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DSA Visualizations /

Overview of TechniquesOverview of Techniques14
❯

  1. 1Overview of DSA Problem Solving Techniques
  2. 2Brute Force Technique in DSA
  3. 3Greedy Algorithm Technique in DSA
  4. 4Divide and Conquer Technique in DSA
  5. 5Dynamic Programming Technique in DSA
  6. 6Backtracking Technique in DSA
  7. 7Recursion Technique in DSA
  8. 8Sliding Window Technique in DSA
  9. 9Two Pointers Technique
  10. 10Binary Search Technique
  11. 11Tree / Graph Traversal Technique in DSA
  12. 12Bit Manipulation Technique in DSA
  13. 13Hashing Technique
  14. 14Heaps Technique in DSA

SortingSorting10
❯

  1. 1Bubble Sort
  2. 2Selection Sort
  3. 3Insertion Sort
  4. 4Merge Sort
  5. 5Quick Sort
  6. 6Heap Sort
  7. 7Radix Sort
  8. 8Counting Sort
  9. 9Bucket Sort
  10. 10Shell Sort

ArraysArrays32
❯

  1. 1Find Maximum and Minimum in Array using Loop
  2. 2Find Second Largest in Array
  3. 3Find Second Smallest in Array
  4. 4Reverse Array using Two Pointers
  5. 5Check if Array is Sorted
  6. 6Remove Duplicates from Sorted Array
  7. 7Left Rotate an Array by One Place
  8. 8Left Rotate an Array by K Places
  9. 9Move Zeroes in Array to End
  10. 10Linear Search in Array
  11. 11Union of Two Arrays
  12. 12Find Missing Number in Array
  13. 13Max Consecutive Ones in Array
  14. 14Find Kth Smallest Element
  15. 15Longest Subarray with Given Sum (Positives)
  16. 16Longest Subarray with Given Sum (Positives and Negatives)
  17. 17Find Majority Element in Array (more than n/2 times)
  18. 18Find Majority Element in Array (more than n/3 times)
  19. 19Maximum Subarray Sum using Kadane's Algorithm
  20. 20Print Subarray with Maximum Sum
  21. 21Stock Buy and Sell
  22. 22Rearrange Array Alternating Positive and Negative Elements
  23. 23Next Permutation of Array
  24. 24Leaders in an Array
  25. 25Longest Consecutive Sequence in Array
  26. 26Count Subarrays with Given Sum
  27. 27Sort an Array of 0s, 1s, and 2s
  28. 28Two Sum Problem
  29. 29Three Sum Problem
  30. 304 Sum Problem
  31. 31Find Length of Largest Subarray with 0 Sum
  32. 32Find Maximum Product Subarray

Arrays - Binary SearchArrays - Binary Search28
❯

  1. 1Binary Search in Array<br><span class="highlight">Using Iteration</span>
  2. 2Find Lower Bound in Sorted Array
  3. 3Find Upper Bound in Sorted Array
  4. 4Search Insert Position in Sorted Array (Lower Bound Approach)
  5. 5Floor and Ceil in Sorted Array
  6. 6First Occurrence in a Sorted Array
  7. 7Last Occurrence in a Sorted Array
  8. 8Count Occurrences in Sorted Array
  9. 9Search Element in a Rotated Sorted Array
  10. 10Search in Rotated Sorted Array with Duplicates
  11. 11Minimum in Rotated Sorted Array
  12. 12Find Rotation Count in Sorted Array
  13. 13Search Single Element in Sorted Array
  14. 14Find Peak Element in Array
  15. 15Square Root using Binary Search
  16. 16Nth Root of a Number using Binary Search
  17. 17Koko Eating Bananas
  18. 18Minimum Days to Make M Bouquets
  19. 19Find the Smallest Divisor Given a Threshold
  20. 20Capacity to Ship Packages within D Days
  21. 21Kth Missing Positive Number
  22. 22Aggressive Cows Problem
  23. 23Allocate Minimum Number of Pages
  24. 24Split Array - Minimize Largest Sum
  25. 25Painter's Partition Problem
  26. 26Minimize Maximum Distance Between Gas Stations
  27. 27Median of Two Sorted Arrays of Different Sizes
  28. 28K-th Element of Two Sorted Arrays

