## Solution

#### Approach 1: Sliding Window

Intuition

We can rephrase this as a problem about the prefix sums of A. Let P[i] = A[0] + A[1] + ... + A[i-1]. We want the smallest y-x such that y > x and P[y] - P[x] >= K.

Motivated by that equation, let opt(y) be the largest x such that P[x] <= P[y] - K. We need two key observations:

• If x1 < x2 and P[x2] <= P[x1], then opt(y) can never be x1, as if P[x1] <= P[y] - K, then P[x2] <= P[x1] <= P[y] - K but y - x2 is smaller. This implies that our candidates x for opt(y) will have increasing values of P[x].

• If opt(y1) = x, then we do not need to consider this x again. For if we find some y2 > y1 with opt(y2) = x, then it represents an answer of y2 - x which is worse (larger) than y1 - x.

Algorithm

Maintain a "monoqueue" 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.

Complexity Analysis

• Time Complexity: , where is the length of A.

• Space Complexity: .

Analysis written by: @awice.