683. K Empty Slots

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Description

There is a garden with N slots. In each slot, there is a flower. The N flowers will bloom one by one in N days. 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:
The given array will be in the range [1, 20000].

Problem URL

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Solution

花园中有N个花池，每天都会有一株花开花然后保持开花状态。flowers数组记录的是第i天开花的花在花园中的位置。给一个k，要求找到在哪一天（多个情况下的最小天）时，正好有两个已经开花的花池，它俩中间有k个空的花池。

We want the day that two flowers have k empty slots between them. So make it a reverse, we use days array to contain the blooming day of the flower at position i. Now it becomes to a problem that we want to find a window with size k. The boundary left & right is the flower blooms at days[left] and days[right]. The flowers I between them have the properity that days[left + I] > days[left] && days[left + I] > days[right].

Based on that, we start iteration, it ends when rights reach the end of days. If days[I] < days[left] or <= days[right], if it is right, calculate a result. After that, update the postion of left and right.

Code

class Solution {
    public int kEmptySlots(int[] flowers, int k) {
        int[] days = new int[flowers.length];
        for (int i = 0; i < flowers.length; i++){
            days[flowers[i] - 1] = i + 1;
        }
        int left = 0, right = k + 1;
        int res = Integer.MAX_VALUE;
        for (int i = 0; right < days.length; i++){
            if (days[i] < days[left] || days[i] <= days[right]){
                if (i == right){
                   res = Math.min(res, Math.max(days[left], days[right])); 
                }
                left = i;
                right = i + k + 1;
            }
        }
        return res == Integer.MAX_VALUE ? -1 : res;
    }
}

Time Complexity: O(n)
Space Complexity: O(n)

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Review

A TLE solution which document new blooming flower position and then calculate empty slots one by one.

class Solution {
    public int kEmptySlots(int[] flowers, int k) {
        int len = flowers.length;
        int[] slots = new int[len];
        for (int i = 0; i < len; i++){
            slots[flowers[i] - 1] = 1;
            if (i < 1){
                continue;
            }
            for (int j = 0; j < len; j++){
                if (slots[j] == 1){
                    int nextFlower = j + 1;
                    while(nextFlower < len && slots[nextFlower] != 1){
                        nextFlower++;
                    }
                    if (nextFlower - j - 1 == k && nextFlower != len){
                        return i + 1;
                    }
                }
            }
        }
        return -1; 
    }
}