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:
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 <= 105
-
2 <= k <= 104
-
s
only contains lower case English letters.
SOLUTION:
class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
stack = []
for c in s:
if len(stack) == 0 or c != stack[-1][0]:
stack.append([c, 1])
else:
stack[-1][1] += 1
if stack[-1][1] == k:
stack.pop()
return "".join([c[0] * c[1] for c in stack])
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