<p>You are given a string <code>s</code>. We want to partition the string into as many parts as possible so that each letter appears in at most one part.</p>

<p>Note that the partition is done so that after concatenating all the parts in order, the resultant string should be <code>s</code>.</p>

<p>Return <em>a list of integers representing the size of these parts</em>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;ababcbacadefegdehijhklij&quot;
<strong>Output:</strong> [9,7,8]
<strong>Explanation:</strong>
The partition is &quot;ababcbaca&quot;, &quot;defegde&quot;, &quot;hijhklij&quot;.
This is a partition so that each letter appears in at most one part.
A partition like &quot;ababcbacadefegde&quot;, &quot;hijhklij&quot; is incorrect, because it splits s into less parts.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;eccbbbbdec&quot;
<strong>Output:</strong> [10]
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 500</code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution:
    def partitionLabels(self, s: str) -> List[int]:
        last = {c:i for i,c in enumerate(s)} #last index
        ans = []
        j = 0
        anchor = 0
        for i, c in enumerate(s):
            j = max(j, last[c])
            if i == j:
                ans.append(i-anchor+1)
                anchor = j + 1
        return ans
16300