<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>

from collections import Counter
import heapq

class Solution:
    def minDeletions(self, s: str) -> int:
        chars = Counter(s)
        freq = []

        for val in chars.values():
            freq.append(val)

        freq.sort(reverse=True)
        i = freq[0]
        index = 0
        visited = set([])
        duplicates = []
        ans = 0

        while index < len(freq):
            if freq[index] not in visited:
                visited.add(freq[index])
                index = index + 1
                continue
            else:
                while i > 0 and duplicates:
                    if i not in visited and i < (-duplicates[0]):
                        max_dup = heapq.heappop(duplicates)
                        max_dup = -max_dup
                        visited.add(i)
                        ans = ans + (max_dup - i)
                    i = i - 1
                heapq.heappush(duplicates, -freq[index])
                index = index + 1

        while i > 0 and duplicates:
            if i not in visited and i < (-duplicates[0]):
                max_dup = heapq.heappop(duplicates)
                max_dup = -max_dup
                visited.add(i)
                ans = ans + (max_dup - i)
            i = i - 1

        if duplicates:
            ans = ans - sum(duplicates)

        return ans
        
98