- Difficulty: Hard
- Tags: LeetCode, Hard, Binary Indexed Tree, Array, Ordered Set, Sliding Window, leetcode-683, O(n)
Problem
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 from1
ton
.0 <= k <= 2 * 104