<p>A string <code>s</code> is called <strong>good</strong> if there are no two different characters in <code>s</code> that have the same <strong>frequency</strong>.</p>

<p>Given a string <code>s</code>, return<em> the <strong>minimum</strong> number of characters you need to delete to make </em><code>s</code><em> <strong>good</strong>.</em></p>

<p>The <strong>frequency</strong> of a character in a string is the number of times it appears in the string. For example, in the string <code>&quot;aab&quot;</code>, the <strong>frequency</strong> of <code>&#39;a&#39;</code> is <code>2</code>, while the <strong>frequency</strong> of <code>&#39;b&#39;</code> is <code>1</code>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;aab&quot;
<strong>Output:</strong> 0
<strong>Explanation:</strong> <code>s</code> is already good.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;aaabbbcc&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> You can delete two &#39;b&#39;s resulting in the good string &quot;aaabcc&quot;.
Another way it to delete one &#39;b&#39; and one &#39;c&#39; resulting in the good string &quot;aaabbc&quot;.</pre>

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

<pre>
<strong>Input:</strong> s = &quot;ceabaacb&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> You can delete both &#39;c&#39;s resulting in the good string &quot;eabaab&quot;.
Note that we only care about characters that are still in the string at the end (i.e. frequency of 0 is ignored).
</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</code>&nbsp;contains only lowercase English letters.</li>
</ul>

class Solution1:
    def minDeletions(self, s: str) -> int:
        freq = {}
        op = 0
        for i in s:
            freq[i] = 1 if i not in freq else freq[i] + 1
                
        freq = sorted(freq.items() , key= lambda el : el[1], reverse=True)
        freq = [el[1] for el in freq]
        
        change = True
        while change:
            change = False
            for i in range(1,len(freq)):
                if freq[i] == freq[i-1] and freq[i] != 0:
                    op += 1
                    freq[i] -= 1
                    change = True
            if change:
                freq.sort(reverse=True)
        return op
    
    
class Solution:
    def minDeletions(self, s: str) -> int:
        freq = {}
        for char in s:
            freq[char] = freq.get(char, 0) + 1
        
        freq_list = list(freq.values())
        freq_set = set()  # To store used frequencies
        deletions = 0
        
        for f in freq_list:
            # Decrease frequency until it's unique
            while f > 0 and f in freq_set:
                f -= 1
                deletions += 1
            freq_set.add(f)
        print(freq_set)
        return deletions

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