<p>Given a string <code>s</code>, remove duplicate letters so that every letter appears once and only once. You must make sure your result is <span data-keyword="lexicographically-smaller-string"><strong>the smallest in lexicographical order</strong></span> among all possible results.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;bcabc&quot;
<strong>Output:</strong> &quot;abc&quot;
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;cbacdcbc&quot;
<strong>Output:</strong> &quot;acdb&quot;
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>4</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Note:</strong> This question is the same as 1081: <a href="https://leetcode.com/problems/smallest-subsequence-of-distinct-characters/" target="_blank">https://leetcode.com/problems/smallest-subsequence-of-distinct-characters/</a></p>

class Solution {
public:
    string removeDuplicateLetters(string s) {
        stack<int> st;
        std::map<char, int> frequency;
        vector<bool> visited(26);
        string answer = "";

        for (int c = 0; c < s.length(); c++){

            if (frequency.find(s[c]) == frequency.end()){
                frequency.insert({s[c], 1});
            }

            else{
                frequency[s[c]]++;
            }
        }

        for (int d = 0; d < s.length(); d++){
            
            while (!st.empty() && s[d] < st.top() && frequency[st.top()] > 0 && !visited[s[d] - 'a']){
                visited[st.top() - 'a'] = false;
                st.pop();
            }

            if (!visited[s[d] - 'a']){
                visited[s[d] - 'a'] = true;
                st.push(s[d]);
            }
            frequency[s[d]]--;
        }

        while (!st.empty()){
            answer.push_back(st.top());
            st.pop();
        }
        reverse(answer.begin(), answer.end());

        return answer;
    }
};
9800