<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:
    def frequencySort(self, s: str) -> str:
        # 1. Count frequency of characters 
        # 2. Create tuple of characters (frequency, char)
        # 3. Sort by frequency 
        # 4. Recreate string 
        # O(nlogn) time complexity as O(n) + O(n) + O(nlogn) + O(n)
        # O(n) space complexity for creating array of tuples 
        # count_c = Counter(s)
        # arr = []
        # for k, v in count_c.items():
        #     arr.append((v, k))

        # arr.sort(key = lambda x: x[0], reverse = True)
        
        # str_arr = []
        # for frequency, char in arr:
        #     str_arr += [char] * frequency 
        
        # return ''.join(str_arr)


        # 1. Sort string
        # 2. Count frequency of characters, once done add them to count array e.g: count 3 for char 'a' -> count_Arr[3].append('a')
        # 3. Walk in reverse from the back of the count array to recreate the string 
        # O(nlogn) time complexity as O(nlogn) + O(n) + O(n) [amortized]
        # O(n) space complexity for creating count array
        s = sorted(s) 
        s += " "

        arr = [[] for _ in range(len(s)+1)]

        count = 1
        for i in range(1, len(s)):
            if s[i-1] != s[i]:
                print(s[i-1])
                arr[count].append(s[i-1])
                count = 1
            else:
                count += 1
        
        str_arr = []
        for i in range(len(arr) -1, -1, -1):
            if not arr[i]:
                continue
            for char in arr[i]:
                str_arr += [char] * i 
        
        return ''.join(str_arr) 

        






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