LeetCode #684 — MEDIUM

Redundant Connection

Move from brute-force thinking to an efficient approach using union-find strategy.

The Problem

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] = [a_i, b_i] indicates that there is an edge between nodes a_i and b_i 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]

Constraints:

n == edges.length
3 <= n <= 1000
edges[i].length == 2
1 <= a_i < b_i <= edges.length
a_i != b_i
There are no repeated edges.
The given graph is connected.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Union-Find

Example 1

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

Example 2

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

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #684: Redundant Connection
class Solution {
    private int[] p;

    public int[] findRedundantConnection(int[][] edges) {
        int n = edges.length;
        p = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
        }
        for (int i = 0;; ++i) {
            int pa = find(edges[i][0] - 1);
            int pb = find(edges[i][1] - 1);
            if (pa == pb) {
                return edges[i];
            }
            p[pa] = pb;
        }
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}

// Accepted solution for LeetCode #684: Redundant Connection
func findRedundantConnection(edges [][]int) []int {
	n := len(edges)
	p := make([]int, n)
	for i := range p {
		p[i] = i
	}
	var find func(int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	for i := 0; ; i++ {
		pa, pb := find(edges[i][0]-1), find(edges[i][1]-1)
		if pa == pb {
			return edges[i]
		}
		p[pa] = pb
	}
}

# Accepted solution for LeetCode #684: Redundant Connection
class Solution:
    def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
        def find(x: int) -> int:
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        p = list(range(len(edges)))
        for a, b in edges:
            pa, pb = find(a - 1), find(b - 1)
            if pa == pb:
                return [a, b]
            p[pa] = pb

// Accepted solution for LeetCode #684: Redundant Connection
use std::cmp::Ordering;

struct UnionFind {
    parent: Vec<usize>,
    rank: Vec<i32>,
}

impl UnionFind {
    fn new(n: usize) -> Self {
        UnionFind {
            parent: (0..(n + 1)).collect(),
            rank: vec![1; n + 1],
        }
    }

    fn find(&mut self, n: usize) -> usize {
        let mut p = self.parent[n];

        while p != self.parent[p] {
            self.parent[p] = self.parent[self.parent[p]];
            p = self.parent[p];
        }

        p
    }

    fn union(&mut self, n1: usize, n2: usize) -> bool {
        let p1 = self.find(n1);
        let p2 = self.find(n2);

        if p1 == p2 {
            return false;
        }
        match self.rank[p1].cmp(&self.rank[p2]) {
            Ordering::Greater => {
                self.parent[p2] = p1;
                self.rank[p1] += self.rank[p2];
            }
            _ => {
                self.parent[p1] = p2;
                self.rank[p2] = self.rank[p1];
            }
        }
        true
    }
}

impl Solution {
    pub fn find_redundant_connection(edges: Vec<Vec<i32>>) -> Vec<i32> {
        let mut union_find = UnionFind::new(edges.len() + 1);

        for edge in edges {
            let (n1, n2) = (edge[0] as usize, edge[1] as usize);
            if !union_find.union(n1, n2) {
                return vec![n1 as i32, n2 as i32];
            }
        }
        unreachable!()
    }
}

// Accepted solution for LeetCode #684: Redundant Connection
function findRedundantConnection(edges: number[][]): number[] {
    const n = edges.length;
    const p: number[] = Array.from({ length: n }, (_, i) => i);
    const find = (x: number): number => {
        if (p[x] !== x) {
            p[x] = find(p[x]);
        }
        return p[x];
    };
    for (let i = 0; ; ++i) {
        const pa = find(edges[i][0] - 1);
        const pb = find(edges[i][1] - 1);
        if (pa === pb) {
            return edges[i];
        }
        p[pa] = pb;
    }
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(n) space

Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.

UNION-FIND

O(α(n)) time

O(n) space

With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.

Shortcut: Union-Find with path compression + rank → O(α(n)) per operation ≈ O(1). Just say “nearly constant.”

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.