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

Problem

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* 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 (Java)

class Solution {
    public int firstDayBeenInAllRooms(int[] nextVisit) {
        int[] dp = new int[nextVisit.length];
        int m = 1000000007;
        for (int i = 1; i < dp.length; i++) {
            int steps = 2 * dp[i - 1] - dp[nextVisit[i - 1]] + 2;
            dp[i] = steps < 0 ? (steps + m) % m : steps % m;
        }
        return dp[dp.length - 1];
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).