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1997. First Day Where You Have Been in All the Rooms
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 a room with a lower or equal room number 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
Solutions

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]; } }

class Solution { public: int firstDayBeenInAllRooms(vector<int>& nextVisit) { int n = nextVisit.size(); vector<long long> f(n); const int mod = 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 f[n  1]; } };

class Solution: def firstDayBeenInAllRooms(self, nextVisit: List[int]) > int: n = len(nextVisit) f = [0] * n mod = 10**9 + 7 for i in range(1, n): f[i] = (f[i  1] + 1 + f[i  1]  f[nextVisit[i  1]] + 1) % mod return f[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] }

function firstDayBeenInAllRooms(nextVisit: number[]): number { const n = nextVisit.length; const mod = 1e9 + 7; const f: number[] = new Array<number>(n).fill(0); for (let i = 1; i < n; ++i) { f[i] = (f[i  1] + 1 + f[i  1]  f[nextVisit[i  1]] + 1 + mod) % mod; } return f[n  1]; }

public class Solution { public int FirstDayBeenInAllRooms(int[] nextVisit) { int n = nextVisit.Length; long[] f = new long[n]; 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]; } }