LeetCode-in-Java

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

Medium

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

public 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];
    }
}

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