Problem Description
There are n rooms labeled from 0 to n - 1. Starting from room 0 on day 0, each day you visit a room. The next room to visit is determined by these rules:
- If you visit room i and it is your odd-numbered visit (including the current one), then on the next day you visit room nextVisit[i] (where 0 <= nextVisit[i] <= i).
- If it is your even-numbered visit, you will visit room (i + 1) mod n. Determine the first day when you have visited every room at least once. Because the number of days can be very large, return the answer modulo 10^9 + 7.
Key Insights
- Use dynamic programming (DP) where dp[i] represents the number of days needed to have visited rooms 0 through i.
- The recurrence relation is derived from the observation that to visit room i for the first time, you come from room i - 1, but you might need to “revisit” previous rooms based on the odd/even visit pattern.
- The recurrence: dp[i] = (2 * dp[i - 1] - dp[nextVisit[i - 1]] + 2) modulo MOD.
- Initialize with dp[0] = 0 since room 0 is visited on day 0.
- The final answer is given by dp[n - 1].
Space and Time Complexity
Time Complexity: O(n), since we compute dp[i] for each room from 1 to n - 1. Space Complexity: O(n), for storing the dp array.
Solution
We use a DP table where each element dp[i] represents the number of days required so that all rooms from 0 to i have been visited at least once. To derive dp[i]:
- When arriving at room i (from room i-1), if you have visited room i-1 an odd number of times, you go to room nextVisit[i-1]. This causes extra days before progressing.
- When the visit count is even, you go directly to room i.
This results in the recurrence: dp[i] = (2 * dp[i - 1] - dp[nextVisit[i - 1]] + 2) modulo MOD This recurrence accounts for the necessary “backtracking” to the room specified by nextVisit[i-1] and then finally moving forward. The twist is that you must visit room i - 1 once more (even visit) before going to room i. Finally return dp[n - 1] modulo MOD.
Data structures used:
- Array: dp[] of length n to store the number of days.
The trick is to carefully handle the recurrence and the modulo operation to avoid overflow.