<p>Given a string <code>s</code>, partition the string into one or more <strong>substrings</strong> such that the characters in each substring are <strong>unique</strong>. That is, no letter appears in a single substring more than <strong>once</strong>.</p>

<p>Return <em>the <strong>minimum</strong> number of substrings in such a partition.</em></p>

<p>Note that each character should belong to exactly one substring in a partition.</p>

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

<pre>
<strong>Input:</strong> s = &quot;abacaba&quot;
<strong>Output:</strong> 4
<strong>Explanation:</strong>
Two possible partitions are (&quot;a&quot;,&quot;ba&quot;,&quot;cab&quot;,&quot;a&quot;) and (&quot;ab&quot;,&quot;a&quot;,&quot;ca&quot;,&quot;ba&quot;).
It can be shown that 4 is the minimum number of substrings needed.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;ssssss&quot;
<strong>Output:</strong> 6
<strong>Explanation:
</strong>The only valid partition is (&quot;s&quot;,&quot;s&quot;,&quot;s&quot;,&quot;s&quot;,&quot;s&quot;,&quot;s&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 English lowercase letters.</li>
</ul>

class Solution:
    def partitionString(self, s: str) -> int:
        
        counts = defaultdict(int)

        for ch in s:
            counts[ch] +=1
        
        # a = 4 -> 3 -> 2 -> 1 -> 0
        # b = 2 -> 1 -> 0
        # c = 1 -> 0
        i = 0
        ans = 0
        seen = set()
        while i <len(s):
            # seen = set()
            if s[i] in seen:
                ans +=1
                seen = set()
                continue
            
            seen.add(s[i])
            i+=1

        return ans + 1
            




        ans = 0
        print(counts)
        # while counts:
        for key, val in counts.items():
            if val > ans:
                ans = val
            #     val -=val
            
            #     if val == 0:
            #         del counts[key]
            # ans +=1
    
        return ans


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