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LeetCode 3493: Properties Graph Solution

Master LeetCode problem 3493 (Properties Graph), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

Problem Explanation

Explanation

To solve this problem, we will first define the intersect function that calculates the number of distinct integers common to two arrays. Then, we will construct an undirected graph based on the given properties array and the intersection condition. Finally, we will find the number of connected components in the graph using Depth First Search (DFS).

Algorithm:

  1. Implement the intersect function to find the number of distinct common elements between two arrays.
  2. Construct an undirected graph where each node represents an index in the properties array.
  3. Connect nodes i and j with an edge if intersect(properties[i], properties[j]) >= k.
  4. Perform DFS to find connected components in the graph.
  5. Count the number of connected components.

Time Complexity:

The time complexity of this algorithm is O(n^2 * m) where n is the number of rows in the properties array and m is the number of elements in each row.

Space Complexity:

The space complexity of this algorithm is O(n^2) for the adjacency matrix representation of the graph.

Solution Code

class Solution {
    public int propertiesGraph(int[][] properties, int k) {
        int n = properties.length;
        boolean[][] graph = new boolean[n][n];
        
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if (intersect(properties[i], properties[j]) >= k) {
                    graph[i][j] = true;
                    graph[j][i] = true;
                }
            }
        }
        
        boolean[] visited = new boolean[n];
        int components = 0;
        
        for (int i = 0; i < n; i++) {
            if (!visited[i]) {
                components++;
                dfs(i, graph, visited);
            }
        }
        
        return components;
    }
    
    private void dfs(int node, boolean[][] graph, boolean[] visited) {
        visited[node] = true;
        
        for (int i = 0; i < graph.length; i++) {
            if (!visited[i] && graph[node][i]) {
                dfs(i, graph, visited);
            }
        }
    }
    
    private int intersect(int[] a, int[] b) {
        Set<Integer> setA = new HashSet<>();
        Set<Integer> setB = new HashSet<>();
        
        for (int num : a) {
            setA.add(num);
        }
        
        for (int num : b) {
            setB.add(num);
        }
        
        setA.retainAll(setB);
        
        return setA.size();
    }
}

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