<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>

#define add insert
#define pb push_back
#define ppb pop_back
#define fir first
#define sec second
#define len length
#define umap unordered_map
#define pq priority_queue
#define pi pair

class Solution {
public:
    vector<int> partitionLabels(string s) {
        vector<pair<int, int>>raw(26, {INT_MAX, INT_MIN});
        for (int i= 0; i < s.length(); i++){
            raw[s[i] - 'a'].first = min(raw[s[i] - 'a'].first, i);
            raw[s[i] - 'a'].second = max(raw[s[i] - 'a'].second, i);
        }
        vector<vector<int>>data;
        for (int i = 0; i < 26; i++){
            if (raw[i].first == INT_MAX){continue;}
            data.pb({raw[i].first, raw[i].second});
        }
        sort(data.begin(), data.end());
        vector<vector<int>>depot;
        depot.pb(data[0]);
        for (int i = 1; i < data.size(); i++){
            if (data[i][0] <= depot.back()[1]){
                depot.back()[1] = max(depot.back()[1], data[i][1]);
            }
            else{depot.pb(data[i]);}
        }
        vector<int>res;
        for (auto& item: depot){
            res.pb(item[1] - item[0] + 1);
        }
        return res;
    }
};
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