LeetCode #952 — HARD

Largest Component Size by Common Factor

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 integer array of unique positive integers nums. Consider the following graph:

There are nums.length nodes, labeled nums[0] to nums[nums.length - 1],
There is an undirected edge between nums[i] and nums[j] if nums[i] and nums[j] share a common factor greater than 1.

Return the size of the largest connected component in the graph.

Example 1:

Input: nums = [4,6,15,35]
Output: 4

Example 2:

Input: nums = [20,50,9,63]
Output: 2

Example 3:

Input: nums = [2,3,6,7,4,12,21,39]
Output: 8

Constraints:

1 <= nums.length <= 2 * 10⁴
1 <= nums[i] <= 10⁵
All the values of nums are unique.

Step 01

Brute Force Baseline

Problem summary: You are given an integer array of unique positive integers nums. Consider the following graph: There are nums.length nodes, labeled nums[0] to nums[nums.length - 1], There is an undirected edge between nums[i] and nums[j] if nums[i] and nums[j] share a common factor greater than 1. Return the size of the largest connected component in the graph.

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

[4,6,15,35]

Example 2

[20,50,9,63]

Example 3

[2,3,6,7,4,12,21,39]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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

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 #952: Largest Component Size by Common Factor
class UnionFind {
    int[] p;

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

    void union(int a, int b) {
        int pa = find(a), pb = find(b);
        if (pa != pb) {
            p[pa] = pb;
        }
    }

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

class Solution {
    public int largestComponentSize(int[] nums) {
        int m = 0;
        for (int v : nums) {
            m = Math.max(m, v);
        }
        UnionFind uf = new UnionFind(m + 1);
        for (int v : nums) {
            int i = 2;
            while (i <= v / i) {
                if (v % i == 0) {
                    uf.union(v, i);
                    uf.union(v, v / i);
                }
                ++i;
            }
        }
        int[] cnt = new int[m + 1];
        int ans = 0;
        for (int v : nums) {
            int t = uf.find(v);
            ++cnt[t];
            ans = Math.max(ans, cnt[t]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #952: Largest Component Size by Common Factor
func largestComponentSize(nums []int) int {
	m := slices.Max(nums)
	p := make([]int, m+1)
	for i := range p {
		p[i] = i
	}
	var find func(int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	union := func(a, b int) {
		pa, pb := find(a), find(b)
		if pa != pb {
			p[pa] = pb
		}
	}
	for _, v := range nums {
		i := 2
		for i <= v/i {
			if v%i == 0 {
				union(v, i)
				union(v, v/i)
			}
			i++
		}
	}
	cnt := make([]int, m+1)
	for _, v := range nums {
		t := find(v)
		cnt[t]++
	}
	return slices.Max(cnt)
}

# Accepted solution for LeetCode #952: Largest Component Size by Common Factor
class UnionFind:
    def __init__(self, n):
        self.p = list(range(n))

    def union(self, a, b):
        pa, pb = self.find(a), self.find(b)
        if pa != pb:
            self.p[pa] = pb

    def find(self, x):
        if self.p[x] != x:
            self.p[x] = self.find(self.p[x])
        return self.p[x]


class Solution:
    def largestComponentSize(self, nums: List[int]) -> int:
        uf = UnionFind(max(nums) + 1)
        for v in nums:
            i = 2
            while i <= v // i:
                if v % i == 0:
                    uf.union(v, i)
                    uf.union(v, v // i)
                i += 1
        return max(Counter(uf.find(v) for v in nums).values())

// Accepted solution for LeetCode #952: Largest Component Size by Common Factor
/**
 * [0952] Largest Component Size by Common Factor
 *
 * You are given an integer array of unique positive integers nums. Consider the following graph:
 *
 * 	There are nums.length nodes, labeled nums[0] to nums[nums.length - 1],
 * 	There is an undirected edge between nums[i] and nums[j] if nums[i] and nums[j] share a common factor greater than 1.
 *
 * Return the size of the largest connected component in the graph.
 *  
 * Example 1:
 * <img alt="" src="https://assets.leetcode.com/uploads/2018/12/01/ex1.png" style="width: 500px; height: 97px;" />
 * Input: nums = [4,6,15,35]
 * Output: 4
 *
 * Example 2:
 * <img alt="" src="https://assets.leetcode.com/uploads/2018/12/01/ex2.png" style="width: 500px; height: 85px;" />
 * Input: nums = [20,50,9,63]
 * Output: 2
 *
 * Example 3:
 * <img alt="" src="https://assets.leetcode.com/uploads/2018/12/01/ex3.png" style="width: 500px; height: 260px;" />
 * Input: nums = [2,3,6,7,4,12,21,39]
 * Output: 8
 *
 *  
 * Constraints:
 *
 * 	1 <= nums.length <= 2 * 10^4
 * 	1 <= nums[i] <= 10^5
 * 	All the values of nums are unique.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/largest-component-size-by-common-factor/
// discuss: https://leetcode.com/problems/largest-component-size-by-common-factor/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

