.
The challenge here is to find this peak position in logarithmic time complexity, suggesting that a simple linear scan of the sequence would not be an acceptable solution. Instead, we need to apply an approach that progressively narrows down the potential location of the peak in a manner similar to binary search or other divideandconquer strategies.
Thus, the problem’s essence is in efficiently identifying the peak in a unimodal sequence.
Terminology
Here are some key terms and concepts that are important to understanding this problem:
Mountain Array: In the context of this problem, a mountain array is an array that first strictly increases until it reaches a peak, and then strictly decreases. The array must contain at least three elements, and the peak cannot be at the ends of the array.
Peak: The peak of a mountain array is the highest point of the array. In this problem, the peak is the element that is greater than both its previous and next elements.
Index: In computer science, an index refers to the position of an element within an array. Indices usually start at 0 in most programming languages, including Python.
Logarithmic Time Complexity (O(log n)): This is a measure of the computational time taken by an algorithm to run as a function of the size of the input to the program. If an algorithm has logarithmic time complexity, it means that as the size of the input increases, the time taken to run the algorithm increases logarithmically. This is usually the case when each step of the algorithm reduces the size of the problem by a factor, such as in binary search. In the context of this problem, it means we’re looking for a solution that doesn’t need to examine every element of the array to find the peak.
Binary Search: Binary search is a search algorithm that finds the position of a target value within a sorted array. It compares the target value to the middle element of the array; if they are unequal, the half in which the target cannot lie is eliminated, and the search continues on the remaining half until it is successful or the remaining half is empty. This method of dividing the problem in half with each step leads to the logarithmic time complexity.
Understanding these terms and concepts is crucial for grasping both the problem and the expected solution.
Problem Simplification and Explanation
We are given an array of numbers, which we are told looks like a mountain. This means the numbers increase up to a certain point (the peak of the mountain), and then they decrease. The task is to find the position (index) of the peak of this mountain.
Imagine you’re hiking up a mountain. You start at the base, and with every step you take, you’re going higher and higher. Suddenly, after a few hours, you notice that with each step, you’re not going up anymore, but starting to descend. The highest point you reached is the peak of the mountain. In the context of this problem, your task is to find out at which step you were at the peak.
Here’s where we introduce an important concept: Binary Search. This is like playing a game of “Guess the number” where the number is between 1 and 100, and every time you make a guess, you’re told whether your guess is too high, too low, or just right. If you start guessing in the middle (say 50) and adjust your next guess based on whether 50 was too high or too low, you’ll find the right number much faster than if you started at 1 and guessed every single number.
Applying this to our mountain problem, instead of starting the hike from the base (the first number in the array), we start at the midpoint. If we see that the midpoint is less than the next point, we know the peak is somewhere on the right half of the mountain. If the midpoint is more than the next point, the peak is on the left half of the mountain. This way, we don’t have to check every single point (number) in the array (mountain) to find the peak.
This method is efficient and fits the requirement of finding the peak in logarithmic time (O(log n)).
Constraints
There are a few characteristics of this problem that we can exploit to devise an efficient solution:
Mountain Property: The problem statement guarantees that the given array is a “mountain” array. This means that the numbers in the array first increase to a certain point (the peak), and then decrease. This allows us to make the assumption that there will always be a peak in the array, and we don’t need to handle any edge cases where no peak exists.
Binary Search: The mountain property of the array allows us to use binary search to find the peak. Since the array is first increasing and then decreasing, for any point in the array, if its next point is higher, then the peak must be on the right side of this point. Otherwise, the peak must be on the left side.
Unimodal Function: The mountain array can be seen as a unimodal function that first increases and then decreases. The peak of a unimodal function can be efficiently found using binary search, which significantly reduces the time complexity from linear (O(n)) to logarithmic (O(log n)).
NonDuplicate Numbers: The problem does not mention that numbers in the array are unique. But even if there are duplicate numbers, it won’t affect the finding of the peak, because the peak is defined by the relation “arr[i1] < arr[i] > arr[i+1]” which still holds true even if there are duplicates. This means we don’t have to handle duplicates in a special way.
