Valid Anagram Solutions in C++

// Source : https://leetcode.com/problems/valid-anagram/
// Author : Hao Chen
// Date   : 2015-08-16




class Solution {
public:

    //stupid way - but easy to understand  - 76ms
    bool isAnagram01(string s, string t) {
        sort(s.begin(), s.end());
        sort(t.begin(), t.end());
        return s == t;
    }
    
    //using a map to count every chars in the string.
    bool isAnagram02(string s, string t) {
        int map[26] ={0} ; //only lowercase alphabets
        //memset(map, 0, sizeof(map));
        
        // count each char for s
        for (int i=0; i<s.size(); i++) {
            map[s[i]-'a']++;
        }
        // decrease the count for t
        for (int i=0; i<t.size(); i++) {
            map[t[i]-'a']--;
        }
        
        //if all alphabets is zero, then they are anagram 
        for (int i=0; i<sizeof(map)/sizeof(map[0]); i++) {
            if (map[i]!=0) return false;
        }
        return true;
    }
    

    
    bool isAnagram(string s, string t) {
        return isAnagram02(s,t); //12ms
        return isAnagram01(s,t); //76ms
    }
    
};

/// Source : https://leetcode.com/problems/valid-anagram/
/// Author : liuyubobobo
/// Time   : 2018-08-13

#include <iostream>

using namespace std;

/// Sorting
/// Time Complexity: O(nlogn)
/// Space Complexity: O(1)
class Solution {
public:
    bool isAnagram(string s, string t) {

        sort(s.begin(), s.end());
        sort(t.begin(), t.end());

        return s == t;
    }
};


int main() {

    cout << Solution().isAnagram("anagram", "nagaram") << endl;
    cout << Solution().isAnagram("rat", "car") << endl;
    cout << Solution().isAnagram("ab", "a") << endl;

    return 0;
}

/// Source : https://leetcode.com/problems/valid-anagram/
/// Author : liuyubobobo
/// Time   : 2017-01-17

#include <iostream>

using namespace std;

/// Using Hashtable
/// Time Complexity: O(n)
/// Space Complexity: O(26)
class Solution {
public:
    bool isAnagram(string s, string t) {

        if( s.size() != t.size() )
            return false;

        int freq[26] = {0};
        for( int i = 0 ; i < s.size() ; i ++ )
            freq[s[i]-'a'] ++;

        for( int i = 0 ; i < t.size() ; i ++ ){
            freq[t[i]-'a'] --;
            if( freq[t[i]-'a'] < 0 )
                return false;
        }

        return true;
    }
};


int main() {

    cout << Solution().isAnagram("anagram", "nagaram") << endl;
    cout << Solution().isAnagram("rat", "car") << endl;
    cout << Solution().isAnagram("ab", "a") << endl;

    return 0;
}