LeetCode862. Shortest Subarray with Sum at Least K

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
Output: 1

Example 2:

Input: A = 4
Output: -1

Example 3:

Input: A = 3
Output: 3

Note:

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

思路

本来以为用dp来做，看了下答案，解法中并没有。使用的解法是滑动窗口，将问题重新定义为和A的前缀和有关，定义

P[i] = A[0] + A[1] + ... + A[i-1]

我们想要求的便是最小的 y-x，y>x 并且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.

opt(y)是使得当 P[x] <= P[y] - K 时 x 能取到的最大值。

1. 如果有 x1<x2 并且 P[x2]<=P[x1]，那么opt(y)一定不是 x1，因为如果有P[x1] <= P[y] - K，那么 P[x2] <= P[x1] <= P[y] - K，但是 y - x2 is smaller。这表明对于opt(y)的候选x应该是在使P(x)递增的区间去找。要注意这里的P[x1]指的是从0到X1的数组元素之和，不是单单指一个x1位置上元素的值。

2. 如果opt(y1)=x, 那么不需要再次考虑x。因为如果我们找到某些y2>y1并且opt(y2)=x，那么这表明这个解答 y2-x 是比之前的解答 y1-x 是更坏的答案。

Calculate prefix sum B of list A.
B[j] - B[i] represents the sum of subarray A[i] ~ A[j-1]
Deque d will keep indexes of increasing B[i].
For every B[i], we will compare B[i] - B[d[0]] with K.

    public int shortestSubarray(int[] A, int K) {
        int N = A.length, res = N + 1;
        int[] B = new int[N + 1];
　　　// 下面利用数组A重新构造了数组B，满足B[i+1]-B[j]=A[i]+A[i-1].....+A[j]
        for (int i = 0; i < N; i++) B[i + 1] = B[i] + A[i];
        Deque<Integer> d = new ArrayDeque<>();
        for (int i = 0; i < N + 1; i++) {　　
            while (d.size() > 0 && B[i] - B[d.getFirst()] >=  K)
                res = Math.min(res, i - d.pollFirst());　// 双端队列存的是索引
            while (d.size() > 0 && B[i] <= B[d.getLast()]) d.pollLast(); // Deque d keep indexes of increasing B[i]
            d.addLast(i);
        }
        return res <= N ? res : -1;
    }

上面的出入队列顺序是这样的：首先对于每个索引i，对应的是B[i]，将这个索引作为y位置来考虑，因为双端队列保持的索引是的B[i]是递增的，为了从最大处逼近K，我们从队头依次取索引出来计算：

B[i] - B[d.getFirst()]

如果比K大，那么则要找这其中距离索引i最近的那一个：

res = Math.min(res, i - d.pollFirst());

然后是队列要keep indexes of increasing B[i]，索引判断当前的B[i]是否大于队列尾部的索引处的

B[i] <= B[d.getLast()

如果不能构成递增，根据之前的分析，当前y所在的位置i的最优解opt(y)一定不会是在前面递增的部分取，所以队列要从后往前一个个弹出队尾直至能和B[i]构成递增序列。

LeetCode862. Shortest Subarray with Sum at Least K

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