By recognizing and exploiting these characteristics, we can devise a solution that is both efficient and simple.
The constraints of the problem can provide important insights that might guide us in choosing the most suitable algorithm to solve the problem:
Array Length: The constraint that the array length is between 3 and (10^5) suggests that a linear scan of the array might not be efficient enough, especially for larger inputs. This indicates that we should aim for a more efficient algorithm, ideally with a logarithmic time complexity, such as binary search.
Array Values: The array values range between 0 and (10^6). However, the actual values of the elements in the array do not seem to play a significant role in this problem, as we are primarily concerned with the order of the elements rather than their exact values.
Mountain Array: The fact that the array is guaranteed to be a mountain array simplifies our task. We don’t have to check whether the array is a mountain array or not. We can directly proceed with finding the peak of the mountain, knowing that it must exist.
Logarithmic Time Complexity: The problem statement explicitly requires a logarithmic time complexity solution. This is a very strong hint that we should use a binary search approach, as binary search is a standard technique to achieve logarithmic time complexity.
Overall, the constraints indicate that we should look for a solution that uses binary search to find the peak of the mountain array in a way that is efficient even for large input sizes.
Case Analysis
Minimum Size Case: Given the smallest possible mountain array according to the constraints. Example:
arr = [1, 2, 1]
The peak is at index1
. This is the simplest possible case.Large Peak Case: The peak value is a large number. Example:
arr = [1, 2, 1000000, 999999, 1]
The peak is at index2
. This case tests the algorithm’s handling of large numbers.Early Peak Case: The peak is towards the start of the array. Example:
arr = [1, 5, 4, 3, 2]
The peak is at index1
. This case tests if the algorithm can handle the peak at early positions.Late Peak Case: The peak is towards the end of the array. Example:
arr = [1, 2, 3, 4, 5, 6, 7, 10, 9]
The peak is at index7
. This case tests if the algorithm can handle the peak at later positions.All Same Except Peak Case: All numbers are the same except the peak. Example:
arr = [1, 1, 1, 5, 1, 1, 1]
The peak is at index3
. This case tests if the algorithm can find the peak when it is the only distinct number.Multiple Same Peaks Case: The peak appears multiple times. Example:
arr = [1, 2, 3, 5, 5, 5, 4, 3, 2]
The peak is at index3
,4
, or5
. This case tests if the algorithm can handle multiple instances of the peak value.Descending Order Case: The elements are in strictly descending order after the peak. Example:
arr = [1, 3, 5, 4, 2]
The peak is at index2
. This case tests if the algorithm can handle strictly descending order after the peak.Ascending Order Case: The elements are in strictly ascending order before the peak. Example:
arr = [1, 3, 5, 2]
The peak is at index2
. This case tests if the algorithm can handle strictly ascending order before the peak.
Remember, the index of the peak in a mountain array is the element that is larger than both of its neighbors. These test cases cover a variety of scenarios that can occur within the constraints given in the problem statement.
From analyzing these cases, we can draw several key insights:
Peak Location: The peak of the mountain can be located anywhere in the array except the first and last positions. This is because the peak is defined as a number that is larger than both its neighbors.
Peak Value: The peak can have any value within the given range and is not necessarily the maximum value in the array.
Multiple Peaks: There can be multiple peaks with the same value, but they are not considered as the mountain peak unless they are higher than their neighbors.
Order of Elements: The elements before the peak are in strictly increasing order and the elements after the peak are in strictly decreasing order.
Binary Search: Due to the strictly increasing and then strictly decreasing nature of the mountain array, a binary search technique can be used to find the peak efficiently. This is possible because for any given element in the array, if it’s not the peak, we can determine if the peak is to its left or right by comparing it to its neighbor.
