683. K Empty Slots

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:

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

这题表述有毛病，day说是1~N，其实应该是0~N-1。
而且flowers[i]定义为第i天，开在x位置。开始以为第i位置，第x天开。
而且还是中间间隔几个空位。
没仔细审题就会出错啊！
奇葩问题描述，做法倒不难，就是理解上容易分歧，耽误时间。
坑！
循环每一天，然后开花，然后检查新开花位置两边距离各多少，更新hashset。

class Solution {
    public int kEmptySlots(int[] flowers, int k) {
        int n = flowers.length;
        boolean[] garden = new boolean[n];
        HashSet<Integer> d = new HashSet<Integer>();
        // iterate through day
        for(int i=0;i<n;i++){
            int x = flowers[i]-1;
            //System.out.println("x: "+x);
            garden[x]=true;
            // iterate through flowers
            int left = 1, right = 1;
            while(x-left>=0){
                if(garden[x-left]) {d.add(left-1);break;}
                left++;
            }
            while(x+right<n){
                if(garden[x+right]) {d.add(right-1);break;}
                right++;
            }
            //System.out.println("left: "+left+" right: "+right);
            if(d.contains(k)) return i+1;
        }
        return -1;
    }
}