Reconstruct Original Digits from English
<p>Given a string <code>s</code> containing an out-of-order English representation of digits <code>0-9</code>, return <em>the digits in <strong>ascending</strong> order</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s = "owoztneoer"
<strong>Output:</strong> "012"
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s = "fviefuro"
<strong>Output:</strong> "45"
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s[i]</code> is one of the characters <code>[&quot;e&quot;,&quot;g&quot;,&quot;f&quot;,&quot;i&quot;,&quot;h&quot;,&quot;o&quot;,&quot;n&quot;,&quot;s&quot;,&quot;r&quot;,&quot;u&quot;,&quot;t&quot;,&quot;w&quot;,&quot;v&quot;,&quot;x&quot;,&quot;z&quot;]</code>.</li>
	<li><code>s</code> is <strong>guaranteed</strong> to be valid.</li>
</ul>
Hash Table; Math; String
class Solution {
public:
    string originalDigits(string s) {
        vector<int> counter(26);
        for (char c : s) ++counter[c - 'a'];
        vector<int> cnt(10);
        cnt[0] = counter['z' - 'a'];
        cnt[2] = counter['w' - 'a'];
        cnt[4] = counter['u' - 'a'];
        cnt[6] = counter['x' - 'a'];
        cnt[8] = counter['g' - 'a'];

        cnt[3] = counter['h' - 'a'] - cnt[8];
        cnt[5] = counter['f' - 'a'] - cnt[4];
        cnt[7] = counter['s' - 'a'] - cnt[6];

        cnt[1] = counter['o' - 'a'] - cnt[0] - cnt[2] - cnt[4];
        cnt[9] = counter['i' - 'a'] - cnt[5] - cnt[6] - cnt[8];

        string ans;
        for (int i = 0; i < 10; ++i)
            for (int j = 0; j < cnt[i]; ++j)
                ans += char(i + '0');
        return ans;
    }
};