<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:
        length = len(s)
        i = 0
        stack = [[s[0], 1]]

        for i in range(1, length):
            if stack == []:
                stack.append([s[i], 1])
                continue

            if stack[-1][0] == s[i]:
                stack[-1][1] += 1
            else:
                stack.append([s[i], 1])

            if stack[-1][1] == k:
                stack.pop()

        return "".join([x[0]*x[1] for x in stack])


        # while i <= (length - k):
        #     print(s, s[i:i+k], i)

        #     if len(set(s[i:i+k])) == 1:
        #         # update array
        #         s = s[:i] + s[i+k:]
        #         length = len(s)
        #         i = 0
        #         continue

        #     i += 1

        # return s 
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