<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 {
public:
    int numSplits(string s) {
      /* int n=s.size();
       int count=0;

       for(int i=0;i<n;i++)
       {
           unordered_set<char> left, right;
        for(int j=0;j<i;j++)
        {
            left.insert(s[j]);
        }
        for(int j=i;j<n;j++){
            right.insert(s[j]);
        }
        if(left.size()==right.size()){
            count++;
        }
       }
       return count;
       */

         int n = s.size();
        vector<int> right(n);
        unordered_set<char> set;
        
        
      for (int i = n - 1; i >= 0; --i) {
            set.insert(s[i]);
            right[i] = set.size();
        }
        
        int count = 0;
        set.clear();
        
        
        for (int i = 0; i < n - 1; i++) {
            set.insert(s[i]);
            if (set.size() == right[i + 1]) {
                count++;
            }
        }
        
        return count;
    }
};
13000