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 <= 1001 <= 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
}
}
}