Write same_team_pairs(pairs) that counts unique unordered pairs: [["a","b"],["b","a"],["a","c"]] → 2.
same_team_pairs([['a', 'b'], ['b', 'a'], ['a', 'c']]) → 2
same_team_pairs([]) → 0
Submit runs 4 tests, including hidden edge cases.
Each hint gives a little more help. Try on your own first.
Run your code to see the output here.