Find Center of Star Graph

📋

Problem

Imagine a social network where one person is the central hub connecting everyone else directly. Finding this person quickly helps in understanding the network's structure.

Given an undirected star graph consisting of n nodes labeled from 1 to n. A star graph is a graph where there is one center node connected to all other nodes, and no other edges exist. You are given a 2D integer array edges where each edges[i] = [ui, vi] indicates that there is an edge between nodes ui and vi. Return the center node of the given star graph.

3 ≤ n ≤ 10^5edges.length == n - 1edges[i].length == 21 ≤ ui, vi ≤ nui != viThe given edges represent a valid star graph

Edge cases: Minimum nodes (n=3) → center is the node connected to the other twoEdges where the first two edges share a common node → that node is centerEdges where the center node is the smallest numbered node

</>

IDE

def findCenter(edges: list[list[int]]) -> int:public int findCenter(int[][] edges)int findCenter(vector<vector<int>>& edges)function findCenter(edges)

def findCenter(edges):
    # Write your solution here
    pass

class Solution {
    public int findCenter(int[][] edges) {
        // Write your solution here
        return 0;
    }
}

#include <vector>
using namespace std;

int findCenter(vector<vector<int>>& edges) {
    // Write your solution here
    return 0;
}

function findCenter(edges) {
    // Write your solution here
}

Common Bugs to Avoid

Wrong: First node of first edgeAssuming the first node in the first edge is always the center without checking the second edge.✅ Return the node common to the first two edges instead of blindly returning the first node.

Wrong: Node with highest frequency but not centerIncorrect frequency counting or not counting both nodes in edges properly.✅ Count both nodes in each edge and return the node with frequency n-1.

Wrong: Smallest node alwaysAssuming center is always the smallest node without checking edges.✅ Check the common node in first two edges regardless of node values.

Wrong: Largest node alwaysAssuming center is always the largest node without checking edges.✅ Check the common node in first two edges regardless of node values.

✓

Test Cases

Try testing your solution on the smallest star graph and cases where center is smallest or largest node.

Consider edge cases where greedy or partial checks fail; verify your logic on node occurrences.

Think about the time complexity of your approach; can you find the center in constant time?

t1_01basic

Input{"edges":[[1,2],[2,3],[4,2]]}

Expected2

⏱ Performance - must finish in 2000ms

Node 2 is connected to nodes 1, 3, and 4, making it the center of the star.

💡 Check which node appears in all edges.

💡 The center node appears in every edge; count node occurrences.

💡 Compare the first two edges; the common node is the center.

Why it failed: Incorrect output means you likely did not identify the node common to all edges. Fix by checking the first two edges and returning their common node.

✓ Correctly identified the center node by checking node occurrences.

t1_02basic

Input{"edges":[[3,4],[3,2],[3,1],[3,5]]}

Expected3

⏱ Performance - must finish in 2000ms

Node 3 connects to all other nodes 1, 2, 4, and 5, so it is the center.

💡 Look for the node that appears in every edge.

💡 Count the frequency of nodes in edges; the center appears n-1 times.

💡 Check the first two edges for a common node; that node is the center.

Why it failed: Wrong output suggests failure to identify the node common to all edges. Fix by returning the node common to the first two edges.

✓ Center node correctly found by identifying the common node in first two edges.

t2_01edge

Input{"edges":[[1,2],[1,3]]}

Expected1

⏱ Performance - must finish in 2000ms

Minimum nodes (n=3). Node 1 connects to nodes 2 and 3, so it is the center.

💡 Test the smallest valid star graph with 3 nodes.

💡 The center node connects to both other nodes.

💡 Check the first two edges for the common node; it is the center.

Why it failed: Failed on minimum nodes test; likely did not handle smallest input correctly. Fix by ensuring code works for n=3 and returns the node common to both edges.

✓ Correctly handled minimum nodes case.

t2_02edge

Input{"edges":[[5,1],[5,2],[5,3],[5,4]]}

Expected5

⏱ Performance - must finish in 2000ms

Center node is the largest numbered node (5) connected to all others.

💡 Center node can be the largest node number.

💡 Do not assume center is smallest node; check edges carefully.

💡 Find the node common to the first two edges; that node is the center.

