862. Shortest Subarray with Sum at Least K

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

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;
    }
};

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
}