How do you solve Process Restricted Friend Requests on LeetCode?

Process Restricted Friend Requests (LeetCode #2076) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * α(n)) time complexity and O(n) space complexity. Key insight: Using Union-Find allows efficient management of friend connections.

What is the brute force solution for Process Restricted Friend Requests?

The brute force approach for Process Restricted Friend Requests is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we check each friend request against all restrictions to see if it can be fulfilled. This is straightforward but inefficient as it involves checking every request against every restriction. The optimal Optimal Solution approach improves this to O(m * α(n)).

What algorithm or data structure is used to solve Process Restricted Friend Requests?

Process Restricted Friend Requests uses the following concepts: Union-Find, Graph Theory. The recommended patterns to study are: Union-Find, Graph Theory.

What are the interview tips for Process Restricted Friend Requests?

Understand the problem constraints and relationships between nodes. Practice implementing union-find as it's a common pattern. Think about how to optimize naive solutions before coding.

What are common mistakes when solving Process Restricted Friend Requests?

Not considering indirect connections when checking restrictions. Failing to properly implement the union-find structure.

#2076

Process Restricted Friend Requests

Hard

Union-FindGraph TheoryUnion-FindGraph Theory

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(m * α(n))
Space	O(1)	O(n)

💡

Intuition

Time O(m * α(n))Space O(n)

Using the Union-Find data structure allows us to efficiently manage and check the connected components of the network. This way, we can quickly determine if two people can become friends without violating restrictions.

⚙️

Algorithm

3 steps

1Step 1: Initialize a Union-Find structure to manage friend groups.
2Step 2: Process each request, checking if adding the friendship would violate any restrictions.
3Step 3: If it doesn't violate, union the two people; if it does, mark the request as unsuccessful.

solution.py23 lines

1class UnionFind:
2    def __init__(self, n):
3        self.parent = list(range(n))
4    def find(self, x):
5        if self.parent[x] != x:
6            self.parent[x] = self.find(self.parent[x])
7        return self.parent[x]
8    def union(self, x, y):
9        rootX = self.find(x)
10        rootY = self.find(y)
11        if rootX != rootY:
12            self.parent[rootY] = rootX
13
14def processRequests(n, restrictions, requests):
15    uf = UnionFind(n)
16    result = [True] * len(requests)
17    for u, v in requests:
18        if any(uf.find(u) == uf.find(x) and uf.find(v) == uf.find(y) for x, y in restrictions):
19            result.append(False)
20        else:
21            uf.union(u, v)
22            result.append(True)
23    return result

ℹ

Complexity note: The time complexity is O(m * α(n)), where m is the number of requests and α is the inverse Ackermann function, which grows very slowly. The space complexity is O(n) for the Union-Find structure.

1Using Union-Find allows efficient management of friend connections.
2Directly checking restrictions for each request is inefficient.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.