Graph Count Connected Components

mediumcodingAlgorithmsGraphsUnion Find~25 min

Asked at Google, Meta

Overview

Given n nodes labeled from 0 to n - 1 and a list of undirected edges, return the number of connected components in the graph.

Constraints

1 <= n <= 2000
0 <= edges.length <= 5000
edges[i].length === 2
0 <= edges[i][0], edges[i][1] < n
No duplicate edges.

Examples

countComponents(5, [[0,1], [1,2], [3,4]]);
// => 2
// Component 1: {0, 1, 2}
// Component 2: {3, 4}

countComponents(5, [[0,1], [1,2], [2,3], [3,4]]);
// => 1  (all connected)

countComponents(4, []);
// => 4  (each node is its own component)

Solution

Reveal solution

function countComponents(n, edges) {
  const parent = Array.from({ length: n }, (_, i) => i);

  function find(x) {
    while (parent[x] !== x) {
      parent[x] = parent[parent[x]]; // path compression
      x = parent[x];
    }
    return x;
  }

  function union(a, b) {
    const rootA = find(a), rootB = find(b);
    if (rootA === rootB) return false;
    parent[rootA] = rootB;
    return true;
  }

  let components = n;
  for (const [a, b] of edges) {
    if (union(a, b)) components--;
  }

  return components;
}

Union-Find with path compression. Start with n components and merge on each edge.

Resources

graph-connected-components.js

Graph Count Connected Components

mediumcodingAlgorithmsGraphs

Overview

Given n nodes labeled from 0 to n - 1 and a list of undirected edges, return the number of connected components in the graph.

Constraints

1 <= n <= 2000
0 <= edges.length <= 5000
edges[i].length === 2
0 <= edges[i][0], edges[i][1] < n
No duplicate edges.

Examples

countComponents(5, [[0,1], [1,2], [3,4]]);
// => 2
// Component 1: {0, 1, 2}
// Component 2: {3, 4}

countComponents(5, [[0,1], [1,2], [2,3], [3,4]]);
// => 1  (all connected)

countComponents(4, []);
// => 4  (each node is its own component)

Solution

Reveal solution

function countComponents(n, edges) {
  const parent = Array.from({ length: n }, (_, i) => i);

  function find(x) {
    while (parent[x] !== x) {
      parent[x] = parent[parent[x]]; // path compression
      x = parent[x];
    }
    return x;
  }

  function union(a, b) {
    const rootA = find(a), rootB = find(b);
    if (rootA === rootB) return false;
    parent[rootA] = rootB;
    return true;
  }

  let components = n;
  for (const [a, b] of edges) {
    if (union(a, b)) components--;
  }

  return components;
}

Union-Find with path compression. Start with n components and merge on each edge.

Resources

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