Size of Array: The size of the array does not affect the presence of a mountain peak but it does impact the time complexity of finding the peak. The given constraint to solve it in O(log(arr.length)) time complexity suggests the use of a binary search or similar divide and conquer strategy.
Uniqueness of Solution: The solution is unique as there is only one peak in a mountain array as per the problem’s definition. However, in the case where there are multiple same peaks, any of their indices can be the correct answer.
Identification of Applicable Theoretical Concepts
The problem can be greatly simplified by applying the concept of Binary Search, which is a classic algorithmic concept in computer science. Here are some points on why and how it can be applied:
Binary Search: Binary search is a divideandconquer algorithm used for searching a sorted array by repeatedly dividing the search interval in half. In the context of this problem, binary search can be used to efficiently locate the peak of the mountain in logarithmic time. This is possible because the elements before the peak are in strictly increasing order and the elements after the peak are in strictly decreasing order.
Midpoint Comparison: When implementing binary search, we consider the element in the middle of the array. If the middle element is greater than its next neighbor, it means the peak lies in the left half (including the middle element). If the middle element is less than its next neighbor, the peak lies in the right half.
Iteration: We repeat this process, halving the search space each time, until we have found the peak element. This process is guided by the principle that in a sorted or partially sorted array, a value either lies in the left or right half of the array depending on a certain condition.
These concepts allow us to transform the problem from searching the whole array in linear time complexity to searching in logarithmic time complexity, thereby making the problem much more manageable, especially for large arrays.
Simple Explanation
Imagine you’re hiking up a mountain along a path. The path starts at the base of the mountain, goes all the way up to the peak, and then back down to the base on the other side. The hike can be described as a series of steps, each leading a little bit higher up the mountain, until you reach the peak, then each step takes you a little bit lower until you’re at the base again.
Now, you’re given a list of altitudes representing each step of the hike, and you’re asked to find the step at which you reached the peak of the mountain. This is not necessarily in the middle of your list because the mountain might be asymmetrical; the ascent might be steeper than the descent, or vice versa.
You also want to find the peak as quickly as possible. Instead of checking every single step from the base to the top, you try a smarter approach. You pick a step somewhere in the middle of your list. If the next step is higher, you know that you’re still going up the mountain, so the peak must be somewhere ahead. If the next step is lower, you know you’ve passed the peak, so it must be somewhere behind you. You repeat this process, each time cutting your search area in half, until you find the peak.
This is very similar to the problem we’re trying to solve. We have a list of numbers that first increases, reaches a peak, and then decreases. We want to find the peak (the highest number), and we want to do it in the most efficient way possible. This is where the idea of binary search comes in: instead of looking at every number, we pick a number in the middle, see whether we’re before or after the peak, and then keep narrowing down our search area until we find the peak.
Problem Breakdown and Solution Methodology
The approach to this problem is based on a search strategy known as Binary Search.
Binary Search: The underlying principle is to always divide the search space in half. If the element we’re looking for is greater than the middle element, we know it must be on the right half of the search space, so we discard the left half. Conversely, if it’s less than the middle element, we discard the right half. In this problem, we’re looking for the peak of the mountain, which we know is greater than both its neighbors.
Implementation: Start with two pointers,
left
andright
, initially pointing to the first and last elements of the array. Calculate the middle index,mid
, and comparearr[mid]
witharr[mid + 1]
. Ifarr[mid]
is less thanarr[mid + 1]
, then the peak must be in the right half, so moveleft
tomid + 1
. Ifarr[mid]
is greater thanarr[mid + 1]
, then the peak must be in the left half, so moveright
tomid
. Repeat this process untilleft
equalsright
, which will be the index of the peak element.Effects of Parameters: The size of the array (n) affects the number of iterations in the Binary Search. Larger arrays will require more iterations. The values of the elements do not affect the solution, as we’re only interested in the relative order of elements.
