Solution For Keys And Rooms

Problem Statement:

You are given an array called rooms which consists of n different rooms, numbered from 0 to n-1. Each room contains a set of keys that are needed to access other rooms. The key set in room i is represented as a list of integers rooms[i], which consists of unique integers in the range [0, n-1]. A key key in rooms[i] allows you to open a door to room key.

You start from room 0 and must visit every room. You can use any keys that are currently in your possession, but you cannot enter a room that you do not have a key for. If you can visit every room, return True, otherwise return False.

Solution:

To solve this problem, we can use a Depth First Search (DFS) algorithm to explore all the rooms. We can start by visiting room 0 and adding all its keys to a set called visited. We can then iterate through the keys of room 0 and recursively visit all the rooms that have not been visited yet, adding their keys to the set as we go along. We can repeat this process until we have visited all the rooms or until we cannot visit any more rooms.

To keep track of which rooms we have visited, we can use a boolean array called visited_rooms where visited_rooms[i] is True if we have visited room i and False otherwise. Initially, all elements of visited_rooms are False. Whenever we visit a new room, we mark its corresponding element in visited_rooms to be True.

Once we have visited all the rooms, we can check if all elements of visited_rooms are True. If they are, we return True, otherwise we return False.

Algorithm:

1. Create a boolean array called visited_rooms of size n and set all its elements to False.
2. Create a set called visited and add 0 to it.
3. Call the DFS function with inputs rooms, 0, visited, and visited_rooms.
4. In the DFS function, iterate through the keys of room u and if the corresponding element in visited_rooms is False, recursively call the DFS function with inputs rooms, key, visited, and visited_rooms.
5. Mark the corresponding element in visited_rooms to be True.
6. Return from the DFS function.
7. Check if all elements of visited_rooms are True. If they are, return True, otherwise return False.

Implementation:

“`
class Solution:
def canVisitAllRooms(self, rooms: List[List[int]]) -> bool:
n = len(rooms)
visited_rooms = [False] * n
visited = set()
visited.add(0)

    def dfs(u):
        for key in rooms[u]:
            if not visited_rooms[key]:
                visited.add(key)
                dfs(key)
        visited_rooms[u] = True

    dfs(0)
    return all(visited_rooms)

“`

Time Complexity:

The time complexity of this algorithm is O(n + m) where n is the number of rooms and m is the total number of keys in all the rooms. This is because we visit each room only once and we iterate through all the keys in all the rooms. Since m can be at most equal to n^2, the time complexity of this algorithm is O(n^2) in the worst case.

Space Complexity:

The space complexity of this algorithm is O(n) because we use a boolean array called visited_rooms of size n to keep track of which rooms we have visited. We also use a set called visited for the same purpose, which can contain at most n elements. In addition, we use the recursive stack of the DFS function, which can contain at most n elements. Therefore, the total space complexity of this algorithm is O(n).

Step by Step Implementation For Keys And Rooms

There are a few different ways to approach this problem. One way would be to use a depth-first search algorithm. Another way would be to use a breadth-first search algorithm.

We can also use a combination of both depth-first and breadth-first search.

1. Depth-First Search

We can use a depth-first search algorithm to find out if all the rooms can be visited. We can keep track of which rooms have been visited in a boolean array.

We can start from the first room and then explore all the rooms that are connected to it. We can mark the rooms as visited once we have explored them.

If we are able to visit all the rooms, then we can return true. Otherwise, we will return false.

public boolean canVisitAllRooms(List rooms) { boolean[] visited = new boolean[rooms.size()]; visited[0] = true; Stack stack = new Stack(); stack.push(0); while(!stack.isEmpty()) { int curr = stack.pop(); List room = rooms.get(curr); for(int i = 0; i < room.size(); i++) { int next = room.get(i); if(!visited[next]) { visited[next] = true; stack.push(next); } } } for(boolean v : visited) { if(!v) return false; } return true; }

2. Breadth-First Search

We can also use a breadth-first search algorithm to find out if all the rooms can be visited. We can keep track of which rooms have been visited in a boolean array.

We can start from the first room and then explore all the rooms that are connected to it. We can mark the rooms as visited once we have explored them.

If we are able to visit all the rooms, then we can return true. Otherwise, we will return false.

public boolean canVisitAllRooms(List rooms) { boolean[] visited = new boolean[rooms.size()]; visited[0] = true; Queue queue = new LinkedList(); queue.add(0); while(!queue.isEmpty()) { int curr = queue.poll(); List room = rooms.get(curr); for(int i = 0; i < room.size(); i++) { int next = room.get(i); if(!visited[next]) { visited[next] = true; queue.add(next); } } } for(boolean v : visited) { if(!v) return false; } return true; }

3. Depth-First Search + Breadth-First Search

We can also use a combination of both depth-first and breadth-first search.

We can keep track of which rooms have been visited in a boolean array.

