# 683. K Empty Slots

## Description

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


Constraints:

• n == bulbs.length
• 1 <= n <= 2 * 104
• 1 <= bulbs[i] <= n
• bulbs is a permutation of numbers from 1 to n.
• 0 <= k <= 2 * 104

## Solutions

Solution 1: Binary Indexed Tree

We can use a Binary Indexed Tree to maintain the prefix sum of the bulbs. Every time we turn on a bulb, we update the corresponding position in the Binary Indexed Tree. Then we check if the $k$ bulbs to the left or right of the current bulb are all turned off and the $(k+1)$-th bulb is already turned on. If either of these conditions is met, we return the current day.

The time complexity is $O(n \times \log n)$ and the space complexity is $O(n)$, where $n$ is the number of bulbs.

• class Solution {
public int kEmptySlots(int[] bulbs, int k) {
int n = bulbs.length;
BinaryIndexedTree tree = new BinaryIndexedTree(n);
boolean[] vis = new boolean[n + 1];
for (int i = 1; i <= n; ++i) {
int x = bulbs[i - 1];
tree.update(x, 1);
vis[x] = true;
int y = x - k - 1;
if (y > 0 && vis[y] && tree.query(x - 1) - tree.query(y) == 0) {
return i;
}
y = x + k + 1;
if (y <= n && vis[y] && tree.query(y - 1) - tree.query(x) == 0) {
return i;
}
}
return -1;
}
}

class BinaryIndexedTree {
private int n;
private int[] c;

public BinaryIndexedTree(int n) {
this.n = n;
this.c = new int[n + 1];
}

public void update(int x, int delta) {
for (; x <= n; x += x & -x) {
c[x] += delta;
}
}

public int query(int x) {
int s = 0;
for (; x > 0; x -= x & -x) {
s += c[x];
}
return s;
}
}

• class BinaryIndexedTree {
public:
int n;
vector<int> c;

BinaryIndexedTree(int _n)
: n(_n)
, c(_n + 1) {}

void update(int x, int delta) {
for (; x <= n; x += x & -x) {
c[x] += delta;
}
}

int query(int x) {
int s = 0;
for (; x; x -= x & -x) {
s += c[x];
}
return s;
}
};

class Solution {
public:
int kEmptySlots(vector<int>& bulbs, int k) {
int n = bulbs.size();
BinaryIndexedTree* tree = new BinaryIndexedTree(n);
bool vis[n + 1];
memset(vis, false, sizeof(vis));
for (int i = 1; i <= n; ++i) {
int x = bulbs[i - 1];
tree->update(x, 1);
vis[x] = true;
int y = x - k - 1;
if (y > 0 && vis[y] && tree->query(x - 1) - tree->query(y) == 0) {
return i;
}
y = x + k + 1;
if (y <= n && vis[y] && tree->query(y - 1) - tree->query(x) == 0) {
return i;
}
}
return -1;
}
};

• class BinaryIndexedTree:
def __init__(self, n):
self.n = n
self.c = [0] * (n + 1)

def update(self, x, delta):
while x <= self.n:
self.c[x] += delta
x += x & -x

def query(self, x):
s = 0
while x:
s += self.c[x]
x -= x & -x
return s

class Solution:
def kEmptySlots(self, bulbs: List[int], k: int) -> int:
n = len(bulbs)
tree = BinaryIndexedTree(n)
vis = [False] * (n + 1)
for i, x in enumerate(bulbs, 1):
tree.update(x, 1)
vis[x] = True
y = x - k - 1
if y > 0 and vis[y] and tree.query(x - 1) - tree.query(y) == 0:
return i
y = x + k + 1
if y <= n and vis[y] and tree.query(y - 1) - tree.query(x) == 0:
return i
return -1


• type BinaryIndexedTree struct {
n int
c []int
}

func newBinaryIndexedTree(n int) *BinaryIndexedTree {
c := make([]int, n+1)
return &BinaryIndexedTree{n, c}
}

func (this *BinaryIndexedTree) update(x, delta int) {
for ; x <= this.n; x += x & -x {
this.c[x] += delta
}
}

func (this *BinaryIndexedTree) query(x int) (s int) {
for ; x > 0; x -= x & -x {
s += this.c[x]
}
return
}

func kEmptySlots(bulbs []int, k int) int {
n := len(bulbs)
tree := newBinaryIndexedTree(n)
vis := make([]bool, n+1)
for i, x := range bulbs {
tree.update(x, 1)
vis[x] = true
i++
y := x - k - 1
if y > 0 && vis[y] && tree.query(x-1)-tree.query(y) == 0 {
return i
}
y = x + k + 1
if y <= n && vis[y] && tree.query(y-1)-tree.query(x) == 0 {
return i
}
}
return -1
}

• class BinaryIndexedTree {
private n: number;
private c: number[];

constructor(n: number) {
this.n = n;
this.c = Array(n + 1).fill(0);
}

public update(x: number, delta: number) {
for (; x <= this.n; x += x & -x) {
this.c[x] += delta;
}
}

public query(x: number): number {
let s = 0;
for (; x > 0; x -= x & -x) {
s += this.c[x];
}
return s;
}
}

function kEmptySlots(bulbs: number[], k: number): number {
const n = bulbs.length;
const tree = new BinaryIndexedTree(n);
const vis: boolean[] = Array(n + 1).fill(false);
for (let i = 1; i <= n; ++i) {
const x = bulbs[i - 1];
tree.update(x, 1);
vis[x] = true;
let y = x - k - 1;
if (y > 0 && vis[y] && tree.query(x - 1) - tree.query(y) === 0) {
return i;
}
y = x + k + 1;
if (y <= n && vis[y] && tree.query(y - 1) - tree.query(x) === 0) {
return i;
}
}
return -1;
}