Website Pair with the Most Common Visitors

Each row of visits is [user, website]. A user may visit one website more than once, but contributes at most once to that website's visitor set.
Among all unordered pairs of distinct websites, return the pair with the largest number of shared visitors. Sort the two names inside the result. Break ties by the lexicographically smallest pair. Return an empty array when fewer than two distinct websites appear.

Function
mostCommonVisitorPair(visits: String[][]) → String[]

Examples
Example 1
visits = [["u1","a"],["u1","b"],["u2","a"],["u2","b"],["u3","c"]]
return = ["a","b"]
Websites a and b share two visitors, more than any other pair.
Example 2
visits = [["u1","a"],["u1","a"],["u1","b"],["u2","b"],["u2","c"]]
return = ["a","b"]
Repeated visits by one user do not inflate an intersection.

Constraints
0 <= visits.length <= 50000
Every row contains exactly two nonempty strings.

The trick is to flip the grouping. Build a map from user to a set of websites, so repeated visits collapse automatically. Then for each user, sort their sites, loop over every pair, and increment a counter keyed by the pair. Track the best count, and on ties keep the lexicographically smaller pair. The pitfall is skipping the set and counting raw rows, or forgetting that pairs must be unordered, so (a,b) and (b,a) must hit the same key. Sorting each user's sites fixes that. Watch the cost too. A user with many sites generates quadratic pairs, but with 50000 rows the total stays manageable. Return an empty array when fewer than two distinct websites exist. If the tie-break or dedup slips under pressure, StealthCoder is the hedge during the live OA, but this is about fifteen lines once you see it.

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