<p>Given a string <code>s</code>, return <em>the </em><span data-keyword="lexicographically-smaller-string"><em>lexicographically smallest</em></span> <span data-keyword="subsequence-string"><em>subsequence</em></span><em> of</em> <code>s</code> <em>that contains all the distinct characters of</em> <code>s</code> <em>exactly once</em>.</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;= 1000</code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

<p>&nbsp;</p>
<strong>Note:</strong> This question is the same as 316: <a href="https://leetcode.com/problems/remove-duplicate-letters/" target="_blank">https://leetcode.com/problems/remove-duplicate-letters/</a>
from collections import Counter
class Solution:
    def smallestSubsequence(self, s: str) -> str:
        counter = Counter(s)
        st = []
        confirmed = set()
        for ch in s:
            # check the condition if it can be popped
            while len(st) > 0 and ch < st[-1] and counter[st[-1]] > 0  and ch not in confirmed:
                confirmed.discard(st.pop(-1))
            if ch not in confirmed:
                st.append(ch)
                confirmed.add(ch)
            counter[ch] -= 1
        return "".join(st)
16400