<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>

# 1636. Sort Array by Increasing Frequency

#Time: O(nlogn) Space O(n)
class Solution:
    def frequencySort(self, s: str) -> str:
        freq = Counter(s)
        
        #sorted(string) -> sorted list
        s1 = sorted(s, key = lambda x : -freq[x])
        
        return "".join(s1)

    
#bucket sort
#Time: O(n) Space O(n)
class Solution:
    def frequencySort(self, s: str) -> str:
        counter = Counter(s)    #"tree" -> Counter({'t': 1, 'r': 1, 'e': 2})
        
        # n = max(counter.values())
        # buckets = [[] for _ in range(n)] 
        # # for char, freq in counter.items():
        # #     buckets[freq - 1].append(char)  #freq 从number变index
        # for key in counter:
        #     buckets[counter[key] - 1].append(key) #freq:[key...]
        
        buckets = defaultdict(list)
        for key in counter:
            buckets[counter[key] - 1].append(key) #freq:[key...]
            
        result = ""
        n = max(counter.values())
                        #[['t', 'r'], ['e']]
        for i in range(n - 1, -1, -1):   #backward, 从最大freq开始
            for char in buckets[i]:
                result += char * (i + 1) #freq 从index变number
                
        return result
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