Scale Engineer

First Day Where You Have Been in All the Rooms

Med
#1819Time: O(D), where D is the final answer (the number of days). Since D can be very large, this approach is too slow for the given constraints.Space: O(n) to store the `counts` and `visited` arrays.1 company
Patterns
Dynamic Programming
Data structures
Array
Companies
ByteDance

Prompt

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

Approaches

2 approaches with complexity analysis and trade-offs.

This approach directly simulates the process described in the problem. We keep track of the current day, the current room, and the number of times each room has been visited. We follow the rules day by day, advancing one day at a time, until we have visited all n rooms. The first day we visit room n-1 is the first day we have visited all rooms.

Algorithm

  • Initialize day = 0, currentRoom = 0.
  • Initialize a counts array of size n to store visit counts for each room.
  • Loop until currentRoom is n-1:
    1. Increment the visit count for the currentRoom in the counts array.
    2. Determine the nextRoom based on the rules:
      • If counts[currentRoom] is odd, nextRoom = nextVisit[currentRoom].
      • If counts[currentRoom] is even, nextRoom = (currentRoom + 1) % n.
    3. Update currentRoom to nextRoom.
    4. Increment day.
  • Once the loop terminates, day holds the value of the first day we arrive at room n-1, which is the answer.

Walkthrough

We can simulate the visiting process step-by-step. We'll need variables to keep track of the currentDay, the currentRoom, and an array visitCounts to store how many times each room has been visited.

The simulation starts on day = 0 in room = 0. In a loop, we determine the next room to visit based on the current room and its visit count. If the number of visits to the current room (including the current one) is odd, we go to nextVisit[currentRoom]. If it's even, we go to (currentRoom + 1) % n. We increment the day and update the current room. This continues until we land on room n-1 for the first time. Since we visit rooms 0, 1, ..., i-1 before visiting room i for the first time, the first day we reach room n-1 is our answer.

While straightforward, this approach is too slow because the number of days can be enormous, and simulating each day individually is not feasible within typical time limits.

// This is a conceptual implementation for illustration.// It will time out for large inputs due to the large number of days.class Solution {    public int firstDayBeenInAllRooms(int[] nextVisit) {        int n = nextVisit.length;        long[] counts = new long[n];        int currentRoom = 0;        long day = 0;        boolean[] visited = new boolean[n];        int visitedCount = 0;         while (true) {            if (!visited[currentRoom]) {                visited[currentRoom] = true;                visitedCount++;                if (visitedCount == n) {                    // We have just arrived at the last unvisited room.                    return (int) day;                }            }             counts[currentRoom]++;            int nextRoom;            if (counts[currentRoom] % 2 == 1) { // Odd visit                nextRoom = nextVisit[currentRoom];            } else { // Even visit                nextRoom = (currentRoom + 1) % n;            }                        currentRoom = nextRoom;            day++;        }    }}

Complexity

Time

O(D), where D is the final answer (the number of days). Since D can be very large, this approach is too slow for the given constraints.

Space

O(n) to store the `counts` and `visited` arrays.

Trade-offs

Pros

  • Simple to understand and implement as it directly models the problem statement.

  • Requires minimal complex logic.

Cons

  • Extremely inefficient for the given constraints.

  • The number of days can be very large (up to 10^9), causing the simulation to run for a very long time and result in a 'Time Limit Exceeded' error on most platforms.

Solutions

public class Solution {    public int FirstDayBeenInAllRooms(int[] nextVisit) {        int n = nextVisit.Length;        long[] f = new long[n];        int mod = (int) 1 e9 + 7;        for (int i = 1; i < n; ++i) {            f[i] = (f[i - 1] + 1 + f[i - 1] - f[nextVisit[i - 1]] + 1 + mod) % mod;        }        return (int) f[n - 1];    }}

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