# K Empty Slots

This page explains Java solution to problem `K Empty Slots` using `TreeSet` data structure.

## Problem Statement

You have `n` bulbs in a row numbered from `1` to `n`. Initially, all the bulbs are turned off. We turn on exactly one bulb every day until all bulbs are on after `n` days.

You are given an array bulbs of length `n` where `bulbs[i] = x` means that on the `(i+1)th` day, we will turn on the bulb at position `x` where `i` is `0-indexed` and `x` is `1-indexed`.

Given an integer `k`, return the minimum day number such that there exists two turned on bulbs that have exactly `k` bulbs between them that are all turned off. If there isn't such day, return `-1`.

Example 1:

Input: bulbs = [1,3,2], k = 1
Output: 2
Explanation:
On the first day: bulbs[0] = 1, first bulb is turned on: [1,0,0]
On the second day: bulbs[1] = 3, third bulb is turned on: [1,0,1]
On the third day: bulbs[2] = 2, second bulb is turned on: [1,1,1]
We return 2 because on the second day, there were two on bulbs with one off bulb between them.

Example 2:

Input: bulbs = [1,2,3], k = 1
Output: -1

## Solution

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``````
package com.vc.hard;

import java.util.*;

class KEmptySlots {
public int kEmptySlots(int[] bulbs, int k) {
TreeSet<Integer> bulbSet = new TreeSet<>();
for(int day = 0; day < bulbs.length; day++) {
int bulbPosition = bulbs[day];
Integer prevBulbPosition = bulbSet.lower(bulbPosition);
Integer nextBulbPosition = bulbSet.higher(bulbPosition);
if((prevBulbPosition != null && bulbPosition - prevBulbPosition - 1 == k) ||
(nextBulbPosition != null && nextBulbPosition - bulbPosition - 1 == k)) {
return day + 1;
}
}
return -1;
}
}
``````

## Time Complexity

O(N * log N) Where
N is total number of bulbs in an input array.

## Space Complexity

O(N) Where
N is total number of bulbs in an input array.