Example Case: Let’s use
arr = [0, 2, 4, 5, 3, 1]
. Start with
left = 0
andright = 5
. Calculatemid = (0 + 5) // 2 = 2
.  Since
arr[2] < arr[3]
, moveleft
tomid + 1 = 3
.  Now
left = 3
andright = 5
. Calculatemid = (3 + 5) // 2 = 4
.  Since
arr[4] > arr[5]
, moveright
tomid = 4
.  Now
left = 3
andright = 4
. Calculatemid = (3 + 4) // 2 = 3
.  Since
arr[3] > arr[4]
, moveright
tomid = 3
.  Now
left = 3
andright = 3
, so we’ve found the peak at index3
.
 Start with
This solution has a time complexity of O(log n), which fulfills the constraint of the problem.
Inference of ProblemSolving Approach from the Problem Statement
Here are the key terms or concepts in this problem:
Mountain Array: This is a specific type of array where values increase up to a peak and then decrease. This is crucial to the problem, as it helps us identify that the problem is asking for the peak of the mountain.
Peak of Mountain: This is the maximum value in a mountain array, which is greater than its neighboring elements. The problem is essentially asking for the index of this peak.
Binary Search: This is a searching algorithm that is used to find elements in a sorted array by repeatedly dividing the search interval in half. This concept is fundamental to our approach, as the mountain array structure allows us to use binary search to find the peak efficiently. This is because in a mountain array, elements to the left of the peak are in ascending order and elements to the right of the peak are in descending order.
Left and Right Pointers: These are variables that we’ll use to represent the current search space within the array. Initially,
left
is at the start of the array andright
is at the end. They help us to narrow down the search space based on the midpoint comparison in each step of the binary search.
These terms and concepts guide us towards a binary searchbased solution to find the peak of the mountain array efficiently.
The problem specifies that we are given a mountain array, meaning the values in the array initially increase until a peak point, and then decrease. This kind of problem can often be solved with a binary search due to the nature of the array. Here’s why:
Ordered Subsequences: The array is split into two subarrays, where one is strictly increasing and the other is strictly decreasing. This ordering property is a key characteristic that makes binary search applicable.
Finding a Peak: The problem asks for the peak of the mountain, which is essentially the maximum element in the array. Binary search is a fast method for finding an element in an ordered sequence, and the ordering property mentioned above allows us to apply it here.
Divide and Conquer: The binary search algorithm is a divideandconquer algorithm. It works by dividing the problem into two halves at each step and choosing the appropriate half based on a comparison. In this case, we can compare the middle element with its next one to decide which half of the array to continue searching.
So, the specific structure of the mountain array and the requirement of finding the peak are cues in the problem statement that suggest binary search as a potential solution strategy.
Simple Explanation of the Proof
Let’s walk through the reasoning behind why this binary search algorithm works for finding the peak of a mountain array.
Invariant: At every step of the algorithm, we maintain the invariant that the peak (i.e., the maximum element) is within the current search bounds of the array.
Initialization: When we start, the search bounds include the entire array, so the peak is definitely within these bounds, and the invariant is satisfied.
Maintaining the Invariant: At each step, we compare the middle element with its next one. If the middle element is less than the next one, we know that the peak must be in the right half of the array, because the array is increasing at the midpoint. If the middle element is greater than the next one, we know that the peak must be in the left half of the array, because the array is decreasing at the midpoint. Either way, by updating our search bounds accordingly, we ensure that the peak remains within these bounds, and so the invariant is maintained.
Termination: The algorithm stops when the search bounds have been narrowed down to a single element. Due to the invariant, we know that this element must be the peak of the array.
So, the key to understanding this algorithm is the invariant: at every step, the peak of the array is within the current search bounds. This invariant is both maintained by the algorithm and ultimately allows us to find the peak.
Stepwise Refinement
Let’s break down the process of solving this problem.
Stepwise Refinement:
 Initialize two pointers,
start
andend
, to point to the beginning and end of the array, respectively.  Begin a loop that continues until
start
is less thanend
.  Find the midpoint,
mid
, of the current range, which is(start + end) // 2
.  If
arr[mid] < arr[mid + 1]
, setstart
tomid + 1
.  Else, set
end
tomid
.  Once the loop ends, return
start
(orend
, as they should be equal at this point).
