Partition Labels
<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. For example, the string <code>&quot;ababcc&quot;</code> can be partitioned into <code>[&quot;abab&quot;, &quot;cc&quot;]</code>, but partitions such as <code>[&quot;aba&quot;, &quot;bcc&quot;]</code> or <code>[&quot;ab&quot;, &quot;ab&quot;, &quot;cc&quot;]</code> are invalid.</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>
Greedy; Hash Table; Two Pointers; String
class Solution {
public:
    vector<int> partitionLabels(string s) {
        int last[26] = {0};
        int n = s.size();
        for (int i = 0; i < n; ++i) {
            last[s[i] - 'a'] = i;
        }
        vector<int> ans;
        int mx = 0, j = 0;
        for (int i = 0; i < n; ++i) {
            mx = max(mx, last[s[i] - 'a']);
            if (mx == i) {
                ans.push_back(i - j + 1);
                j = i + 1;
            }
        }
        return ans;
    }
};