LeetCode-in-Java
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 java.util.List;
public class Solution {
public int minimumSumSubarray(List<Integer> li, int l, int r) {
int n = li.size();
int min = Integer.MAX_VALUE;
int[] a = new int[n + 1];
for (int i = 1; i <= n; i++) {
a[i] = a[i - 1] + li.get(i - 1);
}
for (int size = l; size <= r; size++) {
for (int i = size - 1; i < n; i++) {
int sum = a[i + 1] - a[i + 1 - size];
if (sum > 0) {
min = Math.min(min, sum);
}
}
}
return min == Integer.MAX_VALUE ? -1 : min;
}
}