We can start from the first room and then use a depth-first search to explore all the rooms that are connected to it. We can mark the rooms as visited once we have explored them.

We can then use a breadth-first search to explore all the rooms that are connected to the rooms that have been visited. We can mark the rooms as visited once we have explored them.

If we are able to visit all the rooms, then we can return true. Otherwise, we will return false.

There are N rooms and you start in room 0.  Each room has a distinct number in 0, 1, 2, ..., N-1, and each room may have some keys to access the next room. 

Formally, each room i has a list of keys rooms[i], and each key rooms[i][j] is an integer in [0, 1, ..., N-1] where N = rooms.length.  A key rooms[i][j] = v opens the room with number v.

Initially, all the rooms start locked (except for room 0). 

You can walk back and forth between rooms freely. 

Return true if and only if you can enter every room.

def canVisitAllRooms(self, rooms: List[List[int]]) -> bool:
        # keep track of which rooms have been visited
        visited = [False] * len(rooms)
        # add room 0 to the stack
        stack = [0]
        
        # while there are rooms in the stack
        while stack:
            # pop the first room off the stack
            room = stack.pop(0)
            # if the room has not been visited
            if not visited[room]:
                # mark the room as visited
                visited[room] = True
                # add the room's keys to the stack
                stack.extend(rooms[room])
                
        # return whether all rooms have been visited
        return all(visited)

There are a number of ways to solve this problem, but one approach would be to use a breadth-first search algorithm. The basic idea would be to start at the first room and then explore all of the rooms that are connected to it. Once all of the rooms connected to the first room have been explored, you would then move on to the next room and repeat the process.

One potential issue with this approach is that it is possible to get stuck in a cycle, where you keep exploring the same rooms over and over again. To avoid this, you can keep track of which rooms have been visited and only explore each room once.

function canVisitAllRooms(rooms) {
    // keep track of which rooms have been visited
    const visited = new Set();
    
    // keep track of the rooms that still need to be explored
    const queue = [0];
    
    // while there are still rooms to explore
    while (queue.length > 0) {
        // take the first room off of the queue
        const room = queue.shift();
        
        // mark the room as visited
        visited.add(room);
        
        // get the list of keys for the room
        const keys = rooms[room];
        
        // for each key in the room
        for (const key of keys) {
            // if the key leads to a new room
            if (!visited.has(key)) {
                // add the room to the queue
                queue.push(key);
            }
        }
    }
    
    // if we've visited all of the rooms, return true
    return visited.size === rooms.length;
}

There are N rooms and you start in room 0.  Each room has a distinct number in 0, 1, 2, ..., N-1, and each room may have some keys to access the next room. 

Formally, each room i has a list of keys rooms[i], and each key rooms[i][j] is an integer in [0, 1, ..., N-1] where N = rooms.length.  A key rooms[i][j] = v opens the room with number v.

Initially, all the rooms start locked (except for room 0). 

You can walk back and forth between rooms freely. 

Return true if and only if you can enter every room.

Example 1:

Input: [[1],[2],[3],[]]
Output: true
Explanation:  
We start in room 0, and pick up key 1.
We then go to room 1, and pick up key 2.
We then go to room 2, and pick up key 3.
We then go to room 3.  Since we were able to go to every room, we return true.
Example 2:

Input: [[1,3],[3,0,1],[2],[0]]
Output: false
Explanation: We can't enter the room with number 2.
Note:

1 <= rooms.length <= 1000
0 <= rooms[i].length <= 1000
The number of keys in all rooms combined is at most 3000.

There are a few ways to approach this problem. One way would be to use a depth-first search algorithm. Another way would be to use a breadth-first search algorithm.

We can start by initializing a list of rooms that we have not yet visited. We can then iterate through each room in the list. For each room, we can check if there are any keys in the room. If there are keys in the room, we can add those keys to our list of keys. We can then use those keys to open any doors in the room that are locked. Once we have opened all the doors in the room, we can add the room to our list of visited rooms.

If we reach a room that is already in our list of visited rooms, we can skip it.

Once we have iterated through all the rooms, we can check if our list of visited rooms is equal to the total number of rooms in the input. If it is, then we have visited all the rooms and we can return true. Otherwise, we can return false.

Here is some pseudocode for this approach:

function canVisitAllRooms(rooms): // Initialize a list of rooms that we have not yet visited unvisitedRooms = rooms // Iterate through each room in the list for room in unvisitedRooms: // Check if there are any keys in the room keys = room.getKeys() // If there are keys in the room, add them to our list of keys if (keys.length > 0): for key in keys: addKeyToList(key) // Use the keys to open any doors in the room that are locked openDoors(room, keys) // Add the room to our list of visited rooms addRoomToVisited(room) // If we reach a room that is already in our list of visited rooms, skip it if (room in visitedRooms): continue // Once we have iterated through all the rooms, check if our list of visited rooms // is equal to the total number of rooms in the input if (visitedRooms.length == rooms.length): return true return false