Why it failed: Incorrect output due to assuming center is smallest node. Fix by checking the common node in first two edges regardless of node values.

✓ Correctly identified center node even when it is the largest node.

t2_03edge

Input{"edges":[[2,1],[2,3],[2,4],[2,5]]}

Expected2

⏱ Performance - must finish in 2000ms

Center node is the smallest numbered node (2) connected to all others.

💡 Center node can be the smallest node number.

💡 Check the first two edges for the common node.

💡 Return the node common to the first two edges as center.

Why it failed: Incorrect output due to not handling smallest node as center. Fix by returning the node common to first two edges.

✓ Correctly identified center node when it is the smallest node.

t3_01corner

Input{"edges":[[1,2],[3,2],[4,2],[5,2],[6,2]]}

Expected2

⏱ Performance - must finish in 2000ms

Greedy trap: picking first node of first edge (1) is wrong; center is node 2.

💡 Don't assume the first node of the first edge is center.

💡 Check the common node between first two edges to find center.

💡 Return the node that appears in both first edges, not just the first node.

Why it failed: Failed greedy trap test by returning first node of first edge. Fix by returning the node common to first two edges.

✓ Correctly avoided greedy trap by checking first two edges.

t3_02corner

Input{"edges":[[7,3],[3,1],[3,2],[3,4],[3,5]]}

Expected3

⏱ Performance - must finish in 2000ms

Off-by-one error: center node 3 is not first node in all edges; must check both nodes.

💡 Check both nodes in each edge, not just first node.

💡 Find the node common to first two edges.

💡 Return the node that appears in both first two edges.

Why it failed: Failed by only checking first node of edges. Fix by checking both nodes in first two edges to find common node.

✓ Correctly handled node positions in edges.

t3_03corner

Input{"edges":[[10,1],[10,2],[10,3],[10,4],[10,5],[10,6],[10,7],[10,8],[10,9]]}

Expected10

⏱ Performance - must finish in 2000ms

Hash map frequency confusion: must not count frequencies incorrectly; center appears n-1 times.

💡 Count node frequencies correctly.

💡 Center node frequency equals n-1 edges.

💡 Return node with frequency n-1 from edges.

Why it failed: Incorrect frequency counting caused wrong output. Fix by counting both nodes in each edge and returning node with frequency n-1.

✓ Correctly counted frequencies and identified center.

t4_01performance

Input{"_description":"n=100000 at constraint boundary - executor generates this"}

Expectednull

⏱ Performance - must finish in 2000ms

n=100000 edges; O(n) solution must complete within 2 seconds.

💡 Optimal solution runs in O(n) time.

💡 Avoid nested loops; use constant time checks.

💡 Check only first two edges to find center node.

Why it failed: TLE due to O(n^2) or higher complexity. Fix by using O(n) or O(1) approach checking first two edges only.

✓ Solution runs in O(n) time and passes performance test.

Practice

1. You are given a list of edges representing an undirected graph with n nodes labeled from 0 to n-1. You need to determine if the graph forms a valid tree. Which algorithmic approach guarantees an optimal solution to check both connectivity and absence of cycles efficiently?

easy

A. Use a greedy algorithm that adds edges one by one and checks if the graph remains connected.

B. Apply dynamic programming to count the number of ways to reach each node and detect cycles.

C. Sort edges and use binary search to detect cycles and connectivity.

D. Perform a BFS or DFS traversal to detect cycles and verify that all nodes are connected.

Solution

1. Step 1: Understand problem requirements

   The problem requires checking if the graph is connected and contains no cycles, which defines a tree.

2. Step 2: Identify suitable algorithm

   BFS or DFS traversal can detect cycles by tracking visited nodes and parent relationships, and also verify connectivity by ensuring all nodes are visited.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   BFS/DFS simultaneously checks connectivity and cycles efficiently [OK]

Hint: Trees have n-1 edges and no cycles [OK]

Common Mistakes:

• Assuming greedy edge addition ensures connectivity
• Using DP which is unrelated here

2. Consider the following Union-Find based code snippet to check if a graph is bipartite. Given the graph adjacency list [[1,3],[0,2],[1,3],[0,2]], what is the return value of isBipartite(graph)?

easy

A. true

B. false

C. Raises an exception

D. null (infinite loop)

Solution

1. Step 1: Trace union operations for node 0

   Node 0 neighbors: 1 and 3. Union(1,3) is called. Since 1 and 3 are separate, union succeeds.

