LeetCode #1579 — HARD

Remove Max Number of Edges to Keep Graph Fully Traversable

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

Alice and Bob have an undirected graph of n nodes and three types of edges:

  • Type 1: Can be traversed by Alice only.
  • Type 2: Can be traversed by Bob only.
  • Type 3: Can be traversed by both Alice and Bob.

Given an array edges where edges[i] = [typei, ui, vi] represents a bidirectional edge of type typei between nodes ui and vi, find the maximum number of edges you can remove so that after removing the edges, the graph can still be fully traversed by both Alice and Bob. The graph is fully traversed by Alice and Bob if starting from any node, they can reach all other nodes.

Return the maximum number of edges you can remove, or return -1 if Alice and Bob cannot fully traverse the graph.

Example 1:

Input: n = 4, edges = [[3,1,2],[3,2,3],[1,1,3],[1,2,4],[1,1,2],[2,3,4]]
Output: 2
Explanation: If we remove the 2 edges [1,1,2] and [1,1,3]. The graph will still be fully traversable by Alice and Bob. Removing any additional edge will not make it so. So the maximum number of edges we can remove is 2.

Example 2:

Input: n = 4, edges = [[3,1,2],[3,2,3],[1,1,4],[2,1,4]]
Output: 0
Explanation: Notice that removing any edge will not make the graph fully traversable by Alice and Bob.

Example 3:

Input: n = 4, edges = [[3,2,3],[1,1,2],[2,3,4]]
Output: -1
Explanation: In the current graph, Alice cannot reach node 4 from the other nodes. Likewise, Bob cannot reach 1. Therefore it's impossible to make the graph fully traversable.

Constraints:

  • 1 <= n <= 105
  • 1 <= edges.length <= min(105, 3 * n * (n - 1) / 2)
  • edges[i].length == 3
  • 1 <= typei <= 3
  • 1 <= ui < vi <= n
  • All tuples (typei, ui, vi) are distinct.
Patterns Used
🔗Union-Find→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: Alice and Bob have an undirected graph of n nodes and three types of edges: Type 1: Can be traversed by Alice only. Type 2: Can be traversed by Bob only. Type 3: Can be traversed by both Alice and Bob. Given an array edges where edges[i] = [typei, ui, vi] represents a bidirectional edge of type typei between nodes ui and vi, find the maximum number of edges you can remove so that after removing the edges, the graph can still be fully traversed by both Alice and Bob. The graph is fully traversed by Alice and Bob if starting from any node, they can reach all other nodes. Return the maximum number of edges you can remove, or return -1 if Alice and Bob cannot fully traverse the graph.

Baseline thinking

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

Pattern signal: Union-Find

Example 1

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

Example 2

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

Example 3

4
[[3,2,3],[1,1,2],[2,3,4]]
Step 02

Core Insight

What unlocks the optimal approach

  • Build the network instead of removing extra edges.
  • Suppose you have the final graph (after removing extra edges). Consider the subgraph with only the edges that Alice can traverse. What structure does this subgraph have? How many edges are there?
  • Use disjoint set union data structure for both Alice and Bob.
  • Always use Type 3 edges first, and connect the still isolated ones using other edges.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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
Largest constraint values
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 #1579: Remove Max Number of Edges to Keep Graph Fully Traversable
class UnionFind {
    private int[] p;
    private int[] size;
    public int cnt;

    public UnionFind(int n) {
        p = new int[n];
        size = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
            size[i] = 1;
        }
        cnt = n;
    }

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

    public boolean union(int a, int b) {
        int pa = find(a - 1), pb = find(b - 1);
        if (pa == pb) {
            return false;
        }
        if (size[pa] > size[pb]) {
            p[pb] = pa;
            size[pa] += size[pb];
        } else {
            p[pa] = pb;
            size[pb] += size[pa];
        }
        --cnt;
        return true;
    }
}

class Solution {
    public int maxNumEdgesToRemove(int n, int[][] edges) {
        UnionFind ufa = new UnionFind(n);
        UnionFind ufb = new UnionFind(n);
        int ans = 0;
        for (var e : edges) {
            int t = e[0], u = e[1], v = e[2];
            if (t == 3) {
                if (ufa.union(u, v)) {
                    ufb.union(u, v);
                } else {
                    ++ans;
                }
            }
        }
        for (var e : edges) {
            int t = e[0], u = e[1], v = e[2];
            if (t == 1 && !ufa.union(u, v)) {
                ++ans;
            }
            if (t == 2 && !ufb.union(u, v)) {
                ++ans;
            }
        }
        return ufa.cnt == 1 && ufb.cnt == 1 ? ans : -1;
    }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(α(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.

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