<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 Solution:
    def minDeletions(self, s: str) -> int:
        
        freqCount = defaultdict(int)

        for char in s:
            freqCount[char] += 1
        
        freqMap = defaultdict(list)

        for key,val in freqCount.items():
            freqMap[val].append(key)
        
        # print('freqMap: ', freqMap)

        freqSet = set()
        freqSet.update(list(freqMap.keys()))

        ans = 0

        for key,val in freqMap.items():
            n = len(val)

            if n <= 1:
                continue

            for i in range(1,n):
                # find the next min freq which is unique
                nextFreq = key
                while nextFreq in freqSet:
                    nextFreq -= 1

                ans += (key - nextFreq)

                if nextFreq != 0:
                    freqSet.add(nextFreq)

        # print('freqSet: ', freqSet)

        return ans
159