862. Shortest Subarray with Sum at Least K

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

C++

• 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;
      }
  };

JAVA

• 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;
      }
  }

PYTHON

• from collections import deque
  
  class Solution:
      def shortestSubarray(self, nums, k):
          n = len(nums)
          s = [0] * (n + 1)
          for i in range(n):
              s[i + 1] = s[i] + nums[i]
          q = deque()
          ans = n + 1
          for i in range(n + 1):
              while q and s[i] - s[q[0]] >= k:
                  ans = min(ans, i - q.popleft())
              while q and s[q[-1]] >= s[i]:
                  q.pop()
              q.append(i)
          return -1 if ans > n else ans