LeetCode in Kotlin

561. Array Partition

Easy

Given an integer array nums of 2n integers, group these integers into n pairs (a1, b1), (a2, b2), ..., (an, bn) such that the sum of min(ai, bi) for all i is maximized. Return the maximized sum.

Example 1:

Input: nums = [1,4,3,2]

Output: 4

Explanation: All possible pairings (ignoring the ordering of elements) are:

1. (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3
2. (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3
3. (1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4 So the maximum possible sum is 4.

Example 2:

Input: nums = [6,2,6,5,1,2]

Output: 9

Explanation: The optimal pairing is (2, 1), (2, 5), (6, 6). min(2, 1) + min(2, 5) + min(6, 6) = 1 + 2 + 6 = 9.

Constraints:

• 1 <= n <= 104
• nums.length == 2 * n
• -104 <= nums[i] <= 104

Solution

class Solution {
    fun arrayPairSum(nums: IntArray): Int {
        nums.sort()
        var sum = 0
        var i = 0
        while (i < nums.size - 1) {
            sum += Math.min(nums[i], nums[i + 1])
            i = i + 2
        }
        return sum
    }
}

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