LeetCode 2076: Process Restricted Friend Requests Solution

Master LeetCode problem 2076 (Process Restricted Friend Requests), a hard 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.

2076. Process Restricted Friend Requests

Hard

Union Find Graph

Problem Explanation

Explanation:

To solve this problem, we need to simulate the processing of friend requests while considering the restrictions given. We can create a graph where each node represents a person and then iteratively process each request. For each request, we check if the two people can be friends based on the restrictions and the existing friendships.

We will use a union-find data structure to efficiently maintain the relationships between people and determine if two people can become friends. Additionally, we will keep track of the restrictions to ensure that indirect friendships are not formed.

Create an array parent to represent the parent of each person in the network.
Initialize the parent array such that each person is its own parent initially.
For each restriction in the input, union the two people if they are not already friends.
Process each request, check if the two people can be friends based on the restrictions and existing relationships, and update the friendship status accordingly.

Time Complexity: O(n + R + Q * α(n)), where n is the number of people, R is the number of restrictions, Q is the number of requests, and α(n) is the inverse Ackermann function (almost constant). Space Complexity: O(n) for the parent array.

Solution Code

class Solution {
    public boolean[] processQueries(int n, int[][] restrictions, int[][] requests) {
        int[] parent = new int[n];
        Arrays.fill(parent, -1);
        
        for (int[] restriction : restrictions) {
            int u = find(restriction[0], parent);
            int v = find(restriction[1], parent);
            if (u != v) {
                parent[u] = v;
            }
        }
        
        boolean[] result = new boolean[requests.length];
        
        for (int i = 0; i < requests.length; i++) {
            int u = find(requests[i][0], parent);
            int v = find(requests[i][1], parent);
            if (u == v) {
                result[i] = true;
            } else {
                result[i] = false;
                if (parent[u] != v) {
                    parent[u] = v;
                }
            }
        }
        
        return result;
    }
    
    private int find(int x, int[] parent) {
        if (parent[x] == -1) {
            return x;
        }
        return parent[x] = find(parent[x], parent);
    }
}

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