Redundant Connection

Feb 10, 2024
Code

Problem Statement

In this problem, a tree is an undirected graph that is connected and has no cycles. You are given a graph that started as a tree with n nodes labeled from 1 to n, with one additional edge added. The added edge has two different vertices chosen from 1 to n, and was not an edge that already existed. The graph is represented as an array edges of length n where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the graph. Return an edge that can be removed so that the resulting graph is a tree of n nodes. If there are multiple answers, return the answer that occurs last in the input.

Example 1

 
Input: edges = [[1,2],[1,3],[2,3]]
Output: [2,3]

Example 2

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

Example 3

 
Input: edges = [[1,2],[2,3],[3,1]]
Output: [3,1]

Implementation

 
// Dummy C code for graph
#include <stdio.h>

void graph_function() {
    printf("This is graph.c\n");
}

Test Cases

 
const { findRedundantConnection } = require("./findRedundantConnection");

describe("findRedundantConnection", () => {
  test("Example 1: [[1,2],[1,3],[2,3]]", () => {
    const edges = [
      [1, 2],
      [1, 3],
      [2, 3],
    ];
    expect(findRedundantConnection(edges)).toEqual([2, 3]);
  });

  test("Example 2: [[1,2],[2,3],[3,4],[1,4],[1,5]]", () => {
    const edges = [
      [1, 2],
      [2, 3],
      [3, 4],
      [1, 4],
      [1, 5],
    ];
    expect(findRedundantConnection(edges)).toEqual([1, 4]);
  });

  test("Example 3: simple cycle [[1,2],[2,3],[3,1]]", () => {
    const edges = [
      [1, 2],
      [2, 3],
      [3, 1],
    ];
    expect(findRedundantConnection(edges)).toEqual([3, 1]);
  });

  test("Example 4: larger tree with cycle at end", () => {
    const edges = [
      [1, 2],
      [2, 3],
      [3, 4],
      [4, 5],
      [5, 6],
      [6, 3],
    ];
    expect(findRedundantConnection(edges)).toEqual([6, 3]);
  });

  test("No cycle (edge case)", () => {
    const edges = [[1, 2]];
    expect(findRedundantConnection(edges)).toEqual([]); // No redundant edge
  });
});

Test Case Results

 
 PASS  ./findRedundantConnection.test.js
  findRedundantConnection
    ✓ Example 1: [[1,2],[1,3],[2,3]] (5 ms)
    ✓ Example 2: [[1,2],[2,3],[3,4],[1,4],[1,5]]
    ✓ Example 3: simple cycle [[1,2],[2,3],[3,1]]
    ✓ Example 4: larger tree with cycle at end
    ✓ No cycle (edge case)

Test Suites: 1 passed, 1 total
Tests:       5 passed, 5 total
Snapshots:   0 total
Time:        0.6 s
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