423. Reconstruct Original Digits from English

A different approach to the problem, besides the popular math solution.

class Solution 
{
    vector<int> cnt;
    vector<int> ccnt;
    
    void _handle(char c, int d, string str)
    {
        if (ccnt[c - 'a'] > 0)
        {
            cnt[d] += ccnt[c - 'a'];
            for(auto cc : str) ccnt[cc - 'a'] -= cnt[d];
        }
    }
    
public:

    string originalDigits(string s) {
        cnt.assign(10, 0);
        ccnt.assign(26, 0);
        
        // count char freq
        for (auto c : s)
            ccnt[(c-'a')] += 1;
        
        // pass 1 - 0 2 4 6 8
        _handle('z', 0, "zero");
        _handle('w', 2, "two");
        _handle('u', 4, "four");
        _handle('x', 6, "six");
        _handle('g', 8, "eight");
        
        // pass 2 - 1 3
        _handle('o', 1, "one");
        _handle('r', 3, "three");
        
        // pass 3 - 5 7
        _handle('f', 5, "five");
        _handle('s', 7, "seven");
        
        // pass 4 - 9
        _handle('i', 9, "nine");
        
        string ret;
        for(int i = 0; i < cnt.size(); i ++)
            for(int j = 0; j < cnt[i]; j ++)
            ret += to_string(i);
        return ret;
    }
};