Skip to content

862. Shortest Subarray with Sum at Least K 👍

  • Time: $O(n)$
  • Space: $O(n)$
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
class Solution {
 public:
  int shortestSubarray(vector<int>& nums, int k) {
    const int n = nums.size();
    int ans = n + 1;
    deque<int> q;
    vector<long> prefix{0};

    for (int i = 0; i < n; ++i)
      prefix.push_back(prefix.back() + nums[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;
  }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
class Solution {
  public int shortestSubarray(int[] nums, int k) {
    final int n = nums.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) nums[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;
  }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
class Solution:
  def shortestSubarray(self, nums: List[int], k: int) -> int:
    n = len(nums)
    ans = n + 1
    q = collections.deque()
    prefix = [0] + list(itertools.accumulate(nums))

    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