Graphs and Network Algorithms · Minimum Spanning Trees

Kruskal's Algorithm

Kruskal's algorithm builds a minimum spanning tree by taking edges in increasing weight order while avoiding cycles.

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Student Focus

Students see how a greedy choice and a data structure work together.

Guided Lesson Notes

Understanding Kruskal's Algorithm

Kruskal's Algorithm focuses on relationships, reachability, paths, cycles, connectivity, and optimization over edges. Kruskal's algorithm builds a minimum spanning tree by taking edges in increasing weight order while avoiding cycles.

Model: draw vertices as dots and edges as connections; then decide whether direction, weight, or capacity matters.

Invariant: visited, distance, parent, component, or flow arrays must match what has actually been discovered so far.

Implementation: adjacency lists or matrices, queues, stacks, priority queues, union-find, and repeated edge relaxation.

Use Kruskal's Algorithm when the problem mentions roads, networks, prerequisites, components, shortest paths, spanning costs, or dependency order.

Visual Model

A small picture for Kruskal's Algorithm

Key Ideas

Sorted edges
Cycle avoidance
Union-find support

Practice Prompts

Trace Kruskal on a weighted graph.
Explain why cycle checks are required after sorting edges.

Vocabulary

Terms students should be able to say clearly

Vertex

An object or state in the graph.

Edge

A relationship or move between vertices.

Path

A sequence of edges from one vertex to another.

Cycle

A path that returns to a previous vertex.

Component

A group of vertices connected by reachability.

Relaxation

Trying to improve a known distance or cost through an edge.

Worked Example

Worked example: tracing Kruskal's Algorithm

Trace a small input, focusing on sorted edges.

1Write a small input where Kruskal's Algorithm is clearly useful.
2Label the part of the input related to Sorted edges.
3Perform one operation and explain which invariant is still true afterward.
4Run a second operation that touches an edge case, such as an empty side, duplicate value, boundary index, disconnected vertex, or tie.
5Finish by saying which operation dominates the runtime and why.

Complexity Check

Costs students should be able to explain

Operation	Typical Cost	Reason
Adjacency-list traversal	O(V + E)	DFS and BFS visit vertices and inspect edges.
Adjacency matrix storage	O(V^2)	A matrix reserves space for every possible pair.
Weighted optimization	Varies	MST and shortest-path algorithms depend on sorting, heaps, and graph density.

Common Mistakes

What to watch while practicing

Coding before drawing the structure or state changes.
Forgetting the invariant that makes the algorithm correct.
Testing only the sample input and skipping boundary cases.
Giving Big-O without explaining which operation dominates the work.

Interview-Style Coding Problem

Kruskal's Algorithm interview problem: Minimum Cost to Connect All Cities

Interview medium

Problem

Given possible undirected connections between cities, find the minimum total cost to connect every city.

Input

The first line contains n and m. The next m lines contain u v cost using 1-based city labels.

Output

Print the minimum cost, or -1 if the graph cannot be fully connected.

Sample Input

Sample Output

Why the sample works

The minimum spanning tree uses costs 1, 2, and 3 for a total of 6.

Approach

1Sort all connections by cost.
2Use union-find to avoid cycles.
3Take an edge only if it connects two different components.
4The graph is connected only if exactly n - 1 edges are chosen.

O(m log m) time and O(n) extra space.

Java Solution

import java.util.*;

public class Main {
  static class DSU {
    int[] parent;
    int[] rank;

    DSU(int n) {
      parent = new int[n + 1];
      rank = new int[n + 1];
      for (int i = 1; i <= n; i++) parent[i] = i;
    }

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

    boolean union(int a, int b) {
      int ra = find(a);
      int rb = find(b);
      if (ra == rb) return false;
      if (rank[ra] < rank[rb]) {
        parent[ra] = rb;
      } else if (rank[ra] > rank[rb]) {
        parent[rb] = ra;
      } else {
        parent[rb] = ra;
        rank[ra]++;
      }
      return true;
    }
  }

  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int m = sc.nextInt();
    int[][] edges = new int[m][3];
    for (int i = 0; i < m; i++) {
      edges[i][0] = sc.nextInt();
      edges[i][1] = sc.nextInt();
      edges[i][2] = sc.nextInt();
    }

    Arrays.sort(edges, Comparator.comparingInt(a -> a[2]));
    DSU dsu = new DSU(n);
    int used = 0;
    long cost = 0;

    for (int[] edge : edges) {
      if (dsu.union(edge[0], edge[1])) {
        used++;
        cost += edge[2];
      }
    }

    System.out.println(used == n - 1 ? cost : -1);
  }
}

Python Solution

n, m = map(int, input().split())
edges = [tuple(map(int, input().split())) for _ in range(m)]
edges.sort(key=lambda x: x[2])

parent = list(range(n + 1))
rank = [0] * (n + 1)

def find(x):
    if parent[x] != x:
        parent[x] = find(parent[x])
    return parent[x]

def union(a, b):
    ra, rb = find(a), find(b)
    if ra == rb:
        return False
    if rank[ra] < rank[rb]:
        parent[ra] = rb
    elif rank[ra] > rank[rb]:
        parent[rb] = ra
    else:
        parent[rb] = ra
        rank[ra] += 1
    return True

used = 0
cost = 0
for u, v, w in edges:
    if union(u, v):
        used += 1
        cost += w

print(cost if used == n - 1 else -1)

Practice Challenge

Make the idea your own

Create a two-minute explanation of Kruskal's Algorithm: define it, trace one example, name one edge case, and give the runtime of the main operation.

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