<p>You are given a string <code>s</code>, an integer <code>k</code>, a letter <code>letter</code>, and an integer <code>repetition</code>.</p>

<p>Return <em>the <strong>lexicographically smallest</strong> subsequence of</em> <code>s</code><em> of length</em> <code>k</code> <em>that has the letter</em> <code>letter</code> <em>appear <strong>at least</strong></em> <code>repetition</code> <em>times</em>. The test cases are generated so that the <code>letter</code> appears in <code>s</code> <strong>at least</strong> <code>repetition</code> times.</p>

<p>A <strong>subsequence</strong> is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.</p>

<p>A string <code>a</code> is <strong>lexicographically smaller</strong> than a string <code>b</code> if in the first position where <code>a</code> and <code>b</code> differ, string <code>a</code> has a letter that appears earlier in the alphabet than the corresponding letter in <code>b</code>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;leet&quot;, k = 3, letter = &quot;e&quot;, repetition = 1
<strong>Output:</strong> &quot;eet&quot;
<strong>Explanation:</strong> There are four subsequences of length 3 that have the letter &#39;e&#39; appear at least 1 time:
- &quot;lee&quot; (from &quot;<strong><u>lee</u></strong>t&quot;)
- &quot;let&quot; (from &quot;<strong><u>le</u></strong>e<u><strong>t</strong></u>&quot;)
- &quot;let&quot; (from &quot;<u><strong>l</strong></u>e<u><strong>et</strong></u>&quot;)
- &quot;eet&quot; (from &quot;l<u><strong>eet</strong></u>&quot;)
The lexicographically smallest subsequence among them is &quot;eet&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="example-2" src="https://assets.leetcode.com/uploads/2021/09/13/smallest-k-length-subsequence.png" style="width: 339px; height: 67px;" />
<pre>
<strong>Input:</strong> s = &quot;leetcode&quot;, k = 4, letter = &quot;e&quot;, repetition = 2
<strong>Output:</strong> &quot;ecde&quot;
<strong>Explanation:</strong> &quot;ecde&quot; is the lexicographically smallest subsequence of length 4 that has the letter &quot;e&quot; appear at least 2 times.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;bb&quot;, k = 2, letter = &quot;b&quot;, repetition = 2
<strong>Output:</strong> &quot;bb&quot;
<strong>Explanation:</strong> &quot;bb&quot; is the only subsequence of length 2 that has the letter &quot;b&quot; appear at least 2 times.
</pre>

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

<ul>
	<li><code>1 &lt;= repetition &lt;= k &lt;= s.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
	<li><code>letter</code> is a lowercase English letter, and appears in <code>s</code> at least <code>repetition</code> times.</li>
</ul>

class Solution:
    def smallestSubsequence(self, s: str, k: int, letter: str, repetition: int) -> str:
        stack = []
        count_letter = s.count(letter)
        used_letter = 0
        
        for i, char in enumerate(s):
            # While the stack is not empty, the current character is smaller than the last character in the stack,
            # and removing the last character in the stack does not prevent us from forming a valid subsequence
            while (stack and stack[-1] > char and
                   len(stack) + len(s) - i > k and
                   (stack[-1] != letter or used_letter + count_letter > repetition)):
                if stack.pop() == letter:
                    used_letter -= 1
            
            # If we can add the current character to the stack
            if len(stack) < k:
                if char == letter:
                    stack.append(char)
                    used_letter += 1
                elif k - len(stack) > repetition - used_letter:
                    stack.append(char)
            
            # Decrease the count of the remaining letters
            if char == letter:
                count_letter -= 1
        
        return ''.join(stack)

# Example usage:
# solution = Solution()
# print(solution.smallestSubsequence("leet", 3, "e", 1))  # Output: "eet"
# print(solution.smallestSubsequence("leetcode", 4, "e", 2))  # Output: "ecde"
# print(solution.smallestSubsequence("bb", 2, "b", 2))  # Output: "bb"
17500