Top Mutual-Friend Recommendations

You are given a simple undirected friendship graph as an array friends. Each entry contains the two user names joined by one friendship. You are also given a target user t.
A recommendation candidate must not be t, must not already be a direct friend of t, and must share at least one direct friend with t.
Rank candidates by these rules:
A higher number of mutual friends comes first.
When counts tie, the lexicographically smaller user name comes first.
Return the first three user names in that order, or every candidate when fewer than three exist.

Function
recommendFriends(friends: String[][], t: String) → String[]

Examples
Example 1
friends = [["Alice","Bob"],["Alice","Carol"],["Alice","Dave"],["Bob","Carol"],["Bob","Eve"],["Bob","Frank"],["Carol","Dave"],["Carol","Grace"],["Dave","Grace"],["Dave","Henry"],["Eve","Frank"],["Eve","Grace"],["Frank","Grace"]]
t = "Alice"
return = ["Grace","Eve","Frank"]
Grace shares two friends with Alice. Eve, Frank, and Henry each share one; lexical order selects Eve and Frank for the remaining positions.
Example 2
friends = [["A","B"],["B","C"],["B","D"]]
t = "A"
return = ["C","D"]
C and D each share B with A, so both are returned in lexical order.

Constraints
2 <= friends.length <= 2 * 10^5
Every friends[i] contains exactly two user names.
Every user name has length at most 10 and contains only Latin letters.
The graph has no self-loops or duplicate undirected edges.
The target user t appears in at least one friendship pair.

Build an adjacency map from the edge list, using sets of names. Get t's direct friends. For each friend f of t, loop over f's neighbors. Skip t itself and skip anyone in t's friend set. Increment a counter for everyone else. That counter is the mutual friend count, and a candidate with zero mutuals never appears because you only reach people through a shared friend. Then sort by count descending and name ascending, and take the first three. The pitfall is the naive approach of comparing every user pair, which blows up with 2 * 10^5 edges. Walking only two hops from t keeps it near linear plus the sort. Another trap is string comparison: names are Latin letters only, so the default lexicographic sort works, but keep case as given. If you freeze on the live OA, StealthCoder can surface this two-hop counting approach so you can type it out.

StealthCoder is the hedge for the one pattern you didn't drill. It runs invisibly during the screen share.