Find Center of Star Graph - Practice Coding Problems

Find Center of Star Graph - Problem

Graph Easy

Imagine a star-shaped network where one central hub connects to every other node, but no other connections exist between the outer nodes. This is called a star graph!

You're given an undirected star graph with n nodes labeled from 1 to n. In a star graph, there's exactly one center node that connects to all other n-1 nodes, forming a perfect star pattern ⭐.

Your task is to identify which node is the center of this star graph. You'll receive a 2D array edges where each edges[i] = [ui, vi] represents an undirected edge between nodes ui and vi.

Goal: Return the label of the center node.

Key insight: Since it's guaranteed to be a star graph, the center node will appear in every single edge, while all other nodes appear in exactly one edge!

Input & Output

example_1.py — Simple Star

$ Input: edges = [[1,2],[2,3],[4,2]]

› Output: 2

💡 Note: Node 2 is connected to nodes 1, 3, and 4, forming a perfect star with 2 at the center. All edges contain node 2, making it the center.

example_2.py — Minimum Star

$ Input: edges = [[1,2],[5,1],[1,3],[1,4]]

› Output: 1

💡 Note: Node 1 connects to all other nodes (2, 3, 4, 5). This creates a star pattern with 1 as the central hub.

example_3.py — Large Center ID

$ Input: edges = [[100,1],[100,2],[100,3]]

› Output: 100

💡 Note: Even with a large node ID, node 100 is clearly the center as it appears in every edge, connecting to nodes 1, 2, and 3.

Constraints

3 ≤ n ≤ 10⁵
edges.length == n - 1
edges[i].length == 2
1 ≤ u_i, v_i ≤ n
u_i ≠ v_i
The given input is guaranteed to be a valid star graph

Visualization

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Understanding the Visualization

Recognize the Pattern

In a star graph, exactly one node connects to all others - this is the center

Key Insight

The center node must appear in EVERY single edge of the graph

Smart Check

Instead of counting all edges, just check which node appears in both the first and second edge

Key Takeaway

🎯 Key Insight: The center of a star graph appears in every edge, so we can identify it by checking just the first two edges for the common node!

Asked in

G Google 23 a Amazon 18 f Meta 15 ⊞ Microsoft 12

The optimal solution leverages the key insight that in a star graph, the center node appears in every edge. We only need to check the first two edges - whichever node appears in both edges is the center. This gives us O(1) time complexity instead of counting all edges like the brute force approach.

Common Approaches

Approach	Time	Space	Notes
✓ Smart Two-Edge Check (Optimal)	O(1)	O(1)	Check only the first two edges - the center must appear in both
Count All Connections (Brute Force)	O(E)	O(V)	Count how many edges each node appears in - the center will appear in all n-1 edges

Smart Two-Edge Check (Optimal) — Algorithm Steps

Take the first edge [u1, v1]
Take the second edge [u2, v2]
Check if u1 appears in the second edge (u1 == u2 or u1 == v2)
If yes, return u1; otherwise return v1
This works because the center must appear in both edges

Visualization

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Step-by-Step Walkthrough

Examine First Edge

Look at first edge [u1, v1] - center is either u1 or v1

Check Second Edge

Look at second edge [u2, v2] - center must also be here

Find Common Node

The node that appears in both edges is the center

Code -

solution.c — C

#include <stdio.h>

int findCenter(int** edges, int edgesSize, int* edgesColSize) {
    // In a star graph, center appears in every edge
    // So center must appear in both first and second edge
    int u1 = edges[0][0], v1 = edges[0][1];
    int u2 = edges[1][0], v2 = edges[1][1];
    
    // Check if first node of first edge appears in second edge
    if (u1 == u2 || u1 == v2) {
        return u1;
    } else {
        return v1;
    }
}

int main() {
    // Example usage
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

We only need to examine the first two edges, which is constant time

✓ Linear Growth

Space Complexity

O(1)

No additional data structures needed, just a few variables

✓ Linear Space

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Medium Frequency

~8 min Avg. Time

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Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Smart Two-Edge Check (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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