<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>

class Solution {
public:
    string originalDigits(string s) {
     int nums[10];
     vector<int> freq(26);
     for(int i=0;i<s.size();i++){
        freq[s[i]-'a']+=1;
     }
     nums[0]=freq['z'-'a'];
     nums[2]=freq['w'-'a'];
     nums[4]=freq['u'-'a'];
     nums[5]=freq['f'-'a']-nums[4];
     nums[6]=freq['x'-'a'];
     nums[8]=freq['g'-'a'];
     nums[1]=freq['o'-'a']-nums[0]-nums[2]-nums[4];
     nums[3]=freq['h'-'a']-nums[8];
     nums[7]=freq['s'-'a']-nums[6];
     nums[9]=freq['i'-'a']-nums[6]-nums[8]-nums[5];
     string res;
     for(int i=0;i<10;i++){
        if(nums[i]!=0){
            res+=string(nums[i],'0'+i);
        }
     }
     return res;
    }
};
10700