<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 = []
        stack.append([s[0], 1])
        ret = ""

        for i in range(1, len(s)):
            if stack and stack[-1][0] == s[i]:
                stack[-1][-1] += 1
            else:
                stack.append([s[i], 1])
            
            if stack[-1][-1] == k:
                stack.pop()
        
        for char, val in stack:
            temp = char * val
            ret += temp
        
        return ret
        
98