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Done in O(n*m). For each string, we want to bring all the 'a's (if it has) to the front, then the b's, for example we wanna convert yxbaa-> aabxy. This can be done easily by keeping count of how many occurrence per character. Clearly if 2 strings are anagram of each other then they must have to same converted version.
- sgtrouge January 28, 2016The remaining part is using radix/trie sort to group all the string that has the same converted version together. This can be done in O(n*m).