LeetCode in Kotlin

3364. Minimum Positive Sum Subarray

Easy

You are given an integer array nums and two integers l and r. Your task is to find the minimum sum of a subarray whose size is between l and r (inclusive) and whose sum is greater than 0.

Return the minimum sum of such a subarray. If no such subarray exists, return -1.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [3, -2, 1, 4], l = 2, r = 3

Output: 1

Explanation:

The subarrays of length between l = 2 and r = 3 where the sum is greater than 0 are:

• [3, -2] with a sum of 1
• [1, 4] with a sum of 5
• [3, -2, 1] with a sum of 2
• [-2, 1, 4] with a sum of 3

Out of these, the subarray [3, -2] has a sum of 1, which is the smallest positive sum. Hence, the answer is 1.

Example 2:

Input: nums = [-2, 2, -3, 1], l = 2, r = 3

Output: -1

Explanation:

There is no subarray of length between l and r that has a sum greater than 0. So, the answer is -1.

Example 3:

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

Output: 3

Explanation:

The subarray [1, 2] has a length of 2 and the minimum sum greater than 0. So, the answer is 3.

Constraints:

• 1 <= nums.length <= 100
• 1 <= l <= r <= nums.length
• -1000 <= nums[i] <= 1000

Solution

import kotlin.math.min

class Solution {
    fun minimumSumSubarray(li: List<Int>, l: Int, r: Int): Int {
        val n = li.size
        var min = Int.Companion.MAX_VALUE
        val a = IntArray(n + 1)
        for (i in 1..n) {
            a[i] = a[i - 1] + li[i - 1]
        }
        for (size in l..r) {
            for (i in size - 1..<n) {
                val sum = a[i + 1] - a[i + 1 - size]
                if (sum > 0) {
                    min = min(min, sum)
                }
            }
        }
        return if (min == Int.Companion.MAX_VALUE) {
            -1
        } else {
            min
        }
    }
}

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