<p>You are given a string <code>s</code> and an integer <code>k</code>, a <code>k</code> <strong>duplicate removal</strong> consists of choosing <code>k</code> adjacent and equal letters from <code>s</code> and removing them, causing the left and the right side of the deleted substring to concatenate together.</p>

<p>We repeatedly make <code>k</code> <strong>duplicate removals</strong> on <code>s</code> until we no longer can.</p>

<p>Return <em>the final string after all such duplicate removals have been made</em>. It is guaranteed that the answer is <strong>unique</strong>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;abcd&quot;, k = 2
<strong>Output:</strong> &quot;abcd&quot;
<strong>Explanation: </strong>There&#39;s nothing to delete.</pre>

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

<pre>
<strong>Input:</strong> s = &quot;deeedbbcccbdaa&quot;, k = 3
<strong>Output:</strong> &quot;aa&quot;
<strong>Explanation: 
</strong>First delete &quot;eee&quot; and &quot;ccc&quot;, get &quot;ddbbbdaa&quot;
Then delete &quot;bbb&quot;, get &quot;dddaa&quot;
Finally delete &quot;ddd&quot;, get &quot;aa&quot;</pre>

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

<pre>
<strong>Input:</strong> s = &quot;pbbcggttciiippooaais&quot;, k = 2
<strong>Output:</strong> &quot;ps&quot;
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>2 &lt;= k &lt;= 10<sup>4</sup></code></li>
	<li><code>s</code> only contains lowercase English letters.</li>
</ul>

class Solution:
    def removeDuplicates(self, s: str, k: int) -> str:
        
        stack = []

        cdict = [s[0], 0, 1]
        # d e e e d b b b d c c c 
        # k = 3 
        # [d, 0, 1]
        # stack = [[d, 0, 1]]
        # cdict = [e, 1, 1]
        # cdict = [e, 1, 2]
        # cdict = [e, 1, 3]
        # triggers == k: 
        # stack is currently = [d, 0, 1]
        # so we 'delete' these k letters 
        # by popping the stack so now cdict
        # is [d, 0, 1]
        # then d[]
        for idx, char in enumerate(s[1:]):
            if cdict[0] == None or char == cdict[0]: 
                cdict[0] = char
                cdict[2] += 1
                # if we've reached the right count,
                # we want to remove it
                if cdict[2] == k:
                    if stack: 
                        cdict = stack.pop(-1)
                    else: 
                        cdict = [None, 0, 0]
            else: 
                stack.append(cdict)
                cdict = [char, idx, 1]

        if cdict[0] != None: 
            stack.append(cdict)

        return "".join([tup[0]*tup[2] for tup in stack])
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