[LeetCode] Group Anagrams

The function signature has been updated to return a more intuitive vector<vector<string>>which treats a single string as a group of anagrams consisting of only itself.

The idea is to use an unordered_map to store those strings that are anagrams. We use the sorted string as the key and the string itself as the value. The strings are stored in a multisetsince there may be duplicates. Moreover, multiset will sort them by default as we desire.

The code is as follows.

class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        unordered_map<string, multiset<string>> mp;
        for (string s : strs) {
            string t = s; 
            sort(t.begin(), t.end());
            mp[t].insert(s);
        }
        vector<vector<string>> anagrams;
        for (auto m : mp) { 
            vector<string> anagram(m.second.begin(), m.second.end());
            anagrams.push_back(anagram);
        }
        return anagrams;
    }
};

Update: general sort takes O(nlogn) time. In this problem, since the string only contains lower-case alphabets, we can write a sorting function using counting sort (O(n) time) to speed up the sorting process. I write a string sorting functionstrSort below and using it to sort the string achieves the overall running time 72ms for this problem while the above code takes 76ms.

class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        unordered_map<string, multiset<string>> mp;
        for (string s : strs) {
            string t = strSort(s);
            mp[t].insert(s);
        }
        vector<vector<string>> anagrams;
        for (auto m : mp) { 
            vector<string> anagram(m.second.begin(), m.second.end());
            anagrams.push_back(anagram);
        }
        return anagrams;
    }
private:
    string strSort(string& s) {
        int count[26] = {0}, n = s.length();
        for (int i = 0; i < n; i++)
            count[s[i] - 'a']++;
        int p = 0;
        string t(n, 'a');
        for (int j = 0; j < 26; j++)
            for (int i = 0; i < count[j]; i++)
                t[p++] += j;
        return t;
    } 
};