[Leetcode 54] 49 Anagrams

Problem:

Given an array of strings, return all groups of strings that are anagrams.

Note: All inputs will be in lower-case.

Analysis:

anagrams are those strings that consist of same characters but in different order. A very simple way to solve the problem is try to find each string's anagrams. This requires O(n^2) by scanning through the string vector and comparing the two sorted string whether are the same. But it won't work if the geiven string vector is too big.

A O(n) solution is as follows: use a map structure to keep the pair of <string, vector<string>>, the key is the sorted string the value is the original string's vector. In this way, we only need scan through the string vector and the map to find all anagrams. The complecity is only O(n+m), n is ths string number and m is the map entry number.

Code:

O(n^2) solution, Time Limit Execeed under larger test case

class Solution {
public:
    vector<string> anagrams(vector<string> &strs) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int tag[strs.size()];
        vector<string> mid;
        vector<string> res;
        
        res.clear();
        for (int i=0; i<strs.size(); i++) {
            tag[i] = 0;
            mid.push_back(strs[i]);
            sort(mid[i].begin(), mid[i].end());
        }
            
        for (int i=0; i<strs.size(); i++) {
            bool find = false;
            if (tag[i] == 0) {
                for (int j=i+1; j<strs.size(); j++) {
                    if (tag[j] == 0 && mid[i] == mid[j]) {
                            res.push_back(strs[j]);
                            find = true;
                            tag[j] = 1;
                    }
                }
                
            }
            
            if (find) {
                tag[i] = 1;
                res.push_back(strs[i]);
            }
        }
        
        return res;
    }
};

O(n) solution

class Solution {
public:
    vector<string> anagrams(vector<string> &strs) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<string> res;
        map<string, vector<string> > sm; 
        
        for (int i=0; i<strs.size(); i++) {
            string s = strs[i];
            sort(s.begin(), s.end());
            sm[s].push_back(strs[i]);
        }
        
        for (map<string, vector<string> >::iterator it=sm.begin(); 
                    it!=sm.end(); it++) {
            if ((*it).second.size() >= 2) {
                for (int i=0; i<(*it).second.size(); i++) {
                    res.push_back((*it).second[i]);
                }
            }
        }
        
        return res;
    }
};