Keys and Rooms

▶

Steps

setup

Initialize variables

Initialize the number of rooms (n), the visited array to track visited rooms, and the count of visited rooms set to zero.

💡 Setting up these variables is essential to track progress and avoid revisiting rooms.

Line:

n = len(rooms)
visited = [False] * n
count = 0

💡 We now have a clean slate to start DFS and track which rooms get visited.

traverse

Start DFS at room 0

Begin DFS by calling dfs(0). Mark room 0 as visited and increment count to 1.

💡 Starting at room 0 is the problem's requirement and the root of our traversal.

Line:

def dfs(room):
    if visited[room]:
        return
    visited[room] = True
    count += 1

# call dfs(0)

💡 Room 0 is now the first visited room, and the recursion stack contains room 0.

traverse

Explore key from room 0 to room 1

From room 0, find key to room 1. Since room 1 is unvisited, prepare to DFS into room 1.

💡 Keys in the current room determine which rooms we can visit next.

Line:

for key in rooms[room]:
    dfs(key)

💡 The edge from 0 to 1 will be traversed as part of the DFS tree.

traverse

DFS into room 1

Enter room 1, mark it visited, increment count to 2, and add it to the recursion stack.

💡 Visiting a new room means we have a new node in the DFS tree.

Line:

if visited[room]:
    return
visited[room] = True
count += 1

💡 Room 1 is now processing and counted as visited.

traverse

Explore key from room 1 to room 2

From room 1, find key to room 2. Since room 2 is unvisited, prepare to DFS into room 2.

💡 Each key leads to a new room to explore if not visited.

Line:

for key in rooms[room]:
    dfs(key)

💡 The edge from 1 to 2 will be traversed as part of the DFS tree.

traverse

DFS into room 2

Enter room 2, mark it visited, increment count to 3, and add it to the recursion stack.

💡 Visiting room 2 continues the DFS traversal deeper.

Line:

if visited[room]:
    return
visited[room] = True
count += 1

💡 Room 2 is now processing and counted as visited.

traverse

Explore key from room 2 to room 3

From room 2, find key to room 3. Since room 3 is unvisited, prepare to DFS into room 3.

💡 Keys continue to unlock new rooms to explore.

Line:

for key in rooms[room]:
    dfs(key)

💡 The edge from 2 to 3 will be traversed as part of the DFS tree.

traverse

DFS into room 3

Enter room 3, mark it visited, increment count to 4, and add it to the recursion stack.

💡 Room 3 is the last room and marks completion of visiting all rooms.

Line:

if visited[room]:
    return
visited[room] = True
count += 1

💡 Room 3 is now processing and counted as visited.

prune

Early exit after visiting all rooms

Since count equals total rooms (4), the DFS returns early without exploring further keys.

💡 Early stopping saves time by not exploring unnecessary paths.

Line:

if count == n:
    return

💡 The algorithm efficiently stops recursion once the goal is met.

traverse

Backtrack and mark rooms done

Backtrack from room 3 to room 2, marking room 3 as done, then continue backtracking marking rooms done until returning to room 0.

💡 Backtracking completes the DFS call stack and finalizes visited rooms.

Line:

return from dfs calls after exploring all keys

💡 Rooms are fully processed and no further exploration is needed.

reconstruct

Return final result

After DFS completes, return true because count equals total rooms, meaning all rooms were visited.

💡 The final return confirms if all rooms are accessible.

Line:

return count == n

💡 The problem is solved successfully with all rooms visited.

def canVisitAllRooms(rooms):  # STEP 1
    n = len(rooms)  # STEP 1
    visited = [False] * n  # STEP 1
    count = 0  # STEP 1

    def dfs(room):  # STEP 2
        nonlocal count
        if visited[room]:  # STEP 4,6,8
            return  # STEP 9
        visited[room] = True  # STEP 2,4,6,8
        count += 1  # STEP 2,4,6,8
        if count == n:  # STEP 9
            return  # STEP 9
        for key in rooms[room]:  # STEP 3,5,7
            dfs(key)  # STEP 3,5,7

    dfs(0)  # STEP 2
    return count == n  # STEP 11

if __name__ == '__main__':
    rooms = [[1],[2],[3],[]]
    print(canVisitAllRooms(rooms))  # True

📊

Keys and Rooms - Watch the Algorithm Execute, Step by Step

Watching the algorithm step through each room and key reveals how DFS explores the graph and why early stopping is possible once all rooms are visited.

Visiting: none

0

1

2

3

Visiting: 0

0

1

2

3

Queue / Stack

0

Visiting: 0

Edge: 0 → 1

0

1

2

3

Queue / Stack

0

Visiting: 1

0

1

2

3

Queue / Stack

01

Visiting: 1

Edge: 1 → 2

0

1

2

3

Queue / Stack

01

Visiting: 2

0

1

2

3

Queue / Stack

012

Visiting: 2

Edge: 2 → 3

0

1

2

3

Queue / Stack

012

Visiting: 3

0

1

2

3

Queue / Stack

0123

Visiting: 3

0

1

2

3

Queue / Stack

0123

Visiting: none

0

1

2

3

Visiting: none

0

1

2

3

Key Takeaways

✓ DFS explores rooms by following keys, marking rooms visited to avoid cycles.

Reading code alone can hide how DFS uses recursion to explore connected rooms; visualization shows the call stack and visited states clearly.

✓ Early exit when all rooms are visited saves unnecessary exploration.

The early stopping condition is subtle in code but critical for efficiency; seeing it in action clarifies its impact.

✓ Edges used in DFS form a tree showing the path of exploration.

