LeetCode #3378 — HARD

Count Connected Components in LCM Graph

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

Solve on LeetCode
The Problem

Problem Statement

You are given an array of integers nums of size n and a positive integer threshold.

There is a graph consisting of n nodes with the ith node having a value of nums[i]. Two nodes i and j in the graph are connected via an undirected edge if lcm(nums[i], nums[j]) <= threshold.

Return the number of connected components in this 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.

The term lcm(a, b) denotes the least common multiple of a and b.

Example 1:

Input: nums = [2,4,8,3,9], threshold = 5

Output: 4

Explanation: 

The four connected components are (2, 4), (3), (8), (9).

Example 2:

Input: nums = [2,4,8,3,9,12], threshold = 10

Output: 2

Explanation: 

The two connected components are (2, 3, 4, 8, 9), and (12).

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 109
  • All elements of nums are unique.
  • 1 <= threshold <= 2 * 105
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: You are given an array of integers nums of size n and a positive integer threshold. There is a graph consisting of n nodes with the ith node having a value of nums[i]. Two nodes i and j in the graph are connected via an undirected edge if lcm(nums[i], nums[j]) <= threshold. Return the number of connected components in this 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. The term lcm(a, b) denotes the least common multiple of a and b.

Baseline thinking

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

Pattern signal: Array · Hash Map · Math · Union-Find

Example 1

[2,4,8,3,9]
5

Example 2

[2,4,8,3,9,12]
10
Step 02

Core Insight

What unlocks the optimal approach

  • Use DSU
  • Connect a number to all its multiples less than threshold
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 #3378: Count Connected Components in LCM Graph
class DSU {
    private Map<Integer, Integer> parent;
    private Map<Integer, Integer> rank;

    public DSU(int n) {
        parent = new HashMap<>();
        rank = new HashMap<>();
        for (int i = 0; i <= n; i++) {
            parent.put(i, i);
            rank.put(i, 0);
        }
    }

    public void makeSet(int v) {
        parent.put(v, v);
        rank.put(v, 1);
    }

    public int find(int x) {
        if (parent.get(x) != x) {
            parent.put(x, find(parent.get(x)));
        }
        return parent.get(x);
    }

    public void unionSet(int u, int v) {
        u = find(u);
        v = find(v);
        if (u != v) {
            if (rank.get(u) < rank.get(v)) {
                int temp = u;
                u = v;
                v = temp;
            }
            parent.put(v, u);
            if (rank.get(u).equals(rank.get(v))) {
                rank.put(u, rank.get(u) + 1);
            }
        }
    }
}

class Solution {
    public int countComponents(int[] nums, int threshold) {
        DSU dsu = new DSU(threshold);

        for (int num : nums) {
            for (int j = num; j <= threshold; j += num) {
                dsu.unionSet(num, j);
            }
        }

        Set<Integer> uniqueParents = new HashSet<>();
        for (int num : nums) {
            if (num > threshold) {
                uniqueParents.add(num);
            } else {
                uniqueParents.add(dsu.find(num));
            }
        }

        return uniqueParents.size();
    }
}
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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Related Problems
#952Largest Component Size by Common Factorhard→#128Longest Consecutive Sequencemedium→#149Max Points on a Linehard→#268Missing Numbereasy→#380Insert Delete GetRandom O(1)medium→