<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) {
        unordered_map<char, int> dict;
        for(auto c:s) {
            if(dict.find(c) == dict.end())
                dict[c] = 1;
            else dict[c] ++;
        }

        int counter = 0;
        vector<int> res;
        unordered_map<char, int> currDict;
        for(int i=0; i<s.size(); i++) {
            if(currDict.find(s[i]) == currDict.end())
                currDict[s[i]] = 1;
            else currDict[s[i]] ++;

            counter++;

            if(currDict[s[i]] == dict[s[i]])
                currDict.erase(s[i]);

            if(currDict.empty()) {
                res.push_back(counter);
                counter = 0;
            }

            // cout<<"i = "<<i<<endl;
        // for (auto& [k, v] : dict) cout<<k<<" -> "<<v<<endl;
        }

        // if(count > 0)
        //     res[res.size()-1] += count;

        return res;
    }
};
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