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
• -231 <= nums[i] <= 231 - 1

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

Solution

#include <vector>
using namespace std;

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        int count = 1;
        int majority = nums[0];

        // For Potential Majority Element
        for (size_t i = 1; i < nums.size(); i++) {
            if (nums[i] == majority) {
                count++;
            } else {
                if (count > 1) {
                    count--;
                } else {
                    majority = nums[i];
                }
            }
        }

        // For Confirmation
        count = 0;
        for (int num : nums) {
            if (num == majority) {
                count++;
            }
        }

        if (count >= (nums.size() / 2) + 1) {
            return majority;
        } else {
            return -1;
        }
    }
};

This site is open source. Improve this page.