First Day Where You Have Been in All the Rooms

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ByteDance

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ArraysGreedy AlgorithmsDynamic Programming

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

Try coding it in LeetCode

First Day Where You Have Been in All the Rooms

Medium

ByteDance

3 views

Topics:

ArraysGreedy AlgorithmsDynamic Programming

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the `nextVisit` array? Specifically, what is the maximum value of `nextVisit.length`?
What is the range of values that the elements in the `nextVisit` array can take? Can `nextVisit[i]` ever be out of bounds (i.e., greater than or equal to `nextVisit.length`)?
Is it guaranteed that the `nextVisit` array will always be valid in the sense that we will eventually visit all rooms at least once, or should I handle cases where we might get stuck in a loop and never visit all rooms?
Could you provide an example where `nextVisit[i]` is equal to `i` for some index `i` to clarify the behavior in such scenarios?
What is the expected behavior if the number of days exceeds the maximum representable value for an integer before all rooms are visited (even if unlikely), and is the modulo operation (10^9 + 7) applied at each step of the calculation or only to the final result?

Brute Force Solution

Approach

The brute force approach for visiting all rooms is like trying every possible path. We'll explore each room in every possible order, checking if that order lets us visit all rooms.

Here's how the algorithm would work step-by-step:

Start with a guess: what if we visit the rooms in the order they're numbered?
See if that works. Can we actually get into each room using the keys we find along the way?
If that doesn't work, let's try a different order. What if we visit room 2 first, then room 1, then room 3, and so on?
Again, check if this new order lets us get into every room using the keys we find.
Keep trying every single possible order of visiting the rooms. This means trying every possible combination.
For each order, keep track of whether we were able to visit every room.
Once we've tried every single order, find the earliest day that we successfully visited all rooms.

Code Implementation

from itertools import permutations

def first_day_where_all_rooms_visited_brute_force(rooms):
    number_of_rooms = len(rooms)

    # Try every possible order of visiting rooms
    for room_order in permutations(range(number_of_rooms)):
        days_needed = 0
        keys_available = {0}
        rooms_visited = {0}

        for room_index in room_order:
            days_needed += 1
            current_room = room_index

            # If we can't access the current room, skip to next order
            if current_room not in rooms_visited:
                break

            # Add newly found keys to our collection of keys.
            for key in rooms[current_room]:
                keys_available.add(key)

            # Adds the newly visited room to the rooms visited
            rooms_visited.add(current_room)

            rooms_visited.update(keys_available)

            # We visited every room in this order.
            if len(rooms_visited) == number_of_rooms:
                return days_needed

    return -1

Big(O) Analysis

Time Complexity

O(n! * n) – The brute force approach tries every permutation of room visits. There are n! (n factorial) possible orderings of visiting the n rooms. For each permutation, we simulate the process of visiting the rooms and collecting keys, which takes O(n) time in the worst case to check if all rooms can be visited in that specific order since each room visit may take a constant time to process. Therefore, the total time complexity is O(n! * n).

Space Complexity

O(N * N!) – The brute force approach explores every possible order of visiting rooms, which is N! permutations where N is the total number of rooms. For each permutation, we check if it's a valid path to visit all rooms. Keeping track of a single permutation requires storing an array of size N, representing the order of room visits. Since we store at most N! possible orderings that each consume N space, the space complexity is O(N * N!).

Optimal Solution

Approach

The core idea is to figure out the furthest room we can reach from any given point. We keep track of this reachable range and move forward. We stop when the reachable range covers all rooms.

Here's how the algorithm would work step-by-step:

Keep track of the furthest room we can reach from any given room using the keys we have.
Start at room zero.
Check what new rooms can be opened with the key in the current room and update the furthest reachable room.
Keep advancing to the next room.
If we reach a room beyond our current reachable range, it means we need to go back and revisit the rooms we've skipped to get new keys and extend our reach.
Continue this process of moving forward and expanding our reach until we can reach the last room.
The day we can reach the last room is the answer.

Code Implementation

def first_day_been_in_all_rooms(rooms):
    number_of_rooms = len(rooms)
    visited = [False] * number_of_rooms
    visited[0] = True
    furthest_reachable_room = 0
    current_room = 0
    days = 0

    while furthest_reachable_room < number_of_rooms - 1:
        # Update furthest reachable room.
        furthest_reachable_room = max(furthest_reachable_room, current_room + rooms[current_room])

        # Advance to the next room.
        next_room = current_room + 1

        if next_room > furthest_reachable_room:
            # Need to revisit previously visited rooms.
            current_room = 0
            while visited[current_room]:
                current_room += 1
            
        else:
            current_room = next_room

        if not visited[current_room]:
            # Mark the current room as visited.
            visited[current_room] = True

        days += 1

    return days

Big(O) Analysis

Time Complexity

O(n^2) – The algorithm iterates through rooms (n rooms in total). Inside the loop, it potentially revisits previously skipped rooms to expand the reachable range. In the worst-case scenario, each room's key could potentially require revisiting all preceding rooms to update the reachable range if it extends it, especially if rooms are not visited in an ascending or descending fashion. Therefore, each of the n rooms could trigger a loop of close to n times. This leads to approximately n * n operations, which is O(n^2).

Space Complexity

O(1) – The provided plain English solution primarily focuses on iterating through rooms and updating a reachable range. It doesn't explicitly mention the use of auxiliary data structures like lists, maps, or sets to store intermediate results or visited rooms. The algorithm seems to rely only on variables to keep track of the current room and furthest reachable room, which takes up constant space regardless of the number of rooms N. Therefore, the space complexity is O(1).

Edge Cases

Case	How to Handle
Empty input array nextVisit is null or has zero length	Return 0 as there are no rooms to visit if the input is invalid.
Single-element array nextVisit of length 1	Return 1 because we visit room 0 on day 0, and then go to nextVisit[0] which is another visit to room 0 on day 1, so all rooms have been visited.
nextVisit[i] always equals i, creating a self-loop for each room	The day will keep incrementing and we handle this case by appropriately computing the days, until all rooms are visited at least once.
nextVisit[i] always equals i+1 (except for the last element), creating a linear path	The algorithm should correctly traverse the rooms in sequence, accumulating days for each visit.
Maximum-sized input array (e.g., 10^5 elements) to test for time complexity issues	The solution needs to be optimized to avoid exceeding time limits, perhaps using dynamic programming to avoid recomputation.
Integer overflow in day count calculation	Apply modulo operation (10^9 + 7) after each addition to prevent overflow.
nextVisit[i] is greater than or equal to the length of nextVisit array, which causes out of bounds error when i is odd	Add check `if (nextVisit[i] >= n) nextVisit[i] = n-1;` where n is the length of `nextVisit`
nextVisit array creates cycles leading to infinite loop	Ensure the logic correctly accounts for revisits and correctly increments day.