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

class Solution {
  int majorityElement(List<int> arr) {
    int count = 1;
    int majority = arr[0];

    // Step 1: Find Potential Majority Element
    for (int i = 1; i < arr.length; i++) {
      if (arr[i] == majority) {
        count++;
      } else {
        if (count > 1) {
          count--;
        } else {
          majority = arr[i];
        }
      }
    }

    // Step 2: Confirmation of Majority Element
    count = 0;
    for (int num in arr) {
      if (num == majority) {
        count++;
      }
    }

    return count >= (arr.length / 2).ceil() ? majority : -1;
  }
}

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