Given a string <code>s</code>, return the <strong>maximum</strong> length of a <span data-keyword="substring">substring</span>&nbsp;such that it contains <em>at most two occurrences</em> of each character.
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;bcbbbcba&quot;</span></p>

<p><strong>Output:</strong> <span class="example-io">4</span></p>

<p><strong>Explanation:</strong></p>
The following substring has a length of 4 and contains at most two occurrences of each character: <code>&quot;bcbb<u>bcba</u>&quot;</code>.</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;aaaa&quot;</span></p>

<p><strong>Output:</strong> <span class="example-io">2</span></p>

<p><strong>Explanation:</strong></p>
The following substring has a length of 2 and contains at most two occurrences of each character: <code>&quot;<u>aa</u>aa&quot;</code>.</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= s.length &lt;= 100</code></li>
	<li><code>s</code> consists only of lowercase English letters.</li>
</ul>

class Solution:
    def maximumLengthSubstring(self, s: str) -> int:

        d = {}
        ans = 0
        l=0
        h=0
        n = len(s)

        while h < n:
            if s[h] in d:
                d[s[h]]+=1
            else:
                d[s[h]] = 1
                
            if max(list(d.values())) <= 2:
                ans = max(ans , h-l + 1)
                # print("ans: " ,h, " " , l , " ", ans)
                h +=1
               

            else:
                # print("Yes")
                # print(h , " " , l , " " , ans)
                ans = max(ans , h-l)

                while s[l] != s[h]:
                    d[s[l]]-=1
                    l+=1
                    

                d[s[l]]-=1
                l+=1
                h+=1
            

        return ans

        
52