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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

• Java
• C++
• Python
• Go
• TypeScript
• Javascript

• 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
  }
  

• function shortestSubarray(nums: number[], k: number): number {
      const [n, MAX] = [nums.length, Number.POSITIVE_INFINITY];
      const s = Array(n + 1).fill(0);
      const q: number[] = [];
      let ans = MAX;
  
      for (let i = 0; i < n; i++) {
          s[i + 1] = s[i] + nums[i];
      }
  
      for (let i = 0; i < n + 1; i++) {
          while (q.length && s[i] - s[q[0]] >= k) {
              ans = Math.min(ans, i - q.shift()!);
          }
  
          while (q.length && s[i] <= s[q.at(-1)!]) {
              q.pop();
          }
  
          q.push(i);
      }
  
      return ans === MAX ? -1 : ans;
  }
  
  

• function shortestSubarray(nums, k) {
      const [n, MAX] = [nums.length, Number.POSITIVE_INFINITY];
      const s = Array(n + 1).fill(0);
      const q = [];
      let ans = MAX;
  
      for (let i = 0; i < n; i++) {
          s[i + 1] = s[i] + nums[i];
      }
  
      for (let i = 0; i < n + 1; i++) {
          while (q.length && s[i] - s[q[0]] >= k) {
              ans = Math.min(ans, i - q.shift());
          }
  
          while (q.length && s[i] <= s[q.at(-1)]) {
              q.pop();
          }
  
          q.push(i);
      }
  
      return ans === MAX ? -1 : ans;
  }
  
  

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