2. Step 2: Trace union operations for other nodes

   For node 1 neighbors: 0 and 2. Union(0,2) is called. 0 and 2 are separate, union succeeds. Similarly for node 2 and 3, unions succeed without conflict.

3. Step 3: Check for conflicts

   However, the union-find approach here incorrectly unions neighbors of the same node, which can cause false positives. The code logic is flawed for bipartite checking.

4. Final Answer:

   Option B -> Option B

5. Quick Check:

   The code returns false due to a union conflict, indicating the graph is not bipartite [OK]

Hint: Union-Find returns false only on odd cycle detection [OK]

Common Mistakes:

• Confusing union arguments order
• Assuming union returns true on conflict

3. What is the time complexity of the optimal bidirectional BFS approach for the Word Ladder problem, where N is the number of words in the dictionary and L is the length of each word?

medium

A. O(N^2 * L)

B. O(N * 26^L)

C. O(N * L^2)

D. O(N * L * 26)

Solution

1. Step 1: Identify operations per word

   For each word, we try changing each of its L letters to 26 possible letters, generating up to 26*L neighbors.

2. Step 2: Consider total words processed

   In worst case, all N words are visited, so total operations are O(N * L * 26).

3. Step 3: Consider the cost of checking each generated word

   Checking if a generated word is in the dictionary (word_set) can be done in O(1) average, but generating each neighbor involves string slicing which is O(L), so total complexity is O(N * L^2).

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Each neighbor generation involves O(L) slicing, repeated for O(N * L * 26) neighbors, resulting in O(N * L^2) complexity [OK]

Hint: Each word generates O(L*26) neighbors, each neighbor generation costs O(L) due to slicing [OK]

Common Mistakes:

• Confusing exponential 26^L with linear 26*L
• Assuming quadratic N^2 due to pairwise checks
• Ignoring string slicing cost in complexity

4. Suppose the problem is modified so that you can only move in 4 directions (up, down, left, right) instead of 8. Which of the following statements about the shortest path algorithm is true?

hard

A. Greedy Best-First Search guarantees shortest path with 4-direction moves

B. Bidirectional BFS still finds the shortest path, but the path length may be longer due to restricted moves

C. DFS with backtracking becomes optimal since fewer directions reduce complexity

D. Dynamic Programming is required to handle the reduced movement directions

Solution

1. Step 1: Understand movement restriction impact

   Reducing to 4 directions limits possible paths, potentially increasing shortest path length.

2. Step 2: Evaluate algorithm suitability

   Bidirectional BFS still finds shortest path efficiently; DFS and greedy do not guarantee shortest path.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Bidirectional BFS adapts to movement constraints, still optimal [OK]

Hint: Bidirectional BFS adapts to movement constraints [OK]

Common Mistakes:

• Assuming DFS or greedy become optimal
• Thinking DP is needed for movement changes

5. Suppose the Surrounded Regions problem is modified so that the board can be very large and the number of 'O's connected to the boundary is small. Which of the following modifications to the BFS approach would improve space efficiency without sacrificing correctness?

hard

A. Use DFS with recursion instead of BFS to avoid storing a large queue.

B. Only enqueue boundary 'O's and avoid marking visited cells, relying on board state changes.

C. Use BFS but mark visited cells in-place and prune early when no new 'O's are found.

D. Run BFS from every 'O' cell to ensure all regions are checked independently.

Solution

1. Step 1: Consider space usage of BFS queue

   BFS queue can grow large if many 'O's are connected; marking visited in-place helps avoid revisiting.

2. Step 2: Prune BFS early when no new 'O's are found

   This reduces unnecessary queue growth and improves space efficiency without losing correctness.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   In-place marking plus pruning optimizes BFS space [OK]

Hint: In-place marking and pruning reduce BFS queue size [OK]

Common Mistakes:

• Using DFS recursion can cause stack overflow on large boards.
• Not marking visited cells causes infinite loops or revisits.
• Running BFS from every 'O' is inefficient and unnecessary.