Problem description:

You are given a string s and an integer k, 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:

1
2
3
Input: s = "abcd", k = 2
Output: "abcd"
Explanation:There's nothing to delete.

Example 2:

1
2
3
4
5
6
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:

1
2
Input: s = "pbbcggttciiippooaais", k = 2
Output: "ps"

Constraints:

  • 1 <= s.length <= 105
  • 2 <= k <= 104
  • s only contains lower case English letters.

Solution:

Use stack

1
2
3
4
5
6
7
8
9
10
11
class Solution:
    def removeDuplicates(self, s: str, k: int) -> str:
        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)

time complexity: $O()$
space complexity: $O()$
reference:
related problem: