Skip to content

683. K Empty Slots πŸ”’

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.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
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) {