Welcome to Subscribe On Youtube
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 0indexed 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]; } } ############ class Solution { public int firstDayBeenInAllRooms(int[] nextVisit) { int n = nextVisit.length; long[] f = new long[n]; final int mod = (int) 1e9 + 7; for (int i = 1; i < n; ++i) { f[i] = (f[i  1] + 1 + f[i  1]  f[nextVisit[i  1]] + 1 + mod) % mod; } return (int) f[n  1]; } }

// OJ: https://leetcode.com/problems/firstdaywhereyouhavebeeninalltherooms/ // Time: O(N) // Space: O(N) class Solution { public: int firstDayBeenInAllRooms(vector<int>& A) { long mod = 1e9 + 7, N = A.size(); vector<int> dp(N + 1); for (int i = 1; i < N; ++i) { dp[i] = ((long)2 + 2 * dp[i  1]  dp[A[i  1]] + mod) % mod; } return dp[N  1]; } };

# 1997. First Day Where You Have Been in All the Rooms # https://leetcode.com/problems/firstdaywhereyouhavebeeninalltherooms class Solution: def firstDayBeenInAllRooms(self, nextVisit: List[int]) > int: n = len(nextVisit) dp = [0] * n for i in range(1, n): dp[i] = (2 * dp[i  1]  dp[nextVisit[i  1]] + 2) % (10 ** 9 + 7) return dp[1]

func firstDayBeenInAllRooms(nextVisit []int) int { n := len(nextVisit) f := make([]int, n) const mod = 1e9 + 7 for i := 1; i < n; i++ { f[i] = (f[i1] + 1 + f[i1]  f[nextVisit[i1]] + 1 + mod) % mod } return f[n1] }