Problem Description

In the given problem, we envision a network of people, initially with no friendships established. Each person in this network is uniquely identified by a label, which is a number starting from 0 to n - 1. We also have a list of restrictions, where each restriction specifies two people who cannot become friends, either directly or through mutual friends.

On top of these restrictions, we are given a series of friend requests. Each request is a pair of individuals who wish to become friends. Our task is to determine for each friend request whether it could be successful given the constraints of the restrictions.

The result should be an array of boolean values corresponding to each friend request, indicating true if they can become friends, or false otherwise. Even if two people are already direct friends, subsequent requests between them should be considered successful.

Intuition

To solve this problem, we need to manage friend connections in such a way that we can quickly check if two people can become friends considering current friendships and restrictions.

The Union-Find algorithm, or Disjoint Set Union (DSU), is well-suited for this kind of problem as it efficiently handles grouping elements into disjoint sets and checking whether two elements belong to the same set.

Initially, each person forms a set of their own. The find function is used to find the representative of a person's set; if two people have the same representative, they are in the same connected component and hence indirectly friends.

When processing each friend request, we check if they can be added to each other's group. If they are not already in the same group, we examine all the restrictions. If either person in the request matches a pair in the restrictions with the other person in the request, then the friend request is invalid.

If none of the restrictions apply, the request is successful, and we union the two sets representing the requester and the requestee by updating the representative of one of the sets. This way, both individuals will have the same representative and thus be directly connected.

The union operation is performed by updating the parent of one representative to be the other, effectively merging the sets.

Each friend request's result, whether successful or not, is appended to an ans list, which is returned at the end of the process.

Learn more about Union Find and Graph patterns.

Solution Approach

The solution implements the Union-Find algorithm, which is a vital part of this approach. Here we will break down how the algorithm is applied within the provided solution code.

1. Parent Array Initialization: A parent array p is initialized to contain each element as its own parent, indicating that every individual starts in their own distinct set. This array keeps track of connections.

   p = list(range(n))

2. Find Function: The find() function is a recursive function that finds and returns the representative (or the "root" parent) of the set that an individual belongs to. It implements path compression by updating the parent of each node to its root parent during the traversal. This optimization decreases the time complexity of subsequent find() operations.

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

3. Processing Friend Requests: Friend requests are processed in the given order. For each friend request between u and v, we perform the following:

   • Check if u and v already have the same representative, meaning they are in the same set (directly or indirectly friends). If so, the request is immediately successful.

   • If they are not in the same set, examine all the restrictions. For each restriction pair (x, y), check if u or v is restricted with the other by comparing their representatives with x and y. If a restriction is violated, the request cannot be fulfilled, and it is marked as invalid.

   • If the request does not violate any restrictions, it is considered successful, and we perform the union operation. This means that we update the parent of one person's representative (find(u)) to be the other's representative (find(v)), effectively merging their sets.

   ans = []
   for u, v in requests:
       if find(u) == find(v):
           ans.append(True)
       else:
           valid = True
           for x, y in restrictions:
               if (find(u) == find(x) and find(v) == find(y)) or (
                   find(u) == find(y) and find(v) == find(x)
               ):
                   valid = False
                   break
           ans.append(valid)
           if valid:
               p[find(u)] = find(v)

4. Returning the Result: After iterating through all the requests, the list ans is filled with the results of the friend requests (True for a successful friend request, False otherwise). The list is returned as the final output.

   return ans

By leveraging the Union-Find algorithm with path compression, the solution ensures that we can efficiently determine the outcome of friend requests while respecting the constraints imposed by the restrictions.

Example Walkthrough

Let us consider an example where we have n = 5 people labeled from 0 to 4, and we have a list of restrictions restrictions = [(0, 1), (3, 4)], which indicates that person 0 cannot be friends with person 1 and person 3 cannot be friends with person 4. We also have a series of friend requests: requests = [(0, 2), (1, 2), (0, 1), (3, 4), (2, 3)].

Let's walk through the friend requests using the solution approach:

1. Parent Array Initialization: Initially, the parent array p is [0, 1, 2, 3, 4], with each person as their own parent.

2. Find Function: This will be used to find the representatives of each person's set as we process the requests.

3. Processing Friend Requests:

   • Request (0, 2): Since find(0) == 0 and find(2) == 2, they are not in the same set. There are no restrictions involving 0 and 2, so they can be friends. We perform the union by setting p[find(0)] = find(2), so now p[0] = 2. The parent array is now [2, 1, 2, 3, 4].

   • Request (1, 2): Now, find(1) == 1 and find(2) == 2. They are not restricted and are not in the same set, so they can also be friends. After the union, p[1] = 2. The parent array is [2, 2, 2, 3, 4].

   • Request (0, 1): Both have the same representative (find(0) and find(1) are both 2), hence they can still become friends. Since they are already in the same set, no changes are made to the parent array.

   • Request (3, 4): They're not in the same set (find(3) == 3 and find(4) == 4), but they have a direct restriction, so the request is denied. No change is made to the parent array.

   • Request (2, 3): No direct restriction involves 2 and 3, and since they are not in the same set (find(2) == 2 and find(3) == 3), they can be friends. We perform the union p[find(2)] = find(3), and the parent array becomes [2, 2, 3, 3, 4].

4. Returning the Result: The results for the friend requests are [True, True, True, False, True]. This indicates the outcome of each friend request after considering all the restrictions and the current state of friendships.

By using the Union-Find algorithm with optimizations such as path compression, we've efficiently processed the friend requests and determined the successful and unsuccessful connections in this network of people.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given code is primarily determined by two parts: the union-find operations and the restriction checks for each friend request.

• The find function is a typical union-find operation with path compression. The amortized time complexity for each find call is O(alpha(n)), where alpha is the inverse Ackermann function, which grows very slowly and is considered almost constant in practice for all reasonable values of n.

• Each friend request involves checking all the restrictions. There are r restrictions, where r is the number of elements in the restrictions list. For each request, the code loops through all restrictions, leading to r calls to the find function in the worst case.

• Let q be the number of friend requests. For each request, there could be up to r restriction checks, each of which could result in up to 4 calls to the find function (2 pairs of friends to check for each restriction).

The worst-case time complexity can be expressed as O(q * r * alpha(n)).

Space Complexity

The space complexity is determined by the storage for the parent pointers and the output list.

• The parent array p has length n, so the space complexity for the union-find data structure is O(n).

• Apart from the parent pointers, the space used by the restrictions, requests, and the answer list is determined by the input, which we don't count towards the extra space taken by the algorithm.

• The ans list will contain q boolean values, where q is the number of requests.

Therefore, the overall space complexity of the algorithm is O(n + q), but since the space for the output list ans is required for the result of the function, we generally don't count it as additional space complexity in an analysis. Thus, we can simply say the space complexity is O(n).

Learn more about how to find time and space complexity quickly using problem constraints.