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)

`flowers[i] = x`

should mean that the unique flower that blooms at day`i+1`

(not`i`

) will be at position`x`

.- 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;
}
```

C++ version:

```
int kEmptySlots(vector<int>& flowers, int k) {
vector<int> days(flowers.size());
for(int i=0; i<flowers.size();i++)days[flowers[i] - 1] = i + 1;
int left = 0, right = k + 1, res = INT_MAX;
for(int i = 0; right < days.size(); i++){
if(days[i] < days[left] || days[i] <= days[right]){
if(i == right)res = min(res, max(days[left], days[right])); //we get a valid subarray
left = i, right = k + 1 + i;
}
}
return (res == INT_MAX)?-1:res;
}
```