Visualizing tree edges helps understand the traversal order and how keys unlock new rooms step-by-step.

Practice

1. Consider the following BFS-based cloneGraph function. Given the input graph with nodes 1 connected to 2 and 3, and node 2 connected to 3, what is the value of the neighbors list of the cloned node corresponding to node 1 after the first iteration of the while loop (i.e., after processing node 1)?

easy

A. A list containing cloned nodes with values [2, 3]

B. An empty list []

C. A list containing cloned nodes with values [1]

D. A list containing cloned nodes with values [3, 2]

Solution

1. Step 1: Trace the first iteration of the while loop

   Initially, queue contains node 1. We pop node 1 and iterate over its neighbors: nodes 2 and 3.

2. Step 2: Update visited and neighbors

   Nodes 2 and 3 are not visited, so they are cloned and added to visited and queue. Then, the cloned node 1's neighbors list is appended with cloned nodes 2 and 3 in order.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Neighbors of cloned node 1 after first iteration are clones of nodes 2 and 3 [OK]

Hint: Neighbors appended in order during BFS iteration [OK]

Common Mistakes:

• Assuming neighbors list is empty after first iteration
• Mixing up order of neighbors appended

2. Consider the following Python code implementing Kahn's Algorithm for course scheduling. What is the return value of canFinish(3, [[1,0],[2,1]]) after the while loop completes?

easy

A. false

B. true

C. Raises an error due to empty queue

D. Returns true only if numCourses is 2

Solution

1. Step 1: Initialize graph and in-degree arrays

   Graph: 0 -> [1], 1 -> [2], 2 -> []. In-degree: [0,1,1]. Queue starts with nodes having in-degree 0 -> [0].

2. Step 2: Process queue and update count

   Pop 0, count=1; decrement in-degree of 1 to 0, enqueue 1. Pop 1, count=2; decrement in-degree of 2 to 0, enqueue 2. Pop 2, count=3; queue empty.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   All courses processed, so return true [OK]

Hint: Count equals numCourses means no cycle [OK]

Common Mistakes:

• Off-by-one error in count increment
• Misunderstanding queue initialization
• Confusing in-degree decrement timing

3. You are given a grid of heights and need to find a path from the top-left to the bottom-right cell such that the maximum absolute difference in heights between adjacent cells on the path is minimized. Which algorithmic approach guarantees finding the optimal solution efficiently?

easy

A. Dijkstra's algorithm variant treating edge weights as the maximum effort along the path

B. Greedy algorithm that always moves to the adjacent cell with the smallest height difference

C. Dynamic programming that sums all height differences along paths

D. Depth-first search with backtracking exploring all paths exhaustively

Solution

1. Step 1: Understand the problem's cost metric

   The cost to minimize is the maximum edge effort along the path, not the sum of efforts.

2. Step 2: Identify suitable algorithm

   Dijkstra's algorithm can be adapted to track the maximum edge cost on paths, guaranteeing the minimal maximum effort path.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Dijkstra's variant correctly handles max edge cost minimization [OK]

Hint: Minimize max edge cost -> Dijkstra variant [OK]

Common Mistakes:

• Assuming greedy local choice leads to global optimum
• Using sum of efforts instead of max
• Exhaustive DFS is too slow

4. Examine the following buggy code snippet for bidirectional BFS to find the shortest path in a binary matrix. Which line contains the subtle bug that can cause infinite loops or incorrect results?

def visit(queue, visited, other_visited):
    r, c = queue.popleft()
    for dr, dc in directions:
        nr, nc = r + dr, c + dc
        if 0 <= nr < n and 0 <= nc < n and grid[nr][nc] == 0:
            if (nr, nc) in other_visited:
                return visited[(r, c)] + other_visited[(nr, nc)]
            if (nr, nc) not in visited:
                queue.append((nr, nc))  # Bug here
                # Missing visited marking here
    return None

medium

A. Line where the current cell is popped from the queue

B. Line where the cell is checked if in other_visited

C. Line where neighbors are checked for grid boundaries

D. Line where neighbor is appended to queue without marking visited

Solution

1. Step 1: Identify visited marking usage

   Visited must be marked when neighbor is enqueued to avoid revisiting.

2. Step 2: Locate missing visited marking

   Neighbor is appended to queue but not marked visited, causing possible infinite revisits.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Marking visited on enqueue prevents infinite loops [OK]

Hint: Mark visited when enqueuing neighbors [OK]

Common Mistakes:

• Mark visited only when dequeuing
• Forgetting visited marking entirely

5. Suppose the graph edges can be reused multiple times (multi-edges allowed) and self-loops exist. Which modification to the bipartite check algorithm is necessary to correctly handle this variant?

hard

A. Reject any graph with self-loops immediately as non-bipartite

B. Modify Union-Find to union each edge multiple times without changes

C. Use BFS coloring but skip nodes with self-loops to avoid conflicts

D. Ignore self-loops and treat multi-edges as single edges before running BFS or Union-Find

Solution

1. Step 1: Understand the impact of self-loops on bipartiteness

   A self-loop connects a vertex to itself, making it impossible to assign two different colors, so the graph is not bipartite.

2. Step 2: Recognize that multi-edges do not affect bipartiteness if handled properly

   Multi-edges can be treated as single edges; they don't create odd cycles by themselves.

3. Step 3: Conclude that immediate rejection on self-loops is necessary

   Any presence of self-loops means the graph cannot be bipartite, so the algorithm must detect and reject such graphs upfront.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Self-loops break bipartite property immediately [OK]

Hint: Self-loops always break bipartite property [OK]

Common Mistakes:

• Ignoring self-loops
• Trying to color self-looped nodes