Given a string s, a k duplicate removal consists of choosing k adjacent and equal letters from s and removing them causing the left and the right side of the deleted substring to concatenate together.
We repeatedly make k duplicate removals on s until we no longer can.
Return the final string after all such duplicate removals have been made.
It is guaranteed that the answer is unique.
Example 1:
Input: s = "abcd", k = 2
Output: "abcd"
Explanation: There's nothing to delete.
Example 2:
Input: s = "deeedbbcccbdaa", k = 3
Output: "aa"
Explanation: 
First delete "eee" and "ccc", get "ddbbbdaa"
Then delete "bbb", get "dddaa"
Finally delete "ddd", get "aa"
Example 3:
Input: s = "pbbcggttciiippooaais", k = 2
Output: "ps"
Constraints:
1 <= s.length <= 10^5
2 <= k <= 10^4
s only contains lower case English letters.
class Solution:
    def removeDuplicates(self, s: str, k: int) -> str:
        if not s or k == 0:
            return s
        
        stack = [['#', 0]]
        for c in s:
            if stack[-1][0] == c:
                stack[-1][1] += 1
                if stack[-1][1] == k:
                    stack.pop()
            else:
                stack.append([c, 1])
        

        return ''.join(c*k for c, k in stack)

        
        

class Solution:
    def removeDuplicates(self, s: str, k: int) -> str:
        if not s or k == 0:
            return s
        
        stack = [['#', 0]]
        for c in s:
            if stack[-1][0] == c:
                stack[-1][1] += 1
                if stack[-1][1] == k:
                    stack.pop()
            else:
                stack.append([c, 1])
        

        return ''.join(c*k for c, k in stack)

        
        
