eetcode - 3sum closest(kotlin)

Input: nums = [-1,2,1,-4], target = 1
Output: 2
Explanation: The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

import kotlin.math.abs

class Solution {
    var min = Int.MAX_VALUE
    var sumOfThree = 0
    fun threeSumClosest(nums: IntArray, target: Int): Int {
        nums.sort()
        combination(nums, IntArray(3), 3, 0, 0, target)
        return sumOfThree
    }

    fun combination(arr: IntArray, result: IntArray, r: Int, current: Int, start: Int, target: Int) {
        if(r == current) {
            var sum = 0
            for(r in result) {
                sum += r
            }

            val difference = abs(sum - target)
            if(difference < min) {
                min = difference
                sumOfThree = sum
            }

            return
        }
        
        if(min == 0) {
            return
        }

        for(i in start until arr.size) {
            result[current] = arr[i]
            combination(arr, result, r, current+1, i+1, target)
        }
    }
}

import kotlin.math.abs

class Solution {
    fun threeSumClosest(nums: IntArray, target: Int): Int {
        var difference = Int.MAX_VALUE
        nums.sort()

        for(i in nums.indices) {
            if(difference == 0) {
                break
            }

            var low = i+1
            var high = nums.size-1
            while(low < high) {
                val sum = nums[i] + nums[low] + nums[high]
                if(abs(target-sum) < abs(difference)) {
                    difference = target - sum
                }

                // 배열이 정렬되어 있는것을 잊지말자
                if(sum < target) {	// 합이 target보다 작으니까 아래에서 위로 올라가기
                    ++low
                } else {	// 합이 target보다 크니까 위에서 아래로 내려가기
                    --high
                }
            }
        }

        return target-difference
    }
}