Optimal Partition of String
<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>
Greedy; Hash Table; String
class Solution {
public:
    int partitionString(string s) {
        int ans = 1, mask = 0;
        for (char& c : s) {
            int x = c - 'a';
            if (mask >> x & 1) {
                ++ans;
                mask = 0;
            }
            mask |= 1 << x;
        }
        return ans;
    }
};
