#1997

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

First Day Where You Have Been in All the Rooms

Calculate the first day you have visited all rooms using state transition dynamic programming on the nextVisit array.

Problem Statement

You are given n rooms labeled from 0 to n - 1 and a 0-indexed array nextVisit of length n. On day 0, you start in room 0. Each day, you visit exactly one room. If you have visited the current room an odd number of times, you go to nextVisit[currentRoom]; if even, you move to the next sequential room (mod n).

Your task is to find the first day where you have visited all rooms at least once. It is guaranteed that such a day exists. Because the number may be very large, return the result modulo 10^9 + 7.

Examples

Example 1

Input: nextVisit = [0,0]

Output: 2

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

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

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 Approach

Track Days Using DP Array

Define dp[i] as the earliest day you reach room i. Initialize dp[0] = 0. Use the rule that moving from room i to room i+1 depends on the parity of visits and nextVisit[i]. Update dp[i+1] based on dp[i] and dp[nextVisit[i]] for efficient state transitions.

Apply Modulo for Large Numbers

Since the number of days can grow quickly, apply modulo 10^9 + 7 at each calculation. This prevents overflow and ensures the result stays within integer limits while preserving the correctness of dynamic programming states.

Iterate Linearly Over Rooms

Loop from room 1 to n-1 updating dp[i] using the previous states. Each update only depends on dp[i-1] and dp[nextVisit[i-1]], achieving linear time complexity relative to the number of rooms. This exploits the specific state transition pattern of the problem.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n) since each room is processed once using dynamic programming. Space complexity is O(n) to store the dp array for earliest day calculations.

What Interviewers Usually Probe

Check if candidate identifies the state transition pattern between rooms.
Listen for understanding of parity-based room transitions and DP optimization.
Assess if candidate can correctly handle modulo arithmetic to prevent overflow.

Common Pitfalls or Variants

Common pitfalls

Failing to account for the odd/even visit rule when transitioning rooms.
Not applying modulo 10^9 + 7 consistently, causing incorrect large results.
Attempting brute-force simulation which leads to timeouts for large n.

Follow-up variants

Change nextVisit rules to allow skipping multiple rooms and recompute first day.
Ask for minimum days to visit a subset of target rooms instead of all rooms.
Introduce random room re-entries and analyze adjusted state transition DP.

FAQ

What is the main pattern used in First Day Where You Have Been in All the Rooms?

It is a state transition dynamic programming pattern where each room's next visit depends on previous room visits and their parity.

Why do we need modulo 10^9 + 7 in this problem?

The number of days can become extremely large, so modulo ensures results stay within integer limits without affecting correctness.

Can we solve this problem with brute-force simulation?

No, brute-force will timeout for large n due to repeated visits; using DP with state transitions is required.

How do odd and even visits affect room transitions?

If a room has been visited an odd number of times, the next room is determined by nextVisit; if even, move to the next sequential room modulo n.

What is the time complexity of the DP solution?

The solution runs in O(n) time since each room is processed once using dynamic programming with previous states.

terminal

Solution

Solution 1: Dynamic Programming

We define $f[i]$ as the date number of the first visit to the $i$-th room, so the answer is $f[n - 1]$.

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class Solution:
    def firstDayBeenInAllRooms(self, nextVisit: List[int]) -> int:
        n = len(nextVisit)
        f = [0] * n
        mod = 10**9 + 7
        for i in range(1, n):
            f[i] = (f[i - 1] + 1 + f[i - 1] - f[nextVisit[i - 1]] + 1) % mod
        return f[-1]

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