<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) {
        map<char,int> m; // keep track of frequency of every element in the string
        for(auto c : s) m[c]++;

        set<char> st; // to check if an element already exists in the current partition
        vector<int> res; // the final answer

        int count = 0; // count of elements in the current parition

        for(int i=0; i<s.size(); i++){
            st.insert(s[i]);
            m[s[i]]--;

            // if all the occurences of the letter have been inserted in the set
            if(m[s[i]] == 0){
                // remove it from the set so that we dont have to check for it anymore
                st.erase(s[i]);
            } // meaning we keep adding to the set until we have exhausted all the occurences of all the letters in that partition

            count++;

            if(st.empty()){
                res.push_back(count);
                count = 0; // for the next iteration
            }


            
        }

        return res;
    }
};
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