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 everyday 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
, find out 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
Note:
1 <= N <= 20000
1 <= bulbs[i] <= N
bulbs
is a permutation of numbers from1
toN
.0 <= K <= 20000
[Array] [Binary Indexed Tree] [Sliding Window] [Ordered Set]