Problem statement

Given two strings s and t, write a function to determine if t is an anagram of s.

For example,
s = "anagram", t = "nagaram", return true.
s = "rat", t = "car", return false.

Note:
You may assume the string contains only lowercase alphabets.

Solution one: hash table/array(AC)

General idea:

• if the size of two strings is not equal, return false;
• Enumerate each element of two strings, put each element into hash table/array.
• compare the hash table/array

Time complexity is O(n), space complexity is O(n)

hash table version

class Solution {
public:
    bool isAnagram(string s, string t) {
        if(s.size() != t.size()){
            return false;
        }
        unordered_map<char, int> dict;
        for(string::size_type ix = 0; ix < s.size(); ix++){
            dict[s[ix]]++;
        }
        for(string::size_type ix = 0; ix < t.size(); ix++){
            if(dict.find(t[ix]) != dict.end()){
                dict[t[ix]]--;
                if(dict[t[ix]] == 0){
                    dict.erase(t[ix]);
                }
            }
        }
        return dict.empty();
    }
};

array version

class Solution {
public:
    bool isAnagram(string s, string t) {
        if(s.size() != t.size()){
            return false;
        }
        int size = s.size();
        vector<int> sv(26, 0), tv(26, 0);
        for(int i = 0; i < size; i++){
            sv[s[i] - ‘a‘]++;
            tv[t[i] - ‘a‘]++;
        }
        return sv == tv;
    }
};

Solution two: sort and compare

• Sort two strings
• compare whether they are equal

Time complexity is O(nlgn), space complexity is O(1)

class Solution {
public:
    bool isAnagram(string s, string t) {
        sort(s.begin(), s.end());
        sort(t.begin(), t.end());
        return s == t;
    }
};