Count anagram pairs
Given a list of lowercase words, count how many unordered index pairs (i, j) with i < j hold words that are anagrams of each other — identical letters with identical multiplicities, such as "tab" and "bat". Identical words are trivially anagrams, so repeated words form pairs too. For example, ["tab", "bat", "cat", "act", "atb"] has the anagram family {tab, bat, atb} contributing three pairs and {cat, act} contributing one, for an answer of 4. Comparing all pairs directly is O(n^2) — group first instead.
Implement
anagram_pairs(words: list[str]) → intExamples
in
[["tab","bat","cat","act","atb"]]out4What a strong answer looks like
State your approach and its time/space complexity out loud before you optimize. Handle the edge cases (empty input, duplicates, overflow), and say why you chose this over the brute force. Green tests are the floor, not the grade.
0:00 of about 15 min
solution.py
InputExpectedGot
[["tab","bat","cat","act","atb"]]4not run yetsample