StringsStrings2
❯

  1. 1Remove Outermost Parentheses in String
  2. 2Edit Distance Problem

MatrixMatrix3
❯

  1. 1Set Matrix Zeroes
  2. 2Rotate Matrix by 90 Degrees Clockwise
  3. 3Print Matrix in Spiral Manner

Matrix - Binary SearchMatrix - Binary Search5
❯

  1. 1Row with Maximum Number of 1's
  2. 2Search in a Sorted 2D Matrix
  3. 3Search in Row and Column-wise Sorted Matrix
  4. 4Find Target in 2D Matrix
  5. 5Matrix Median

Binary TreesBinary Trees36
❯

  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Postorder Traversal of a Binary Tree Using Recursion
  4. 4Postorder Traversal of a Binary Tree using Iteration
  5. 5Level Order Traversal of a Binary Tree using Recursion
  6. 6Level Order Traversal of a Binary Tree using Iteration
  7. 7Reverse Level Order Traversal of a Binary Tree using Iteration
  8. 8Reverse Level Order Traversal of a Binary Tree using Recursion
  9. 9Find Height of a Binary Tree
  10. 10Find Diameter of a Binary Tree
  11. 11Find Mirror of a Binary Tree - Todo
  12. 12Inorder Traversal of a Binary Tree using Recursion
  13. 13Inorder Traversal of a Binary Tree using Iteration
  14. 14Left View of a Binary Tree
  15. 15Right View of a Binary Tree
  16. 16Top View of a Binary Tree
  17. 17Bottom View of a Binary Tree
  18. 18Zigzag Traversal of a Binary Tree
  19. 19Check if a Binary Tree is Balanced
  20. 20Diagonal Traversal of a Binary Tree
  21. 21Boundary Traversal of a Binary Tree
  22. 22Construct a Binary Tree from a String with Bracket Representation
  23. 23Convert a Binary Tree into a Doubly Linked List
  24. 24Convert a Binary Tree into a Sum Tree
  25. 25Find Minimum Swaps Required to Convert a Binary Tree into a BST
  26. 26Check if a Binary Tree is a Sum Tree
  27. 27Check if All Leaf Nodes are at the Same Level in a Binary Tree
  28. 28Lowest Common Ancestor (LCA) in a Binary Tree
  29. 29Solve the Tree Isomorphism Problem
  30. 30Check if a Binary Tree Contains Duplicate Subtrees of Size 2 or More
  31. 31Check if Two Binary Trees are Mirror Images
  32. 32Calculate the Sum of Nodes on the Longest Path from Root to Leaf in a Binary Tree
  33. 33Print All Paths in a Binary Tree with a Given Sum
  34. 34Find the Distance Between Two Nodes in a Binary Tree
  35. 35Find the kth Ancestor of a Node in a Binary Tree
  36. 36Find All Duplicate Subtrees in a Binary Tree

Binary Search TreesBinary Search Trees14
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  1. 1Find a Value in a Binary Search Tree
  2. 2Delete a Node in a Binary Search Tree
  3. 3Find the Minimum Value in a Binary Search Tree
  4. 4Find the Maximum Value in a Binary Search Tree
  5. 5Find the Inorder Successor in a Binary Search Tree
  6. 6Find the Inorder Predecessor in a Binary Search Tree
  7. 7Check if a Binary Tree is a Binary Search Tree
  8. 8Find the Lowest Common Ancestor of Two Nodes in a Binary Search Tree
  9. 9Convert a Binary Tree into a Binary Search Tree
  10. 10Balance a Binary Search Tree
  11. 11Merge Two Binary Search Trees
  12. 12Find the kth Largest Element in a Binary Search Tree
  13. 13Find the kth Smallest Element in a Binary Search Tree
  14. 14Flatten a Binary Search Tree into a Sorted List

Find Peak Element in Array - Optimal Binary Search Approach

Topic Contents

ProblemExamplesSolutionVisualizationCodeTime ComplexitySpace Complexity

Problem Statement

Given an array of integers, your task is to find the index of any peak element in the array.

