LC 683. K Empty Slots 【lock，hard】

There is a garden with N slots. In each slot, there is a flower. The N flowers will bloom one by one in Ndays. In each day, there will be exactly one flower blooming and it will be in the status of blooming since then.

Given an array flowers consists of number from 1 to N. Each number in the array represents the place where the flower will open in that day.

For example, flowers[i] = x means that the unique flower that blooms at day i will be at position x, where i and x will be in the range from 1 to N.

Also given an integer k, you need to output in which day there exists two flowers in the status of blooming, and also the number of flowers between them is k and these flowers are not blooming.

If there isn't such day, output -1.

Example 1:

Input:
flowers: [1,3,2]
k: 1
Output: 2
Explanation: In the second day, the first and the third flower have become blooming.

Example 2:

Input:
flowers: [1,2,3]
k: 1
Output: -1

Note:

1. The given array will be in the range [1, 20000].

Runtime: 128 ms, faster than 89.38% of C++ online submissions for K Empty Slots.

网上的解法。

class Solution {
public:
    int kEmptySlots(vector<int>& flowers, int k) {
        int n = flowers.size();
        vector<int> days(n, );
        for (int i = ; i < flowers.size(); i++) {
            days[flowers[i] - ] = i + ;
        }
        int left = , right = k+, ret = INT_MAX;
        for (int i = ; right < n; i++) {
            if (days[i] < days[left] || days[i] <= days[right]) {
                if (i == right) ret = min(ret, max(days[left], days[right]));
                left = i;
                right = i + k + ;
            }
        }
        return ret == INT_MAX ? - : ret;
    }
};