LeetCode-in-All

169. Majority Element

Easy

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.

Example 1:

Input: nums = [3,2,3]

Output: 3

Example 2:

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

Output: 2

Constraints:

• n == nums.length
• 1 <= n <= 5 * 104
• -109 <= nums[i] <= 109

Follow-up: Could you solve the problem in linear time and in O(1) space?

Solution

#include <stdio.h>

// Function to find the majority element in an array
int majorityElement(int* arr, int arrSize) {
    int count = 1;
    int majority = arr[0];
    
    // First pass to find a potential majority element using Boyer-Moore Voting Algorithm
    for (int i = 1; i < arrSize; i++) {
        if (arr[i] == majority) {
            count++;
        } else {
            if (count > 1) {
                count--;
            } else {
                majority = arr[i];
                count = 1;
            }
        }
    }

    // Second pass to confirm if the candidate is indeed the majority element
    count = 0;
    for (int i = 0; i < arrSize; i++) {
        if (arr[i] == majority) {
            count++;
        }
    }

    // Check if the majority element appears more than n/2 times
    if (count >= (arrSize / 2) + 1) {
        return majority;
    } else {
        return -1;
    }
}

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