<p>You are given a string <code>s</code>.</p>

<p>A split is called <strong>good</strong> if you can split <code>s</code> into two non-empty strings <code>s<sub>left</sub></code> and <code>s<sub>right</sub></code> where their concatenation is equal to <code>s</code> (i.e., <code>s<sub>left</sub> + s<sub>right</sub> = s</code>) and the number of distinct letters in <code>s<sub>left</sub></code> and <code>s<sub>right</sub></code> is the same.</p>

<p>Return <em>the number of <strong>good splits</strong> you can make in <code>s</code></em>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;aacaba&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are 5 ways to split <code>&quot;aacaba&quot;</code> and 2 of them are good. 
(&quot;a&quot;, &quot;acaba&quot;) Left string and right string contains 1 and 3 different letters respectively.
(&quot;aa&quot;, &quot;caba&quot;) Left string and right string contains 1 and 3 different letters respectively.
(&quot;aac&quot;, &quot;aba&quot;) Left string and right string contains 2 and 2 different letters respectively (good split).
(&quot;aaca&quot;, &quot;ba&quot;) Left string and right string contains 2 and 2 different letters respectively (good split).
(&quot;aacab&quot;, &quot;a&quot;) Left string and right string contains 3 and 1 different letters respectively.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;abcd&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> Split the string as follows (&quot;ab&quot;, &quot;cd&quot;).
</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> consists of only lowercase English letters.</li>
</ul>

class Solution:
    def numSplits(self, s: str) -> int:
        # n = len(s)
        # cnt = 0
        # for i in range(1, n):
        #     leftDict = Counter(s[:i])
        #     rightDict = Counter(s[i:])
        #     if len(leftDict.keys()) == len(rightDict.keys()):
        #         cnt +=1
        # return cnt

        myS1=collections.Counter() #s1 is now empty
        myS2=collections.Counter(s) #s2 is now has all string 
        ways=0 #when len of myS1= len of myS2 --> ways+=1

        for letter in s:    
            myS1[letter]+=1
            myS2[letter]-=1
            
            if myS2[letter]==0:
                myS2.pop(letter)
                
            if len(myS1)==len(myS2):
                ways+=1 
        return ways    
16800