#0296

Redundant Connection

MediumDSA28 min20 XP

Problem

Given undirected edges that form a tree plus one extra edge, return the extra edge that creates a cycle.

Why This Matters

This is union-find cycle detection in its simplest interview form.

Function Signature

def find_redundant_connection(edges):

Examples

Example 1

Inputedges = [[1,2],[1,3],[2,3]]

Output[2, 3]

1-2 and 1-3 already connect 2 and 3, so edge 2-3 creates a cycle.

Constraints

Return the exact requested value.
Handle edge cases cleanly.
Use the intended DSA pattern when brute force would scale poorly.

CodePython

def find_redundant_connection(edges):
    """
    Input example: edges = [[1,2],[1,3],[2,3]]
    Return: [2, 3]
    Return the value from the function so the tests can compare it. Do not print unless the prompt asks.
    """
    # write your code here
    return []

def find_redundant_connection(edges):
    """
    Input example: edges = [[1,2],[1,3],[2,3]]
    Return: [2, 3]
    Return the value from the function so the tests can compare it. Do not print unless the prompt asks.
    """
    # write your code here
    return []

Visible browser tests run here when available.

Testcases2 visible / 3 hidden categories

Triangle

Input[[1,2],[1,3],[2,3]]

Expected[2, 3]

The third edge closes the cycle.

Later cycle

Input[[1,2],[2,3],[3,4],[1,4],[1,5]]

Expected[1, 4]

The edge 1-4 connects nodes already in the same component.

Hidden Test Categories

cycle at final edgelarger labelschain plus one closing edge