// Credit: https://leetcode.com/problems/largest-component-size-by-common-factor/solutions/1592428/rust-unionfind-solution/

struct UnionFind {
    parent: Vec<usize>,
}

impl UnionFind {
    fn new(n: usize) -> Self {
        Self {
            parent: (0..n).collect(),
        }
    }
    fn union(&mut self, x: usize, y: usize) {
        let x = self.find(x);
        let y = self.find(y);
        self.parent[y] = x
    }
    fn find(&mut self, x: usize) -> usize {
        if x != self.parent[x] {
            self.parent[x] = self.find(self.parent[x]);
        }
        self.parent[x]
    }
}

impl Solution {
    pub fn largest_component_size(nums: Vec<i32>) -> i32 {
        let mut sieve = (0..100_001).collect::<Vec<_>>();
        for i in (2..).take_while(|&i| i * i < 100_001) {
            if sieve[i] == i as i32 {
                for j in (i..100_001).step_by(i) {
                    sieve[j] = sieve[j].min(i as i32);
                }
            }
        }
        let mut hm = std::collections::HashMap::new();
        for (i, &num) in nums.iter().enumerate() {
            let mut n = num;
            while n > 1 {
                let p = sieve[n as usize];
                hm.entry(p).or_insert_with(Vec::new).push(i);
                n /= p as i32;
            }
        }
        let mut uf = UnionFind::new(nums.len());
        for v in hm.values_mut() {
            v.dedup();
            v.windows(2).for_each(|w| uf.union(w[0], w[1]));
        }
        let mut counts = vec![0; nums.len()];
        for i in 0..nums.len() {
            counts[uf.find(i)] += 1;
        }
        *counts.iter().max().unwrap() as i32
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_0952_example_1() {
        let nums = vec![4, 6, 15, 35];
        let result = 4;

        assert_eq!(Solution::largest_component_size(nums), result);
    }

    #[test]
    fn test_0952_example_2() {
        let nums = vec![20, 50, 9, 63];
        let result = 2;

        assert_eq!(Solution::largest_component_size(nums), result);
    }

    #[test]
    fn test_0952_example_3() {
        let nums = vec![2, 3, 6, 7, 4, 12, 21, 39];
        let result = 8;

        assert_eq!(Solution::largest_component_size(nums), result);
    }
}

// Accepted solution for LeetCode #952: Largest Component Size by Common Factor
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #952: Largest Component Size by Common Factor
// class UnionFind {
//     int[] p;
// 
//     UnionFind(int n) {
//         p = new int[n];
//         for (int i = 0; i < n; ++i) {
//             p[i] = i;
//         }
//     }
// 
//     void union(int a, int b) {
//         int pa = find(a), pb = find(b);
//         if (pa != pb) {
//             p[pa] = pb;
//         }
//     }
// 
//     int find(int x) {
//         if (p[x] != x) {
//             p[x] = find(p[x]);
//         }
//         return p[x];
//     }
// }
// 
// class Solution {
//     public int largestComponentSize(int[] nums) {
//         int m = 0;
//         for (int v : nums) {
//             m = Math.max(m, v);
//         }
//         UnionFind uf = new UnionFind(m + 1);
//         for (int v : nums) {
//             int i = 2;
//             while (i <= v / i) {
//                 if (v % i == 0) {
//                     uf.union(v, i);
//                     uf.union(v, v / i);
//                 }
//                 ++i;
//             }
//         }
//         int[] cnt = new int[m + 1];
//         int ans = 0;
//         for (int v : nums) {
//             int t = uf.find(v);
//             ++cnt[t];
//             ans = Math.max(ans, cnt[t]);
//         }
//         return ans;
//     }
// }

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.

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.