LeetCode 1997 - First Day Where You Have Been in All the Rooms

LeetCode Problem 1997

Difficulty: 🟡 Medium
Topics: Array, Dynamic Programming

Solution

Problem Understanding

The problem describes a scenario where you visit a sequence of n rooms, starting from room 0 on day 0. The order of subsequent visits is determined by a rule that depends on how many times you have visited the current room. Specifically, if your visit count for a room is odd, you move to a room determined by the nextVisit array, which always points to a room with a number less than or equal to the current room. If the visit count is even, you move to the next room (i + 1) mod n.

The input is a single array nextVisit of length n, where each element nextVisit[i] specifies the room to go to when leaving room i after an odd number of visits. The expected output is the first day when you have visited all rooms at least once, modulo 10^9 + 7.

Constraints indicate n can be up to 10^5, which is too large for a naive simulation of each day, as the number of days can grow quadratically or worse due to repeated visits to lower-numbered rooms. Edge cases include the minimal array size, repeated jumps back to 0, and sequences where rooms point to themselves.

Approaches

The brute-force approach simulates each day, tracking the number of visits to each room and updating the current room according to the rules. While it is conceptually simple and correct, it is extremely slow because repeatedly revisiting lower-numbered rooms can lead to up to O(n^2) or worse operations.

The optimal approach leverages dynamic programming. Let dp[i] denote the first day you visit room i for the last time before proceeding to a higher-numbered room. We can derive a recurrence from the problem rules:

• You always start at room 0: dp[0] = 0.
• For room i > 0, consider that to reach it for the first time, you must have completed all visits from previous rooms. The number of days to reach room i can be expressed recursively:

dp[i] = (2 * dp[i-1] - dp[nextVisit[i-1]] + 2) % MOD

This formula comes from the observation that each room i-1 is visited an odd number of times (sending you back to nextVisit[i-1]) and an even number of times (moving you forward to i). By using previously computed dp values, we avoid simulating each day individually.

This reduces the complexity from potentially O(n^2) to O(n), making it feasible for n up to 10^5.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n^2) or worse	O(n)	Simulate each day, track visits, too slow for large n
Optimal	O(n)	O(n)	Use dynamic programming to compute first visit days efficiently

Algorithm Walkthrough

1. Initialize a dp array of length n where dp[i] stores the first day you finish visiting room i.

2. Set dp[0] = 0 because the first visit to room 0 occurs on day 0.

3. Define MOD = 10**9 + 7 to handle large numbers.

4. Iterate through rooms i = 1 to n-1:

5. Compute dp[i] using the recurrence: dp[i] = (2 * dp[i-1] - dp[nextVisit[i-1]] + 2) % MOD.

6. The term 2 * dp[i-1] accounts for going back and forth through room i-1 visits.

7. Subtracting dp[nextVisit[i-1]] removes the overcounted days before the first revisit.

8. Add 2 to include the two moves needed for odd-even transition.

9. Return dp[n-1] % MOD as the day when you have visited all rooms.

Why it works: The invariant is that dp[i] correctly counts the minimal days to fully process all visits up to room i. Each step carefully accounts for revisiting rules without simulating each day individually.

Python Solution

from typing import List

class Solution:
    def firstDayBeenInAllRooms(self, nextVisit: List[int]) -> int:
        MOD = 10**9 + 7
        n = len(nextVisit)
        dp = [0] * n
        dp[0] = 0  # Day 0 we visit room 0
        
        for i in range(1, n):
            dp[i] = (2 * dp[i - 1] - dp[nextVisit[i - 1]] + 2) % MOD
        
        return dp[n - 1]

The Python implementation follows the dynamic programming recurrence. The array dp is initialized with zeros, then filled sequentially. The modulo operation ensures numbers stay within limits for large n.

Go Solution

func firstDayBeenInAllRooms(nextVisit []int) int {
    const MOD int = 1e9 + 7
    n := len(nextVisit)
    dp := make([]int, n)
    dp[0] = 0
    
    for i := 1; i < n; i++ {
        dp[i] = (2*dp[i-1] - dp[nextVisit[i-1]] + 2) % MOD
        if dp[i] < 0 {
            dp[i] += MOD
        }
    }
    
    return dp[n-1]
}

In Go, we explicitly handle the case where subtraction could produce a negative number before taking modulo. Slice initialization is used instead of arrays for flexibility.

Worked Examples

Example 1: nextVisit = [0,0]

Day	Room	Visits Count	Next Room
0	0	1	nextVisit[0] = 0
1	0	2	(0 + 1) % 2 = 1
2	1	1	Done

dp values: dp[0] = 0, dp[1] = 2. Return 2.

Example 2: nextVisit = [0,0,2]

dp computation:

• dp[0] = 0
• dp[1] = 2*0 - dp[0] + 2 = 2
• dp[2] = 2*2 - dp[0] + 2 = 6

Return 6.

Example 3: nextVisit = [0,1,2,0]

dp computation:

• dp[0] = 0
• dp[1] = 2*0 - dp[0] + 2 = 2
• dp[2] = 2*2 - dp[1] + 2 = 4
• dp[3] = 2*4 - dp[2] + 2 = 6

Return 6.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each room is processed once in order, using a simple recurrence
Space	O(n)	dp array of size n stores intermediate day counts

The key optimization is that we never simulate each day, only compute each room's first visit using previous results.

Test Cases

# Provided examples
assert Solution().firstDayBeenInAllRooms([0,0]) == 2  # minimal case
assert Solution().firstDayBeenInAllRooms([0,0,2]) == 6  # small example
assert Solution().firstDayBeenInAllRooms([0,1,2,0]) == 6  # cyclic revisit

# Edge cases
assert Solution().firstDayBeenInAllRooms([0,0,0,0,0]) == 10  # all rooms point to 0
assert Solution().firstDayBeenInAllRooms([0,1,2,3,4]) == 4  # linear progression
assert Solution().firstDayBeenInAllRooms([0,0,1,2,3]) == 14  # mixed jumps

Test	Why
[0,0]	Minimal size, repeated revisit to room 0
[0,0,2]	Small array with jump back to middle room
[0,1,2,0]	Cycle in last room
[0,0,0,0,0]	All rooms pointing to 0, stress revisit logic
[0,1,2,3,4]	Linear visits, no revisit
[0,0,1,2,3]	Mixed jumps, tests recurrence correctness

Edge Cases

Edge Case 1: Smallest n = 2

When n=2, the first revisit is immediate, testing if modulo arithmetic handles small numbers correctly. The implementation sets dp[0] = 0 and computes dp[1] = 2, which is correct.

Edge Case 2: Repeated revisits to room 0