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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
❯

  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 Majority Element in Array Using Boyer-Moore Voting Algorithm - Optimal Solution

Topic Contents

ProblemExamplesSolutionVisualizationCodeTime ComplexitySpace Complexity

Problem Statement

Given an array of integers, your task is to find the majority element, which is defined as the element that appears more than ⌊n/2⌋ times in the array (where n is the size of the array).

If a majority element exists, return it. If no such element exists, return -1.

This problem assumes that the array may or may not contain a valid majority element.

Examples❯

Input ArrayExpected OutputDescription
[3, 3, 4, 2, 3, 3, 3]33 appears 5 times in an array of size 7 → majority element
[1, 2, 3, 4, 5]-1No element appears more than n/2 = 2.5 times
[2, 2, 1, 1, 1, 2, 2]22 appears 4 times → majority in size 7 array
[7]7Single element is always the majority (n=1)
[]-1Empty array, so no majority element
[5, 5, 5, 5, 5, 1, 2]55 appears more than half the time
[1, 1, 2, 2, 3, 3]-1No element appears more than n/2 = 3 times

Solution❯

To solve the problem of finding the majority element (if it exists), we need to understand what 'majority' means in this context. A majority element is the one that occurs more than n/2 times in the array, where n is the total number of elements.

Different Situations to Consider

  • Case 1: A clear majority element exists
    For example, in the array [3, 3, 3, 2, 3, 3], the number 3 appears more than half the time. So it is the majority element and should be returned.
  • Case 2: No element appears enough times
    In an array like [1, 2, 3, 4], each number appears only once. Since none appear more than n/2 times, the result should be -1.
  • Case 3: Single-element array
    If the array has just one element, like [7], that element is automatically the majority because it satisfies the condition (appearing more than n/2 = 0.5 → at least once).
  • Case 4: Empty array
    An empty array has no elements, so clearly there is no majority. We return -1.
  • Case 5: Two or more elements with equal counts
    If the array has balanced occurrences (e.g., [1, 1, 2, 2, 3, 3]), no one element crosses the majority threshold. Again, return -1.

How Does the Boyer-Moore Voting Algorithm Help?

The Boyer-Moore Voting Algorithm is a clever trick that helps us find the majority element in just one pass using constant space. Here's the core idea:

We keep a candidate and a count. Initially, count is 0. As we scan the array:

  • If count is 0, we choose the current element as a new candidate.
  • If the current element equals the candidate, we increase the count.
  • If it does not match, we decrease the count.

By the end of the loop, the candidate is the majority element if one exists. (You can optionally verify it in a second pass.)

Why This Works

When there is a majority element, the increment and decrement of the count cancel out the influence of non-majority elements. The real majority will always survive this cancellation and remain as the candidate.

This method works in O(n) time and O(1) space, which is optimal for this problem.

Visualization

Algorithm Steps

  1. Given an array arr of size n.
  2. Initialize count = 0 and candidate = None.
  3. Traverse the array:
  4. → If count == 0, set candidate = current element.
  5. → If current element == candidate, increment count, else decrement count.
  6. After the loop, candidate is the majority element.
  7. (Optional: Verify candidate by counting its actual occurrences if required.)

Code

Python
JavaScript
Java
C++
C
def majority_element(arr):
    count = 0
    candidate = None

    for num in arr:
        if count == 0:
            candidate = num
        count += (1 if num == candidate else -1)

    return candidate

# Sample Input
arr = [2, 2, 1, 1, 1, 2, 2]
print("Majority Element:", majority_element(arr))

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(n)The algorithm always traverses the entire array once, even if the majority element appears early.
Average CaseO(n)Regardless of element distribution, a single pass over the array is required to maintain count and candidate.
Average CaseO(n)In the worst case, the candidate may flip multiple times, but the loop still only runs once over the array.

Space Complexity

O(1)

Explanation: Only two variables—count and candidate—are used throughout the algorithm, making the space constant.

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