Valid anagram solution may need to be retweaked for C++

[Malcolmt129]

After doing some debugging I found that the last for loop is not necessary in the C++ implementation. This is because it doesn't check the hashmap by index as the code may suggest. Instead it just adds another value into the hashmap. So instead of:

class Solution {
public:
    bool isAnagram(string s, string t) {
        if(s.size() != t.size()) return false;
        
        unordered_map<char,int> smap;
        unordered_map<char,int> tmap;
        
        for(int i = 0; i < s.size(); i++){
            smap[s[i]]++;
            tmap[t[i]]++;
        }
        
        for(int i = 0; i < smap.size(); i++){
            if(smap[i] != tmap[i]) return false;
        }
        return true;
    } 

Maybe use:

class Solution {
public:
    bool isAnagram(string s, string t) {
     if (s.size() != t.size()) return false;
     
     unordered_map<char,int> smap;
     unordered_map<char,int> tmap;

     for (int i =0; i < s.size(); i++){
        smap[s[i]]++;
        tmap[t[i]]++; 
     }

    if (smap != tmap) return false;
     return true;

    }
};

This solution has been accepted also but it is important that new programmers don't think that you can access an unordered map in a similar way to an array in other languages.