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First Day Where You Have Been in All the Rooms

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 a room with a lower or equal room number specified by nextVisit[i] where 0 <= 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 109 + 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 <= 105
  • 0 <= nextVisit[i] <= i

Solution Explanation: 1997. First Day Where You Have Been in All the Rooms

This problem asks for the first day when you've visited all rooms, given a rule set for room visitation. The solution presented uses dynamic programming for an efficient approach.

Approach:

The core idea is to use dynamic programming to track the first day you visit each room. We build an array f where f[i] represents the day number (0-indexed) when room i is first visited. The final answer is f[n-1], the day you first visit the last room (room n-1).

The recurrence relation is derived from the rules:

  1. Odd visits: If you've visited room i an odd number of times (including the current visit), the next room is nextVisit[i].
  2. Even visits: If you've visited room i an even number of times, the next room is (i + 1) % n.

Let's analyze how to calculate f[i] based on f[i-1]:

  • We reach room i-1 on day f[i-1].
  • From i-1, if we've visited i-1 an odd number of times, we go to nextVisit[i-1] next. This involves going back to a potentially earlier room. The number of days spent is f[i-1] - f[nextVisit[i-1]].
  • Then, it takes one more day to go back to i-1 and one more day to go to i.

This leads to the recurrence relation: f[i] = f[i-1] + 1 + f[i-1] - f[nextVisit[i-1]] + 1

Code Explanation (Python):

class Solution:
    def firstDayBeenInAllRooms(self, nextVisit: List[int]) -> int:
        n = len(nextVisit)
        f = [0] * n  # Initialize DP array
        mod = 10**9 + 7  # Modulo for large results
 
        for i in range(1, n):
            f[i] = (f[i - 1] + 1 + f[i - 1] - f[nextVisit[i - 1]] + 1) % mod #Apply recurrence relation
 
        return f[-1]  # Return the first day to visit the last room

The code iterates through the rooms, calculating the first visit day for each using the recurrence relation. The modulo operation (% mod) prevents integer overflow.

Time and Space Complexity:

  • Time Complexity: O(n) - The loop iterates through the rooms once.
  • Space Complexity: O(n) - We use an array f of size n to store the first visit day for each room.

The provided solutions in other languages (Java, C++, Go, TypeScript, C#) follow the same logic and differ only in syntax. They all achieve the same time and space complexity.