This problem is similar to 844. Backspace String Compare. We’re given a string and asked to remove all asterisks and adjacent characters. We return the string after all asterisks are removed.
The solution is as follows:
class Solution:
def removeStars(self, s: str) -> str:
ans = []
for c in s:
if c == '*' and ans:
ans.pop()
else:
ans.append(c)
return "".join(ans)
_ Time Complexity:
O(n) - We inspect all characters in the string.
_ Space Complexity:
O(n) - We maintain a stack to create the result.