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

It seems that this question has some mistakes. I think there are two places that might lead to misunderstandings: (please feel free to tell me if I'm incorrect)

1. flowers[i] = x should mean that the unique flower that blooms at day i+1 (not i) will be at position x.
2. If you can get multiple possible results, then you need to return the minimum one.

The idea is to use an array days[] to record each position's flower's blooming day. That means days[i] is the blooming day of the flower in position i+1. We just need to find a subarray days[left, left+1,..., left+k-1, right] which satisfies: for any i = left+1,..., left+k-1, we can have days[left] < days[i] && days[right] < days[i]. Then, the result is max(days[left], days[right]).

Java version:

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, 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]));   //we get a valid subarray
                left = i; 
                right = k + 1 + i;
            }
        }
        return (res == Integer.MAX_VALUE)?-1:res;
    }