Most Visited Hotel in a Time Window

Each index describes one hotel visit by userIds[i] to hotels[i] at integer timestamp timestamps[i].
Among visits whose timestamp is in the inclusive interval [startTime, endTime], return the hotel visited by the largest number of distinct users. Break a tie by returning the lexicographically smallest hotel name. Return the empty string when the interval contains no visit.

Function
mostVisitedHotel(userIds: String[], hotels: String[], timestamps: int[], startTime: int, endTime: int) → String

Examples
Example 1
userIds = ["u1","u2","u1","u3"]
hotels = ["A","A","A","B"]
timestamps = [10,20,25,30]
startTime = 10
endTime = 30
return = "A"
Hotel A has two distinct users; repeated visits by u1 count once.
Example 2
userIds = ["u1","u2"]
hotels = ["Beta","Alpha"]
timestamps = [5,6]
startTime = 0
endTime = 10
return = "Alpha"
Both hotels have one user, so the lexicographically smaller name wins.

Constraints
All three arrays have the same length from 0 through 100000.
User IDs and hotel names are non-empty ASCII strings.
-1000000000 <= timestamps[i], startTime, endTime <= 1000000000 and startTime <= endTime.

Walk the arrays once. For each index where startTime <= timestamps[i] <= endTime, add userIds[i] to a set stored under hotels[i] in a hash map. After the pass, scan the map and track the best hotel by set size, replacing it when the size is larger or when the size ties and the name is lexicographically smaller. If the map is empty, return the empty string. That's O(n) time plus the final scan. The common pitfall is counting visits instead of distinct users, which breaks Example 1 where u1 visits A twice. Another is using an exclusive bound on the interval. Also watch the tie-break: compare strings directly, not by length. With up to 100000 entries, don't sort every hotel's users. If you freeze during the live OA, StealthCoder can hand you this hash-map-of-sets shape fast, but you should be able to write it from memory.

Memorize the pattern. If you can't, run StealthCoder. The proctor sees the IDE. They don't see what's behind it.