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LeetCode Solutions

862. Shortest Subarray with Sum at Least K

Time: $O(n)$

Space: $O(n)$

			


class Solution {
 public:
  int shortestSubarray(vector<int>& A, int K) {
    const int n = A.size();
    int ans = n + 1;
    deque<int> q;
    vector<long> prefix(n + 1);

    for (int i = 0; i < n; ++i)
      prefix[i + 1] = prefix[i] + A[i];

    for (int i = 0; i < n + 1; ++i) {
      while (!q.empty() && prefix[i] - prefix[q.front()] >= K)
        ans = min(ans, i - q.front()), q.pop_front();
      while (!q.empty() && prefix[i] <= prefix[q.back()])
        q.pop_back();
      q.push_back(i);
    }

    return ans <= n ? ans : -1;
  }
};




			


class Solution {
  public int shortestSubarray(int[] A, int K) {
    final int n = A.length;
    int ans = n + 1;
    Deque<Integer> q = new ArrayDeque<>();
    long[] prefix = new long[n + 1];

    for (int i = 0; i < n; ++i)
      prefix[i + 1] = (long) A[i] + prefix[i];

    for (int i = 0; i < n + 1; ++i) {
      while (!q.isEmpty() && prefix[i] - prefix[q.getFirst()] >= K)
        ans = Math.min(ans, i - q.pollFirst());
      while (!q.isEmpty() && prefix[i] <= prefix[q.getLast()])
        q.pollLast();
      q.addLast(i);
    }

    return ans <= n ? ans : -1;
  }
}




			


class Solution:
  def shortestSubarray(self, A: List[int], K: int) -> int:
    n = len(A)
    ans = n + 1
    q = deque()
    prefix = [0] + list(itertools.accumulate(A))

    for i in range(n + 1):
      while q and prefix[i] - prefix[q[0]] >= K:
        ans = min(ans, i - q.popleft())
      while q and prefix[i] <= prefix[q[-1]]:
        q.pop()
      q.append(i)

    return ans if ans <= n else -1