<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<pair<int,int>> occurences;
        for(int i=0;i<26;i++) {
            char curr = 'a'+i;
            int first = -1, last = -1;
            for(int i=0;i<s.length();i++) {
                if(s[i] == curr) {
                    if(first == -1) {
                        first = i;
                    }
                    last = i;
                }
            }
            if(first != -1) {
                occurences.push_back({first, last});    
            }
        }
        sort(occurences.begin(), occurences.end());
        stack<pair<int,int>>st;
        for(auto [start, end]: occurences) {
            if(st.empty() || st.top().second < start) {
                st.push({start, end});
            } else {
                start = min(start, st.top().first);
                end = max(end, st.top().second);
                st.pop();
                st.push({start, end});
            }
        }
        vector<int>ans;
        while(!st.empty()) {
            ans.push_back(st.top().second - st.top().first+1);
            st.pop();
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
};

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