<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 {
public:
    vector<int> partitionLabels(string s) {
        vector<int> res;
        int i=0;
        while(i<s.size())
        {
            vector<int> start(26, INT_MAX);
            vector<int> end(26, -1);
            int id=i;
            for(int j=i; j<s.size(); j++)
            {
                start[s[j]-'a']=min(start[s[j]-'a'], j);
                end[s[j]-'a']=max(end[s[j]-'a'], j);
                if(start[s[j]-'a']>=start[s[i]-'a'] && id>=start[s[j]-'a'])
                {
                    id=max(id, end[s[j]-'a']);
                }
            }
            res.push_back(id-i+1);
            i=id+1;
        }
        return res;
    }
};
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