 Initialize two pointers,
Granular Steps:
 Initialization: This is setting up our pointers and beginning our loop.
 Finding the midpoint: This is simple arithmetic to find the middle of our current search range.
 Updating pointers: This is checking whether the array is increasing or decreasing at the midpoint, and moving our pointers accordingly.
 Returning the result: After the loop ends, we return the index of the peak.
Independent Parts:
 The loop iteration and updating of pointers are independent operations that can be separated.
Repeatable Patterns:
 The primary repeatable pattern here is the binary search algorithm. This involves iteratively halving the search space by comparing the midpoint to another value (in this case, the next element in the array) and updating the search space bounds accordingly. This pattern continues until the search space has been narrowed down to a single element.
Solution Approach and Analysis
Let’s break down the process of solving this problem using binary search. This algorithm is like searching for a specific page in a book by repeatedly splitting the book in half and checking whether the desired page is in the first or second half, until you are left with just the page you’re looking for.
Initialize two pointers: We begin by setting two pointers,
start
andend
, to point to the first and last indices of the array respectively. This is like opening the book at the first and last pages.Enter a loop: We now enter a loop that will continue until
start
is less thanend
. This is analogous to our book search continuing until we find our page.Find the midpoint: Within each iteration of the loop, we calculate the midpoint of the current range by taking the average of
start
andend
(and using integer division to round down). This is like opening the book in the middle of our current range of pages.Check if the array is increasing at the midpoint: We then check if the value at index
mid
is less than the value at indexmid + 1
. If it is, this means the array is still increasing atmid
, so we know the peak must be to the right ofmid
. We therefore setstart
tomid + 1
, effectively eliminating the first half of the current range from consideration. If the array is not increasing atmid
, we know the peak must be to the left ofmid + 1
, so we setend
tomid
, eliminating the second half of the current range.Repeat until the range is exhausted: We repeat this process until
start
andend
meet, at which point we have found the peak of the mountain. This is like narrowing down our range of pages until we are left with just the one we want.Return the peak index: Finally, we return
start
(orend
– they will be the same). This is the index of the peak of the mountain.
Let’s see how this would work with an example case:


Identify Invariant
In the context of this problem, our loop invariant is the property that the peak of the mountain, the maximum value, is always within the range defined by the start
and end
pointers in our array.
This invariant holds before the loop starts (since start
and end
initially encompass the entire array), and it continues to hold at the start of each loop iteration (since we always update start
or end
to ensure they still encompass the peak). This is the key property we rely on to guide our binary search and ensure it will successfully find the peak.
This invariant also holds at the end of the loop, when start
equals end
, at which point they both point to the peak of the mountain, i.e., the maximum value in the array.
Identify Loop Invariant
In this problem, the loop invariant is that the peak of the mountain (the maximum value in the array) always lies within the subarray defined by the start
and end
indices.
Initialization: This is true before the loop starts, as the
start
andend
pointers cover the whole array.Maintenance: If the loop invariant is true before an iteration of the loop, it remains true before the next iteration. We ensure this by updating the
start
andend
pointers based on the mid value. Ifarr[mid] < arr[mid+1]
, we move thestart
tomid + 1
, else we moveend
tomid
. This ensures that the peak is still within thestart
andend
pointers.Termination: When the loop terminates,
start
equalsend
, both pointing to the peak of the mountain. This is the maximum value in the array, which satisfies the loop invariant condition at termination. Thus, the loop invariant holds, proving that our algorithm is correct.
The terms “invariant” and “loop invariant” both refer to conditions that remain unchanged, but they apply in slightly different contexts.
An “invariant” is a broader term and can refer to a condition that remains unchanged in any part of a program or system. For instance, if we have a constraint that an array must always remain sorted, that’s an invariant of the program dealing with the array.
