REST API: Highest International Students

The original task retrieves every page from a global university API. For deterministic practice, all fetched records are supplied in universityRecords. Each row has exactly three fields:
university: the university name
city: the university city
internationalStudents: a nonnegative decimal count that may contain comma separators
First consider records whose city exactly equals firstCity. If at least one exists, return the university name with the largest parsed international-student count. Otherwise, perform the same selection for secondCity.
If several universities in the selected city have the same maximum count, return the lexicographically smallest university name.

Function
highestInternationalStudents(firstCity: String, secondCity: String, universityRecords: String[][]) → String

Examples
Example 1
firstCity = "London"
secondCity = "Boston"
universityRecords = [["King's College London","London","15,075"],["University College London","London","21,500"],["Boston Tech","Boston","8,000"]]
return = "University College London"
London has records, so Boston is not considered. 21,500 is larger than 15,075, so University College London is returned.
Example 2
firstCity = "Seattle"
secondCity = "Boston"
universityRecords = [["Beta University","Boston","12,000"],["Alpha University","Boston","12,000"],["Gamma College","Chicago","25,000"]]
return = "Alpha University"
Seattle has no record, so the search falls back to Boston. The two Boston universities tie at 12,000, and Alpha University is lexicographically smaller.

Constraints
For this exercise, assume 1 <= universityRecords.length <= 100000.
For this exercise, assume every row has exactly three nonempty strings in the order [university, city, internationalStudents].
For this exercise, assume each parsed international-student count is between 0 and 10^9, inclusive.
For this exercise, assume at least one record matches firstCity or secondCity.
City matching is exact and case-sensitive.

The pattern is a single pass over the array with a running best. Filter rows where city equals firstCity. If none match, rerun for secondCity. Parse the count by removing commas, then converting to a number. The max is 10^9, so use a 64-bit integer if your language needs one. Track the best count and best name. Replace the best when the count is strictly higher, or when it's equal and the name is lexicographically smaller. The pitfalls are real. Don't compare counts as strings. Don't merge both cities into one pool, because the fallback only triggers when the first city has no records at all. Don't lowercase cities, since matching is case-sensitive. You can track both cities in one loop with two best slots, then pick at the end. If you freeze on the tie rule or the fallback logic, StealthCoder can give you a working solution in the live OA.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.