LeetCode in Kotlin

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 {
    fun majorityElement(arr: IntArray): Int {
        var count = 1
        var majority = arr[0]
        // For Potential Majority Element
        for (i in 1 until arr.size) {
            if (arr[i] == majority) {
                count++
            } else {
                if (count > 1) {
                    count--
                } else {
                    majority = arr[i]
                }
            }
        }
        // For Confirmation
        count = 0
        for (j in arr) {
            if (j == majority) {
                count++
            }
        }
        return if (count >= arr.size / 2 + 1) {
            majority
        } else {
            -1
        }
    }
}

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