99클럽 코테 스터디 42일차 TIL < First Day Where You Have Been in All the Rooms >

오늘의 문제

 

 

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

 

학습 키워드

 

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동적계획법

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문제 풀이

 

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코드

 

class Solution {
    public int firstDayBeenInAllRooms(int[] nextVisit) {
         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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