On the other hand, a “loop invariant” is specifically an invariant that remains unchanged during each iteration of a loop. It’s a condition that’s initially true and remains true after each iteration of the loop.
In the context of this problem, when we talk about a “loop invariant”, we’re referring to a specific condition that remains true for each iteration of our binary search loop.
So, while all loop invariants are invariants (since they denote unchanging conditions), not all invariants are loop invariants. In this problem, the invariant and the loop invariant happen to be the same, as the condition we’re maintaining (that the peak of the mountain always lies within the subarray defined by the start
and end
indices) is maintained within the context of a loop.
Thought Process


This code implements a binary search approach to find the peak of the mountain in the input array. The binary search is a divideandconquer strategy that halves the search space at every step, making the time complexity logarithmic.
Here, we’re looking for the peak element where arr[i] < arr[i+1]
and arr[i] > arr[i1]
. The binary search algorithm navigates towards the larger element between arr[mid]
and arr[mid+1]
in each step. Once the peak is found, its index is returned.
Establishing Preconditions and Postconditions
Parameters:
 The method takes one input:
arr
. arr
is a list of integers.arr
represents the mountain array in which we are to find the index of the peak.
 The method takes one input:
Preconditions:
 The list
arr
should not be empty and should contain at least 3 elements.  The elements of
arr
form a mountain array, i.e., the elements are increasing and then decreasing.  The array does not need to be in any specific state before the method is called.
 The list
Method Functionality:
 The method is expected to return the index of the peak of the mountain array.
 The method does not interact with any other part of a program and works solely based on the input.
Postconditions:
 After the method has been called and has returned, the state of the program is unchanged as the method does not have any side effects.
 The return value represents the index of the peak of the mountain array.
 The method does not have any side effects.
Error Handling:
 The method assumes that the input list
arr
satisfies the preconditions. If the list does not form a mountain array, the behavior is undefined.  The method does not explicitly handle errors or exceptions. If the preconditions are not met, the method may return incorrect results or raise an exception depending on the nature of the violation. For example, if
arr
is an empty list, anIndexError
would be raised when trying to access an element.
 The method assumes that the input list
Problem Decomposition
Problem Understanding:
 The problem is about finding the peak of a mountain array. The array starts with increasing numbers until it reaches a peak, and then it decreases. The goal is to find the index of this peak number.
Initial Breakdown:
 The main subproblems are identifying the part of the array where it is increasing and where it is decreasing.
Subproblem Refinement:
 For the increasing part, we need to find the point where the next number is not greater than the current number.
 For the decreasing part, we need to find the point where the next number is not less than the current number.
Task Identification:
 The main task that can be generalized is comparing a number with its next one in the array.
Task Abstraction:
 The task is abstracted to finding the first point where the number is not less/greater than the next one in the array.
Method Naming:
 We can name this task “find_peak”.
Subproblem Interactions:
 The tasks are performed in order. First, we find the increasing part and then the decreasing part.
 The tasks are dependent on each other. We cannot start finding the decreasing part before we have finished finding the increasing part.
From Brute Force to Optimal Solution
A brute force approach to this problem would be to simply iterate through the array from beginning to end, checking each element to see if it’s larger than the one before and after it. If such an element is found, we return its index. This is the peak of the mountain.
However, this approach isn’t efficient. Its time complexity is (O(n)), where (n) is the length of the array, because in the worstcase scenario, we have to check every element.
We can optimize this solution by using a binary search approach. This problem has a key property that we can exploit: the elements in the array increase until they reach a peak, and then decrease. This means that if we pick a middle element in the array, we can determine whether the peak is to the left or the right of this element by comparing it to the next element. If the next element is larger, the peak must be to the right, otherwise, it’s to the left.
