# 862. Shortest Subarray with Sum at Least K

## Description

Given an integer array nums and an integer k, return the length of the shortest non-empty subarray of nums with a sum of at least k. If there is no such subarray, return -1.

A subarray is a contiguous part of an array.

Example 1:

Input: nums = [1], k = 1
Output: 1


Example 2:

Input: nums = [1,2], k = 4
Output: -1


Example 3:

Input: nums = [2,-1,2], k = 3
Output: 3


Constraints:

• 1 <= nums.length <= 105
• -105 <= nums[i] <= 105
• 1 <= k <= 109

## Solutions

• class Solution {
public int shortestSubarray(int[] nums, int k) {
int n = nums.length;
long[] s = new long[n + 1];
for (int i = 0; i < n; ++i) {
s[i + 1] = s[i] + nums[i];
}
Deque<Integer> q = new ArrayDeque<>();
int ans = n + 1;
for (int i = 0; i <= n; ++i) {
while (!q.isEmpty() && s[i] - s[q.peek()] >= k) {
ans = Math.min(ans, i - q.poll());
}
while (!q.isEmpty() && s[q.peekLast()] >= s[i]) {
q.pollLast();
}
q.offer(i);
}
return ans > n ? -1 : ans;
}
}

• class Solution {
public:
int shortestSubarray(vector<int>& nums, int k) {
int n = nums.size();
vector<long> s(n + 1);
for (int i = 0; i < n; ++i) s[i + 1] = s[i] + nums[i];
deque<int> q;
int ans = n + 1;
for (int i = 0; i <= n; ++i) {
while (!q.empty() && s[i] - s[q.front()] >= k) {
ans = min(ans, i - q.front());
q.pop_front();
}
while (!q.empty() && s[q.back()] >= s[i]) q.pop_back();
q.push_back(i);
}
return ans > n ? -1 : ans;
}
};

• class Solution:
def shortestSubarray(self, nums: List[int], k: int) -> int:
s = list(accumulate(nums, initial=0))
q = deque()
ans = inf
for i, v in enumerate(s):
while q and v - s[q[0]] >= k:
ans = min(ans, i - q.popleft())
while q and s[q[-1]] >= v:
q.pop()
q.append(i)
return -1 if ans == inf else ans


• func shortestSubarray(nums []int, k int) int {
n := len(nums)
s := make([]int, n+1)
for i, x := range nums {
s[i+1] = s[i] + x
}
q := []int{}
ans := n + 1
for i, v := range s {
for len(q) > 0 && v-s[q[0]] >= k {
ans = min(ans, i-q[0])
q = q[1:]
}
for len(q) > 0 && s[q[len(q)-1]] >= v {
q = q[:len(q)-1]
}
q = append(q, i)
}
if ans > n {
return -1
}
return ans
}