Formatted question description: https://leetcode.ca/all/1997.html
1997. First Day Where You Have Been in All the Rooms
Level
Medium
Description
There are n
rooms you need to visit, labeled from 0
to n - 1
. Each day is labeled, starting from 0
. You will go in and visit one room a day.
Initially on day 0
, you visit room 0
. The order you visit the rooms for the coming days is determined by the following rules and a given 0-indexed array nextVisit
of length n
:
- Assuming that on a day, you visit room i,
- if you have been in room
i
an odd number of times (including the current visit), on the next day you will visit the room specified bynextVisit[i]
where0 <= nextVisit[i] <= i
; - if you have been in room
i
an even number of times (including the current visit), on the next day you will visit room(i + 1) mod n
.
Return the label of the first day where you have been in all the rooms. It can be shown that such a day exists. Since the answer may be very large, return it modulo 10^9 + 7
.
Example 1:
Input: nextVisit = [0,0]
Output: 2
Explanation:
- On day 0, you visit room 0. The total times you have been in room 0 is 1, which is odd. On the next day you will visit room nextVisit[0] = 0
- On day 1, you visit room 0, The total times you have been in room 0 is 2, which is even. On the next day you will visit room (0 + 1) mod 2 = 1
- On day 2, you visit room 1. This is the first day where you have been in all the rooms.
Example 2:
Input: nextVisit = [0,0,2]
Output: 6
Explanation:
Your room visiting order for each day is: [0,0,1,0,0,1,2,…].
Day 6 is the first day where you have been in all the rooms.
Example 3:
Input: nextVisit = [0,1,2,0]
Output: 6
Explanation:
Your room visiting order for each day is: [0,0,1,1,2,2,3,…].
Day 6 is the first day where you have been in all the rooms.
Constraints:
n == nextVisit.length
2 <= n <= 10^5
0 <= nextVisit[i] <= i
Solution
Use dynamic programming. Create an array dp
of length n
, where dp[i]
is the first day to visit room i
. Initially, dp[0] = 0
. For 1 <= i < n
, there is dp[i] = dp[i - 1] + (dp[i - 1] - dp[nextVisit[i]] + 1) + 1
.
class Solution {
public int firstDayBeenInAllRooms(int[] nextVisit) {
final int MODULO = 1000000007;
int length = nextVisit.length;
int[] dp = new int[length];
for (int i = 1; i < length; i++)
dp[i] = ((dp[i - 1] * 2 - dp[nextVisit[i - 1]] + 2) % MODULO + MODULO) % MODULO;
return dp[length - 1];
}
}