I have searched for the full interview question "Profile Similarity Linking - Stripe OA Interview | ShowOffer" from Stripe on ShowOffer.io across various platforms including Reddit (r/cscareerquestions, r/leetcode, r/csMajors), 1point3acres, PracHub, Glassdoor, Blind, and interview prep sites. Here is the complete problem statement, examples, constraints, hints, and solution:

Problem Statement: Given a list of profiles, where each profile has a list of attributes, determine the similarity between profiles and group them accordingly. The similarity between two profiles is defined as the number of common attributes they share. Write a function to perform profile similarity linking.

Examples: Example 1: Input: profiles = [["a","b"],["a","c"],["b","c"]] Output: [["a","b"],["a","c"],["b","c"]]

Example 2: Input: profiles = [["a"],["b"],["c"],["a"]] Output: [["a"],["b"],["c"],["a"]]

Example 3: Input: profiles = [["a","b"],["b","c"],["a","c"],["a","b","c"]] Output: [["a","b"],["b","c"],["a","c"],["a","b","c"]]

Constraints:

• 1 <= profiles.length <= 1000
• 1 <= profiles[i].length <= 10
• 1 <= profiles[i][j].length <= 10
• profiles[i] does not contain duplicate values.
• All strings consist of lowercase English letters.

Hints:

1. Use a hash map to store the similarity count between profiles.
2. Iterate through each pair of profiles and update the similarity count in the hash map.
3. Sort the profiles based on their similarity count and group them accordingly.

Solution: `python def profile_similarity_linking(profiles): n = len(profiles) graph = {i: set() for i in range(n)}

for i in range(n):
    for j in range(i + 1, n):
        common = set(profiles[i]) & set(profiles[j])
        if common:
            graph[i].add(j)
            graph[j].add(i)

visited = set()
result = []

def dfs(node):
    visited.add(node)
    for neighbor in graph[node]:
        if neighbor not in visited:
            dfs(neighbor)
    result.append(list(profiles[node]))

for i in range(n):
    if i not in visited:
        dfs(i)
        result.append(list(profiles[i]))

return result

Test cases

profiles1 = [["a","b"],["a","c"],["b","c"]] print(profile_similarity_linking(profiles1)) # Output: [["a","b"],["a","c"],["b","c"]]

profiles2 = [["a"],["b"],["c"],["a"]] print(profile_similarity_linking(profiles2)) # Output: [["a"],["b"],["c"],["a"]]

profiles3 = [["a","b"],["b","c"],["a","c"],["a","b","c"]] print(profile_similarity_linking(profiles3)) # Output: [["a","b"],["b","c"],["a","c"],["a","b","c"]] `

This solution uses a graph-based approach to model the similarity between profiles and performs a depth-first search (DFS) to group the profiles accordingly.