Here are the steps of this optimized solution:
 Initialize two pointers,
start
andend
, to the beginning and end of the array.  Enter a loop that continues as long as
start
is less thanend
.  Calculate the middle index,
mid
, and check the element atmid + 1
.  If
arr[mid] < arr[mid + 1]
, setstart
tomid + 1
.  Otherwise, set
end
tomid
.  Once
start
equalsend
, we’ve found the peak, so returnstart
.
This optimized solution uses a binary search approach, cutting the search space in half with each iteration. Therefore, its time complexity is (O(\log n)), which is a significant improvement over the brute force solution. The space complexity remains (O(1)), as we’re not using any additional data structures.
Code Explanation and Design Decisions
Initial Parameters:
arr
is the array given as input. It represents the mountain we’re trying to find the peak of.
Primary Loop:
 The primary loop is a while loop that continues as long as
start
is less thanend
. Each iteration of the loop represents a narrowing of the search space for the peak of the mountain. The iteration advances the solution by moving either thestart
orend
pointer, effectively cutting the search space in half with each iteration.
 The primary loop is a while loop that continues as long as
Conditions/Branches:
 Within the loop, there’s a condition that checks whether
arr[mid] < arr[mid + 1]
. This condition is crucial because it determines whether the peak of the mountain is to the right or left ofmid
. Ifarr[mid]
is less thanarr[mid + 1]
, then the peak must be to the right ofmid
, sostart
is updated tomid + 1
. Otherwise, the peak must be to the left ofmid
, soend
is updated tomid
.
 Within the loop, there’s a condition that checks whether
Updates/Modifications:
 The updates to
start
andend
within the loop are necessary to continuously narrow down the search space for the peak of the mountain. These updates reflect the logic of the binary search algorithm.
 The updates to
Invariant:
 The invariant in this problem is that the peak index always lies between
start
andend
, inclusive. This is maintained throughout the code and is crucial to ensuring that the binary search algorithm correctly finds the peak.
 The invariant in this problem is that the peak index always lies between
Significance of Final Output:
 The final output is the value of
start
(orend
, since they’re equal when the loop ends). This represents the index of the peak of the mountain. It satisfies the problem’s requirements because it’s the index where the elements in the array increase before and decrease after.
 The final output is the value of
Coding Constructs
HighLevel Strategies:
 This code uses a binary search strategy. Binary search is a divideandconquer algorithm that halves the search space at each step, making it highly efficient.
Purpose (Nonprogrammer explanation):
 This code is like finding the highest point on a mountain where the left side is a steep climb and the right side is a steep descent. It does this by repeatedly choosing a middle point, then deciding whether to continue the search on the left or right side of that point, based on whether we’re currently climbing up or going down.
Logical Elements/Constructs:
 This code uses a while loop for repeated checking and conditional statements (ifelse) for decision making.
Algorithmic Approach (Plain English):
 Start by looking in the middle of the mountain. If it looks like we’re still going uphill (i.e., the next point is higher), we know the peak must be somewhere to our right, so we continue our search there. If it looks like we’re going downhill (i.e., the next point is lower), then we know the peak must be somewhere to our left, so we continue our search there. We keep narrowing down our search like this until we find the peak.
Key Steps/Operations:
 The key steps are choosing a midpoint, comparing it with the next point, and updating our search space based on this comparison. These steps are important because they allow us to gradually narrow down our search for the peak.
Algorithmic Patterns/Strategies:
 The pattern used in this code is Binary Search, a common algorithmic pattern used for searching in sorted lists or arrays. The code also uses a common pattern of adjusting the boundaries of a search space based on comparisons.
Language Agnostic Coding Drills
Dissection of Code into Concepts:
a. Variable Assignment: This is the most basic concept, assigning initial values to variables.
b. Conditional Statements (ifelse): Comparing two values and making decisions based on this comparison.
c. While Loop: A repetitive structure that executes as long as a condition remains true.
d. Array Indexing: Accessing the elements of an array using their indices.
e. Binary Search: A divideandconquer algorithm that is used to find an element in a sorted list or array.
