Find Center of Star Graph

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Steps

setup

Initialize frequency map and calculate n

Create an empty frequency map to count node appearances. Calculate n as the number of nodes, which is edges count plus one.

💡 Initializing frequency map and n sets up the data structures needed to count node connections.

Line:

freq = {}
n = len(edges) + 1

💡 We know the graph has n nodes, so the center node must appear n-1 times in edges.

fill_cells

Process first edge [1, 2]: increment frequency for node 1

Start processing the first edge. Increment frequency count for node 1 by 1.

💡 Counting node 1's appearance in the first edge begins building frequency data.

Line:

freq[u] = freq.get(u, 0) + 1

💡 Each node's frequency increases by one for each edge it appears in.

fill_cells

Process first edge [1, 2]: increment frequency for node 2

Continue processing the first edge. Increment frequency count for node 2 by 1.

💡 Counting node 2's appearance in the first edge begins building frequency data.

Line:

freq[v] = freq.get(v, 0) + 1

💡 Node 2 now has frequency 1, reflecting one connection.

fill_cells

Process second edge [2, 3]: increment frequency for node 2

Start processing the second edge. Increment frequency count for node 2 by 1 (now 2).

💡 Node 2 appears again, increasing its frequency, indicating multiple connections.

Line:

freq[u] = freq.get(u, 0) + 1

💡 Node 2's frequency is increasing faster than others, hinting it may be the center.

fill_cells

Process second edge [2, 3]: increment frequency for node 3

Continue processing the second edge. Increment frequency count for node 3 by 1.

💡 Node 3 appears once, starting its frequency count.

Line:

freq[v] = freq.get(v, 0) + 1

💡 Node 3 has frequency 1, less than node 2's frequency so far.

fill_cells

Process third edge [4, 2]: increment frequency for node 4

Start processing the third edge. Increment frequency count for node 4 by 1.

💡 Node 4 appears once, starting its frequency count.

Line:

freq[u] = freq.get(u, 0) + 1

💡 Node 4 has frequency 1, same as nodes 1 and 3 so far.

fill_cells

Process third edge [4, 2]: increment frequency for node 2

Continue processing the third edge. Increment frequency count for node 2 by 1 (now 3).

💡 Node 2 appears again, increasing its frequency to 3, which is n-1.

Line:

freq[v] = freq.get(v, 0) + 1

💡 Node 2's frequency equals n-1, indicating it connects to all other nodes.

compare

Iterate over frequency map to find center node

Check each node's frequency to find the one equal to n-1 (3).

💡 This step identifies the node connected to all others by its frequency.

Line:

for node, count in freq.items():
    if count == n - 1:
        return node

💡 The center node is the one with frequency exactly n-1.

reconstruct

Confirm center node selection

Confirm node 2 as the center node since its frequency equals n-1.

💡 This step finalizes the selection before returning the answer.

Line:

return node

💡 The center node is uniquely identified by its frequency count.

reconstruct

Return center node as answer

Return node 2 as the center of the star graph because its frequency equals n-1.

💡 Returning the node with maximum frequency completes the algorithm.

Line:

return node

💡 The algorithm successfully finds the center node by frequency counting.

from typing import List

def findCenter(edges: List[List[int]]) -> int:
    freq = {}  # STEP 1: Initialize frequency map
    n = len(edges) + 1  # STEP 1: Calculate number of nodes
    for u, v in edges:  # STEP 2-7: Count frequencies for each node in edges
        freq[u] = freq.get(u, 0) + 1  # STEP 2,4,6: Increment freq for u
        freq[v] = freq.get(v, 0) + 1  # STEP 3,5,7: Increment freq for v
    for node, count in freq.items():  # STEP 8
        if count == n - 1:
            return node  # STEP 9

if __name__ == '__main__':
    edges = [[1,2],[2,3],[4,2]]
    print(findCenter(edges))  # Output: 2

📊

Find Center of Star Graph - Watch the Algorithm Execute, Step by Step

Watching the frequency counting and final selection step-by-step reveals how the center node is uniquely connected to all others in a star graph.

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Key Takeaways

✓ The center node in a star graph appears in all edges, so its frequency equals n-1.

This is hard to see from code alone because frequency counting is implicit; visualization makes it explicit.

✓ Incrementing frequency counts for both nodes in each edge accumulates connection data efficiently.

Visualizing each increment clarifies how the algorithm builds the frequency map step-by-step.

✓ The final comparison step selects the node with frequency n-1 as the center, confirming the star structure.

Seeing the decision branch in the trace helps understand why only one node qualifies as center.

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 buggy code snippet for the 01 Matrix problem. Which line contains the subtle bug that can cause incorrect distances or runtime errors?

medium

A. Line initializing dist with float('inf')

B. Line setting dist[i][j] = 0 when mat[i][j] == 0

C. Line with 'if i >= 0:' condition

D. Line with 'if j < n - 1:' condition in backward pass

Solution

1. Step 1: Analyze the condition 'if i >= 0:'

   This condition is always true for i in range(m), so dist[i-1][j] access is unsafe when i=0 (index -1).

2. Step 2: Identify the bug

   Accessing dist[-1][j] causes incorrect behavior or runtime error; correct condition should be 'if i > 0:'.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Boundary check off-by-one leads to invalid index [OK]

Hint: Check boundary conditions carefully to avoid invalid indices [OK]

Common Mistakes:

• Off-by-one in boundary checks
• Incorrect initialization
• Misplaced conditions

3. 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

4. Suppose the graph may contain cycles. Which modification to the optimal algorithm is necessary to correctly find the minimal set of vertices to reach all nodes?

hard

A. Run the same topological sort algorithm; cycles do not affect the result.

B. Use BFS from each node to find reachability since topological sort fails.

C. Ignore in-degree and pick nodes with maximum out-degree instead.

D. Detect cycles first and return an error or handle strongly connected components.

Solution

1. Step 1: Understand cycle impact

   Topological sort requires DAGs; cycles prevent valid ordering and break the algorithm.

2. Step 2: Modify approach

   Detect cycles using DFS or union-find. For cyclic graphs, identify strongly connected components and treat them as single nodes or return error.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Cycle detection is essential before topological sort [OK]

Hint: Topological sort requires DAG; detect cycles first [OK]

Common Mistakes:

• Assuming topological sort works on cyclic graphs
• Ignoring cycles and returning incorrect minimal sets
• Using BFS from each node which is inefficient

5. Suppose the graph edges can have negative weights but no negative cycles. Which algorithm modification or choice correctly computes the shortest network delay times from a start node?

hard

A. Use DFS with path relaxation and memoization to handle negative weights efficiently.

B. Use Dijkstra's algorithm with a priority queue as usual; it handles negative weights correctly.

C. Use BFS since negative weights can be ignored for shortest paths.

D. Use Bellman-Ford algorithm to relax edges repeatedly, detecting negative cycles if any.

Solution

1. Step 1: Recognize that Dijkstra's algorithm fails with negative weights because it assumes non-negative edge costs

   BFS cannot handle weighted edges properly, and DFS with memoization is inefficient and may not terminate correctly.

2. Step 2: Bellman-Ford algorithm is designed to handle negative weights and detect negative cycles by repeated edge relaxation

   It guarantees correct shortest paths if no negative cycles exist.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Bellman-Ford handles negative weights safely [OK]

Hint: Bellman-Ford handles negative weights; Dijkstra does not [OK]

Common Mistakes:

• Using Dijkstra with negative weights incorrectly
• Assuming BFS works for weighted graphs
• Believing DFS with memoization is efficient here