LeetCode in Kotlin

410. Split Array Largest Sum

Hard

Given an integer array nums and an integer k, split nums into k non-empty subarrays such that the largest sum of any subarray is minimized.

Return the minimized largest sum of the split.

A subarray is a contiguous part of the array.

Example 1:

Input: nums = [7,2,5,10,8], k = 2

Output: 18

Explanation: There are four ways to split nums into two subarrays. The best way is to split it into [7,2,5] and [10,8], where the largest sum among the two subarrays is only 18.

Example 2:

Input: nums = [1,2,3,4,5], k = 2

Output: 9

Explanation: There are four ways to split nums into two subarrays. The best way is to split it into [1,2,3] and [4,5], where the largest sum among the two subarrays is only 9.

Constraints:

• 1 <= nums.length <= 1000
• 0 <= nums[i] <= 106
• 1 <= k <= min(50, nums.length)

Solution

class Solution {
    fun splitArray(nums: IntArray, m: Int): Int {
        var maxVal = 0
        var minVal = nums[0]
        for (num in nums) {
            maxVal += num
            minVal = Math.max(minVal, num)
        }
        while (minVal < maxVal) {
            val midVal = minVal + (maxVal - minVal) / 2
            // if we can split, try to reduce the midVal so decrease maxVal
            if (canSplit(midVal, nums, m)) {
                maxVal = midVal
            } else {
                // if we can't split, then try to increase midVal by increasing minVal
                minVal = midVal + 1
            }
        }
        return minVal
    }

    private fun canSplit(maxSubArrSum: Int, nums: IntArray, m: Int): Boolean {
        var currSum = 0
        var currSplits = 1
        for (num in nums) {
            currSum += num
            if (currSum > maxSubArrSum) {
                currSum = num
                currSplits++
                // if maxSubArrSum was TOO high that we can split the array into more that 'm' parts
                // then its not ideal
                if (currSplits > m) {
                    return false
                }
            }
        }
        return true
    }
}

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