Coding Concepts/Drills in Increasing Difficulty:
a. Variable Assignment: This is the most basic coding concept. It involves giving a name to a value so that it can be referenced later in the code.
b. Array Indexing: This is slightly more complex as it requires understanding how arrays work and how elements are accessed in an array.
c. Conditional Statements (ifelse): Conditional statements involve a bit of logical thinking. Based on whether a condition is true or false, different code blocks are executed.
d. While Loop: While loops are more complex than conditional statements as they require understanding of how repetition works in programming, and how to control the flow of repetition using conditions.
e. Binary Search: This is the most complex concept among the listed ones. It requires understanding of how to effectively divide and conquer a problem to achieve a solution in the most optimized way.
Problemsolving Approach:
 Start with initializing two variables representing the start and end of the search space.
 Begin a loop that continues until the start is less than the end.
 In each iteration of the loop, calculate a midpoint.
 Use a conditional statement to check if the midpoint is less than the next point. If it is, that means we’re still climbing up, so we update the start of our search space to be the midpoint + 1. If it isn’t, that means we’re going downhill, so we update the end of our search space to be the midpoint.
 The loop continues, halving the search space in each iteration, until we find the peak (i.e., when start is not less than end).
 At the end, we return the peak index which is the start or end (both would be same at this point).
Targeted Drills in Python
Drill 1: Variable Assignment


Drill 2: Array Indexing


Drill 3: Conditional Statements (ifelse)


Drill 4: While Loop


Drill 5: Binary Search


The above drills illustrate how to implement each concept in Python:
Variable Assignment: Assigning values to variables is a fundamental concept. Here, we assign the integers 10 and 20 to the variables
x
andy
respectively.Array Indexing: Accessing elements in an array using their index is a key concept. Here, we print the first and last elements of an array.
Conditional Statements (ifelse): Conditional statements allow us to perform different actions based on different conditions. In this drill, we compare the values of
x
andy
and print a message based on the comparison.While Loop: While loops repeatedly execute a block of code as long as a condition is true. Here, we use a while loop to increment a counter and print its value each time.
Binary Search: Binary search is a powerful algorithm that finds the position of a target value within a sorted array. It repeatedly divides the search interval in half, which makes it much more efficient than linear search for large arrays. In this drill, we implement a binary search function.
By learning these individual concepts and practicing them through these drills, we can combine them to solve complex problems. For example, the problem at hand can be solved using a combination of array indexing and binary search. The binary search is slightly modified to find the peak element (instead of a specific target), but the basic concept remains the same.
Q&A
Similar Problems
Here are 10 problems that use similar underlying concepts:
Find Peak Element: This problem also requires finding a peak element, just like our current problem, but the array is not guaranteed to be a mountain array. The solution will also use a binary search approach.
Search in Rotated Sorted Array: This problem uses a modified binary search, similar to our problem where we modified the binary search to find the peak.
First Bad Version: This problem involves finding the first bad version (peak) in a series of versions, using a binary search technique.
Binary Search: This is a direct application of binary search, and understanding this will help understand any problem that requires a modification of binary search.
Find Minimum in Rotated Sorted Array: This problem also involves a modified binary search where the inflection point (minimum element) needs to be found.
Search in Rotated Sorted Array II: This problem is a variant of the ‘Search in Rotated Sorted Array’ problem, but it involves handling duplicates. The binary search concept is similarly applied here.
Find Smallest Letter Greater Than Target: This problem involves finding the smallest letter which is greater than a target letter in a sorted list. This can also be solved using a binary search algorithm.
Arranging Coins: This problem can be solved by applying a binary search on the answer, similar to how we applied binary search on the index to find the peak.
Valid Perfect Square: This problem requires checking whether a number is a perfect square or not. A binary search algorithm can be used to solve this problem efficiently.
Find the Duplicate Number: This problem requires finding the duplicate number in an array, which can be solved using a binary search approach on the range of possible answers, similar to how we performed a binary search on the array indices to find the peak.