LeetCode #2685 — MEDIUM

Count the Number of Complete Components

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

The Problem

Problem Statement

You are given an integer n. There is an undirected graph with n vertices, numbered from 0 to n - 1. You are given a 2D integer array edges where edges[i] = [a_i, b_i] denotes that there exists an undirected edge connecting vertices a_i and b_i.

Return the number of complete connected components of the graph.

A connected component is a subgraph of a graph in which there exists a path between any two vertices, and no vertex of the subgraph shares an edge with a vertex outside of the subgraph.

A connected component is said to be complete if there exists an edge between every pair of its vertices.

Example 1:

Input: n = 6, edges = [[0,1],[0,2],[1,2],[3,4]]
Output: 3
Explanation: From the picture above, one can see that all of the components of this graph are complete.

Example 2:

Input: n = 6, edges = [[0,1],[0,2],[1,2],[3,4],[3,5]]
Output: 1
Explanation: The component containing vertices 0, 1, and 2 is complete since there is an edge between every pair of two vertices. On the other hand, the component containing vertices 3, 4, and 5 is not complete since there is no edge between vertices 4 and 5. Thus, the number of complete components in this graph is 1.

Constraints:

1 <= n <= 50
0 <= edges.length <= n * (n - 1) / 2
edges[i].length == 2
0 <= a_i, b_i <= n - 1
a_i != b_i
There are no repeated edges.

Step 01

Brute Force Baseline

Problem summary: You are given an integer n. There is an undirected graph with n vertices, numbered from 0 to n - 1. You are given a 2D integer array edges where edges[i] = [ai, bi] denotes that there exists an undirected edge connecting vertices ai and bi. Return the number of complete connected components of the graph. A connected component is a subgraph of a graph in which there exists a path between any two vertices, and no vertex of the subgraph shares an edge with a vertex outside of the subgraph. A connected component is said to be complete if there exists an edge between every pair of its vertices.

Baseline thinking

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

Pattern signal: Union-Find

Example 1

6
[[0,1],[0,2],[1,2],[3,4]]

Example 2

6
[[0,1],[0,2],[1,2],[3,4],[3,5]]

Core Insight

What unlocks the optimal approach

Find the connected components of an undirected graph using depth-first search (DFS) or breadth-first search (BFS).
For each connected component, count the number of nodes and edges in the component.
A connected component is complete if and only if the number of edges in the component is equal to m*(m-1)/2, where m is the number of nodes in the component.

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 #2685: Count the Number of Complete Components
class Solution {
    private List<Integer>[] g;
    private boolean[] vis;

    public int countCompleteComponents(int n, int[][] edges) {
        g = new List[n];
        vis = new boolean[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (int[] e : edges) {
            int a = e[0], b = e[1];
            g[a].add(b);
            g[b].add(a);
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (!vis[i]) {
                int[] t = dfs(i);
                if (t[0] * (t[0] - 1) == t[1]) {
                    ++ans;
                }
            }
        }
        return ans;
    }

    private int[] dfs(int i) {
        vis[i] = true;
        int x = 1, y = g[i].size();
        for (int j : g[i]) {
            if (!vis[j]) {
                int[] t = dfs(j);
                x += t[0];
                y += t[1];
            }
        }
        return new int[] {x, y};
    }
}

// Accepted solution for LeetCode #2685: Count the Number of Complete Components
func countCompleteComponents(n int, edges [][]int) (ans int) {
	g := make([][]int, n)
	vis := make([]bool, n)
	for _, e := range edges {
		a, b := e[0], e[1]
		g[a] = append(g[a], b)
		g[b] = append(g[b], a)
	}
	var dfs func(int) (int, int)
	dfs = func(i int) (int, int) {
		vis[i] = true
		x, y := 1, len(g[i])
		for _, j := range g[i] {
			if !vis[j] {
				a, b := dfs(j)
				x += a
				y += b
			}
		}
		return x, y
	}
	for i := range vis {
		if !vis[i] {
			a, b := dfs(i)
			if a*(a-1) == b {
				ans++
			}
		}
	}
	return
}

# Accepted solution for LeetCode #2685: Count the Number of Complete Components
class Solution:
    def countCompleteComponents(self, n: int, edges: List[List[int]]) -> int:
        def dfs(i: int) -> (int, int):
            vis[i] = True
            x, y = 1, len(g[i])
            for j in g[i]:
                if not vis[j]:
                    a, b = dfs(j)
                    x += a
                    y += b
            return x, y

        g = defaultdict(list)
        for a, b in edges:
            g[a].append(b)
            g[b].append(a)
        vis = [False] * n
        ans = 0
        for i in range(n):
            if not vis[i]:
                a, b = dfs(i)
                ans += a * (a - 1) == b
        return ans

// Accepted solution for LeetCode #2685: Count the Number of Complete Components
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2685: Count the Number of Complete Components
// class Solution {
//     private List<Integer>[] g;
//     private boolean[] vis;
// 
//     public int countCompleteComponents(int n, int[][] edges) {
//         g = new List[n];
//         vis = new boolean[n];
//         Arrays.setAll(g, k -> new ArrayList<>());
//         for (int[] e : edges) {
//             int a = e[0], b = e[1];
//             g[a].add(b);
//             g[b].add(a);
//         }
//         int ans = 0;
//         for (int i = 0; i < n; ++i) {
//             if (!vis[i]) {
//                 int[] t = dfs(i);
//                 if (t[0] * (t[0] - 1) == t[1]) {
//                     ++ans;
//                 }
//             }
//         }
//         return ans;
//     }
// 
//     private int[] dfs(int i) {
//         vis[i] = true;
//         int x = 1, y = g[i].size();
//         for (int j : g[i]) {
//             if (!vis[j]) {
//                 int[] t = dfs(j);
//                 x += t[0];
//                 y += t[1];
//             }
//         }
//         return new int[] {x, y};
//     }
// }

// Accepted solution for LeetCode #2685: Count the Number of Complete Components
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2685: Count the Number of Complete Components
// class Solution {
//     private List<Integer>[] g;
//     private boolean[] vis;
// 
//     public int countCompleteComponents(int n, int[][] edges) {
//         g = new List[n];
//         vis = new boolean[n];
//         Arrays.setAll(g, k -> new ArrayList<>());
//         for (int[] e : edges) {
//             int a = e[0], b = e[1];
//             g[a].add(b);
//             g[b].add(a);
//         }
//         int ans = 0;
//         for (int i = 0; i < n; ++i) {
//             if (!vis[i]) {
//                 int[] t = dfs(i);
//                 if (t[0] * (t[0] - 1) == t[1]) {
//                     ++ans;
//                 }
//             }
//         }
//         return ans;
//     }
// 
//     private int[] dfs(int i) {
//         vis[i] = true;
//         int x = 1, y = g[i].size();
//         for (int j : g[i]) {
//             if (!vis[j]) {
//                 int[] t = dfs(j);
//                 x += t[0];
//                 y += t[1];
//             }
//         }
//         return new int[] {x, y};
//     }
// }

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(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.