LeetCode #3493 — MEDIUM

Properties Graph

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

The Problem

Problem Statement

You are given a 2D integer array properties having dimensions n x m and an integer k.

Define a function intersect(a, b) that returns the number of distinct integers common to both arrays a and b.

Construct an undirected graph where each index i corresponds to properties[i]. There is an edge between node i and node j if and only if intersect(properties[i], properties[j]) >= k, where i and j are in the range [0, n - 1] and i != j.

Return the number of connected components in the resulting graph.

Example 1:

Input: properties = [[1,2],[1,1],[3,4],[4,5],[5,6],[7,7]], k = 1

Output: 3

Explanation:

The graph formed has 3 connected components:

Example 2:

Input: properties = [[1,2,3],[2,3,4],[4,3,5]], k = 2

Output: 1

Explanation:

The graph formed has 1 connected component:

Example 3:

Input: properties = [[1,1],[1,1]], k = 2

Output: 2

Explanation:

intersect(properties[0], properties[1]) = 1, which is less than k. This means there is no edge between properties[0] and properties[1] in the graph.

Constraints:

1 <= n == properties.length <= 100
1 <= m == properties[i].length <= 100
1 <= properties[i][j] <= 100
1 <= k <= m

Step 01

Brute Force Baseline

Problem summary: You are given a 2D integer array properties having dimensions n x m and an integer k. Define a function intersect(a, b) that returns the number of distinct integers common to both arrays a and b. Construct an undirected graph where each index i corresponds to properties[i]. There is an edge between node i and node j if and only if intersect(properties[i], properties[j]) >= k, where i and j are in the range [0, n - 1] and i != j. Return the number of connected components in the resulting graph.

Baseline thinking

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

Pattern signal: Array · Hash Map · Union-Find

Example 1

[[1,2],[1,1],[3,4],[4,5],[5,6],[7,7]]
1

Example 2

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

Example 3

[[1,1],[1,1]]
2

Step 02

Core Insight

What unlocks the optimal approach

How can we optimally find the intersection of two arrays? One way is to use <code>len(set(a) & set(b))</code>.
For connected components, think about using DFS, BFS, or DSU.

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 #3493: Properties Graph
class Solution {
    private List<Integer>[] g;
    private boolean[] vis;

    public int numberOfComponents(int[][] properties, int k) {
        int n = properties.length;
        g = new List[n];
        Set<Integer>[] ss = new Set[n];
        Arrays.setAll(g, i -> new ArrayList<>());
        Arrays.setAll(ss, i -> new HashSet<>());
        for (int i = 0; i < n; ++i) {
            for (int x : properties[i]) {
                ss[i].add(x);
            }
        }
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                int cnt = 0;
                for (int x : ss[i]) {
                    if (ss[j].contains(x)) {
                        ++cnt;
                    }
                }
                if (cnt >= k) {
                    g[i].add(j);
                    g[j].add(i);
                }
            }
        }

        int ans = 0;
        vis = new boolean[n];
        for (int i = 0; i < n; ++i) {
            if (!vis[i]) {
                dfs(i);
                ++ans;
            }
        }
        return ans;
    }

    private void dfs(int i) {
        vis[i] = true;
        for (int j : g[i]) {
            if (!vis[j]) {
                dfs(j);
            }
        }
    }
}

// Accepted solution for LeetCode #3493: Properties Graph
func numberOfComponents(properties [][]int, k int) (ans int) {
	n := len(properties)
	ss := make([]map[int]struct{}, n)
	g := make([][]int, n)

	for i := 0; i < n; i++ {
		ss[i] = make(map[int]struct{})
		for _, x := range properties[i] {
			ss[i][x] = struct{}{}
		}
	}

	for i := 0; i < n; i++ {
		for j := 0; j < i; j++ {
			cnt := 0
			for x := range ss[i] {
				if _, ok := ss[j][x]; ok {
					cnt++
				}
			}
			if cnt >= k {
				g[i] = append(g[i], j)
				g[j] = append(g[j], i)
			}
		}
	}

	vis := make([]bool, n)
	var dfs func(int)
	dfs = func(i int) {
		vis[i] = true
		for _, j := range g[i] {
			if !vis[j] {
				dfs(j)
			}
		}
	}

	for i := 0; i < n; i++ {
		if !vis[i] {
			dfs(i)
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #3493: Properties Graph
class Solution:
    def numberOfComponents(self, properties: List[List[int]], k: int) -> int:
        def dfs(i: int) -> None:
            vis[i] = True
            for j in g[i]:
                if not vis[j]:
                    dfs(j)

        n = len(properties)
        ss = list(map(set, properties))
        g = [[] for _ in range(n)]
        for i, s1 in enumerate(ss):
            for j in range(i):
                s2 = ss[j]
                if len(s1 & s2) >= k:
                    g[i].append(j)
                    g[j].append(i)
        ans = 0
        vis = [False] * n
        for i in range(n):
            if not vis[i]:
                dfs(i)
                ans += 1
        return ans

// Accepted solution for LeetCode #3493: Properties Graph
// 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 #3493: Properties Graph
// class Solution {
//     private List<Integer>[] g;
//     private boolean[] vis;
// 
//     public int numberOfComponents(int[][] properties, int k) {
//         int n = properties.length;
//         g = new List[n];
//         Set<Integer>[] ss = new Set[n];
//         Arrays.setAll(g, i -> new ArrayList<>());
//         Arrays.setAll(ss, i -> new HashSet<>());
//         for (int i = 0; i < n; ++i) {
//             for (int x : properties[i]) {
//                 ss[i].add(x);
//             }
//         }
//         for (int i = 0; i < n; ++i) {
//             for (int j = 0; j < i; ++j) {
//                 int cnt = 0;
//                 for (int x : ss[i]) {
//                     if (ss[j].contains(x)) {
//                         ++cnt;
//                     }
//                 }
//                 if (cnt >= k) {
//                     g[i].add(j);
//                     g[j].add(i);
//                 }
//             }
//         }
// 
//         int ans = 0;
//         vis = new boolean[n];
//         for (int i = 0; i < n; ++i) {
//             if (!vis[i]) {
//                 dfs(i);
//                 ++ans;
//             }
//         }
//         return ans;
//     }
// 
//     private void dfs(int i) {
//         vis[i] = true;
//         for (int j : g[i]) {
//             if (!vis[j]) {
//                 dfs(j);
//             }
//         }
//     }
// }

// Accepted solution for LeetCode #3493: Properties Graph
function numberOfComponents(properties: number[][], k: number): number {
    const n = properties.length;
    const ss: Set<number>[] = Array.from({ length: n }, () => new Set());
    const g: number[][] = Array.from({ length: n }, () => []);

    for (let i = 0; i < n; i++) {
        for (const x of properties[i]) {
            ss[i].add(x);
        }
    }

    for (let i = 0; i < n; i++) {
        for (let j = 0; j < i; j++) {
            let cnt = 0;
            for (const x of ss[i]) {
                if (ss[j].has(x)) {
                    cnt++;
                }
            }
            if (cnt >= k) {
                g[i].push(j);
                g[j].push(i);
            }
        }
    }

    let ans = 0;
    const vis: boolean[] = Array(n).fill(false);

    const dfs = (i: number) => {
        vis[i] = true;
        for (const j of g[i]) {
            if (!vis[j]) {
                dfs(j);
            }
        }
    };

    for (let i = 0; i < n; i++) {
        if (!vis[i]) {
            dfs(i);
            ans++;
        }
    }
    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(n^2 × m)

Space

O(n × m)

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.