862

862. Shortest Subarray with Sum at Least K (Hard)

Return the length of the shortest, non-empty, contiguous subarray of A with sum at least K.

If there is no non-empty subarray with sum at least K, return -1.

 

Example 1:

Input: A = [1], K = 1
Output: 1

Example 2:

Input: A = [1,2], K = 4
Output: -1

Example 3:

Input: A = [2,-1,2], K = 3
Output: 3

 

Note:

1. 1 <= A.length <= 50000
2. -10 ^ 5 <= A[i] <= 10 ^ 5
3. 1 <= K <= 10 ^ 9

Companies:
Facebook, Goldman Sachs

Related Topics:
Binary Search, Queue

Solution 1. Sliding Window + Mono-deque

Let P[i] = A[0] + ... A[i - 1] where i ∈ [1, N]. Our goal is to find the smallest y - x such that P[y] - P[x] >= K.

Let opt(y) be the largest x such that P[y] - P[x] >= K. Two key observations:

1. If x1 < x2 and P[x1] >= P[x2], then we don't need to consider x1 because if P[y] - P[x1] >= K then P[y] - P[x2] must >= K as well, and y - x2 < y - x1.
2. If opt(y1) = x, then we do not need to consider this x again. If we find some y2 > y1 with opt(y2) = x, then it represents an answer y2 - x which is worse (larger) than y1 - x.

Rule 1 tells us that we just need to keep a strictly increasing sequence P[a] < P[b] < P[c]....

Rule 2 tells us that we can further shrink the sequence from the front whenever the front element P[x] has been used as opt(y).

Algorithm

Maintain a mono-deque of indices of P: a deque of indices x_0, x_1, ... such that P[x_0], P[x_1], ... is increasing.

When adding a new index y, we'll pop x_i from the end of the deque so that P[x_0], P[x_1], ..., P[y] will be increasing.

If P[y] >= P[x_0] + K, then (as previously described) we don't need to consider this x_0 again, and we can pop it from the front of the deque.

// OJ: https://leetcode.com/problems/shortest-subarray-with-sum-at-least-k/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
// Ref: https://leetcode.com/articles/shortest-subarray-with-sum-atleast-k/
class Solution {
public:
    int shortestSubarray(vector<int>& A, int K) {
        int N = A.size(), ans = INT_MAX;
        vector<long> P(N + 1);
        for (int i = 0; i < N; ++i) P[i + 1] = P[i] + A[i];
        deque<int> q;
        for (int y = 0; y < P.size(); ++y) {
            while (q.size() && P[y] <= P[q.back()]) q.pop_back();
            while (q.size() && P[y] >= P[q.front()] + K) {
                ans = min(ans, y - q.front());
                q.pop_front();
            }
            q.push_back(y);
        }
        return ans == INT_MAX ? -1 : ans;
    }
};