<p>Given a string <code>s</code>, sort it in <strong>decreasing order</strong> based on the <strong>frequency</strong> of the characters. The <strong>frequency</strong> of a character is the number of times it appears in the string.</p>

<p>Return <em>the sorted string</em>. If there are multiple answers, return <em>any of them</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;tree&quot;
<strong>Output:</strong> &quot;eert&quot;
<strong>Explanation:</strong> &#39;e&#39; appears twice while &#39;r&#39; and &#39;t&#39; both appear once.
So &#39;e&#39; must appear before both &#39;r&#39; and &#39;t&#39;. Therefore &quot;eetr&quot; is also a valid answer.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;cccaaa&quot;
<strong>Output:</strong> &quot;aaaccc&quot;
<strong>Explanation:</strong> Both &#39;c&#39; and &#39;a&#39; appear three times, so both &quot;cccaaa&quot; and &quot;aaaccc&quot; are valid answers.
Note that &quot;cacaca&quot; is incorrect, as the same characters must be together.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;Aabb&quot;
<strong>Output:</strong> &quot;bbAa&quot;
<strong>Explanation:</strong> &quot;bbaA&quot; is also a valid answer, but &quot;Aabb&quot; is incorrect.
Note that &#39;A&#39; and &#39;a&#39; are treated as two different characters.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 5 * 10<sup>5</sup></code></li>
	<li><code>s</code> consists of uppercase and lowercase English letters and digits.</li>
</ul>

class Solution {
public:

   /* static bool comp(pair<int,char> p1, pair<int,char> p2)
    {
        if(p1.first >= p2.first)
        return true;

        else
        return false;
    }*/

    string frequencySort(string s) {

        vector<pair<int,char>> hash('z'+1,{0,0});

        for(auto it:s)
        {
            hash[it] = {hash[it].first+1,it};
        }

        sort(hash.begin(),hash.end());

        string res = "";

        for(int i = 122; i>=0; i--)
        {
            if(hash[i].first==0)
            break;

            else
            {
                res = res + string(hash[i].first,hash[i].second);
            }
        }

        return res;
        
    }
};
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