A peak element is defined as an element that is not smaller than its neighbors. For the edge elements (first or last), we only compare with their one existing neighbor.

If there are multiple peaks, return the index of any one of them. If the array is empty, return -1.

Examples❯

Input ArrayPeak IndexPeak ValueDescription
[1, 3, 20, 4, 1]22020 is greater than both 3 and 4, so it's a peak
[10, 20, 15, 2, 23, 90, 67]1 or 520 or 90Both 20 and 90 are peaks; either is valid
[5, 10, 20]220Last element is greater than its left neighbor
[20, 10, 5]020First element is greater than its right neighbor
[10]010Single element is always a peak
[1, 2]12Last element is greater than its only neighbor
[2, 1]02First element is greater than its only neighbor
[]-1-Empty array, no peak exists

Solution❯

To solve the problem of finding a peak element in an array, we need to understand what a peak means. A peak is simply an element that is not smaller than its neighbors. That means for any index i, arr[i] is a peak if:

  • It is the first element and arr[i] ≥ arr[i+1]
  • It is the last element and arr[i] ≥ arr[i-1]
  • It is in the middle and arr[i] ≥ arr[i-1] and arr[i] ≥ arr[i+1]

So how do we find such an element efficiently? Let's walk through different scenarios:

📍 Case 1: The array has only one element

Any single element is trivially a peak because it has no neighbors to compare with. So, the only element is the peak.

📍 Case 2: The peak is at the beginning or end

If the first element is greater than the second, it's a peak. Similarly, if the last element is greater than the second last, it's also a peak.

📍 Case 3: The peak is somewhere in the middle

We can use a binary search approach. At each step, we look at the middle element arr[mid] and compare it with its neighbors:

  • If arr[mid] is greater than or equal to both neighbors, then it is a peak.
  • If arr[mid] is smaller than its right neighbor, then there must be a peak on the right side (because the values are increasing).
  • If arr[mid] is smaller than its left neighbor, then there must be a peak on the left side (because the values are increasing in that direction).

By following this process, we can efficiently reach a peak without scanning every element. The time complexity is O(log n), which is optimal for this problem.

📍 Case 4: The array is empty

If the array is empty, there are no elements to evaluate, so we return -1 to indicate that no peak exists.

Since the problem only asks for any one peak, and not necessarily the greatest one, we can stop as soon as we find a valid peak index using this logic.

Visualization

Algorithm Steps

  1. Initialize low = 0 and high = n - 1.
  2. While low ≤ high, calculate mid = (low + high) / 2.
  3. Check if arr[mid] is a peak:
    • It is greater than or equal to its neighbors (consider edge conditions).
  4. If it is a peak, return mid.
  5. If arr[mid] < arr[mid + 1], move to right half: low = mid + 1.
  6. Else, move to left half: high = mid - 1.

Code

Java
Python
JavaScript
C
C++
C#
Kotlin
Swift
Go
Php
public class PeakElementFinder {
  public static int findPeakElement(int[] arr) {
    int low = 0, high = arr.length - 1;

    while (low <= high) {
      int mid = low + (high - low) / 2;

      boolean leftOK = (mid == 0 || arr[mid] >= arr[mid - 1]);
      boolean rightOK = (mid == arr.length - 1 || arr[mid] >= arr[mid + 1]);

      if (leftOK && rightOK) return mid;
      else if (mid < arr.length - 1 && arr[mid] < arr[mid + 1]) low = mid + 1;
      else high = mid - 1;
    }

    return -1; // Should never reach here if at least one peak exists
  }

  public static void main(String[] args) {
    int[] arr = {1, 3, 20, 4, 1, 0};
    int peakIndex = findPeakElement(arr);
    System.out.println("Peak Index: " + peakIndex + ", Peak Element: " + arr[peakIndex]);
  }
}

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(1)The middle element is a peak in the first check.
Average CaseO(log n)At each step, the search space is halved, similar to binary search.
Average CaseO(log n)In the worst case, binary search continues until the peak is found at the edge.

Space Complexity

O(1)

Explanation: The algorithm uses constant extra space regardless of input size.

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