<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:
        n = len(s)
        freq = [0]*(n+1)
        count = {}
        for i in s:
            count[i] = count.get(i,0)+1
        empty = []
        t = 0
        for k,v in count.items():
            t = max(t,v)
            freq[v] +=1
        for i in range(t,0,-1):
            if not freq[i]:
                empty.append(i)
        res = 0
        for i in range(t,-1,-1):
            if freq[i] > 1:
                for j in range(freq[i]-1,0,-1):
                    k = 0
                    flag = False
                    while empty and k < len(empty):
                        if empty[k] < i:
                            res += i-empty[k]
                            flag = True
                            break
                        k += 1
                    if not flag:
                        res += i
                    while k > -1 and empty:
                        empty.pop(0)
                        k -= 1        
        return res

        

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