Majority Element

Majority Element - Problem

Given an array nums of size n, return the majority element.

The majority element is the element that appears more than ⌊n / 2⌋ times. You may assume that the majority element always exists in the array.

Input & Output

Example 1 — Basic Case

$ Input: nums = [3,2,3]

› Output: 3

💡 Note: Element 3 appears 2 times, which is more than ⌊3/2⌋ = 1, so 3 is the majority element

Example 2 — Larger Array

$ Input: nums = [2,2,1,1,1,2,2]

› Output: 2

💡 Note: Element 2 appears 4 times, which is more than ⌊7/2⌋ = 3, so 2 is the majority element

Example 3 — Single Element

$ Input: nums = [1]

› Output: 1

💡 Note: Only one element exists, so it's automatically the majority element

Constraints

n == nums.length
1 ≤ n ≤ 5 × 10⁴
-10⁹ ≤ nums[i] ≤ 10⁹

Visualization

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Asked in

🍎 Apple 15 G Google 12 a Amazon 10 M Microsoft 8

The key insight is that the majority element appears more than n/2 times, so it will always survive pairwise cancellation. Best approach is Boyer-Moore Voting Algorithm with optimal Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Hash Map Counting	O(n)	O(n)	Use hash map to count frequencies, then find majority
Sorting Approach	O(n log n)	O(1)	Sort array, majority element will be at middle position
Boyer-Moore Voting Algorithm	O(n)	O(1)	Optimal O(1) space solution using voting algorithm
Brute Force	O(n²)	O(1)	Count occurrences of each element by nested loops

Hash Map Counting — Algorithm Steps

Count frequency of each element using hash map
Iterate through hash map to find element with count > n/2
Return the majority element

Visualization

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Step-by-Step Walkthrough

Build Frequency Map

Count occurrences of each element

Check Counts

Find element with count > n/2

Result

Return majority element

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int numsSize) {
    // Simple approach for C - check each element
    for (int i = 0; i < numsSize; i++) {
        int count = 0;
        for (int j = 0; j < numsSize; j++) {
            if (nums[j] == nums[i]) {
                count++;
            }
        }
        if (count > numsSize / 2) {
            return nums[i];
        }
    }
    return nums[0];
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    int nums[10000];
    int size = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[size++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    printf("%d\n", solution(nums, size));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass to build hash map, then iterate through map entries

✓ Linear Growth

Space Complexity

O(n)

Hash map stores frequency count for each unique element

⚡ Linearithmic Space

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High Frequency

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Input & Output

Constraints

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Related Problems

Common Approaches

Hash Map Counting — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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