<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) {
        vector<int> count(26);

        for( char ch: s){
            count[ch-'a']++;
        }

        vector<int> nums(10);
        //unique cases
        nums[0]=count['z'-'a'];
        nums[2]=count['w'-'a'];
        nums[4]=count['u'-'a'];
        nums[6]=count['x'-'a'];
        nums[8]=count['g'-'a'];
        //derived cases
        nums[1]=count['o'-'a']-nums[0]-nums[2]-nums[4];
        nums[3]=count['h'-'a']-nums[8];
        nums[5]=count['f'-'a']-nums[4];
        nums[7]=count['s'-'a']-nums[6];
        nums[9]=count['i'-'a']-nums[6]-nums[8]-nums[5];

        string ans="";
        for(int i=0;i<10;i++){
            while(nums[i]--){
                ans+=to_string(i);
            }
        }
        return ans;
    }
};
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