Today’s problem

https://leetcode.com/problems/clear-digits/description/

Intuition

We need to pop the character before the digit in this question.

Approach

Therefore, all we need is just to utilize a stack.

Complexity

• Time complexity: $O(N)$, N is the length of the string.

• Space complexity: $O(N)$, N is the length of the string.

Code

class Solution:
    def clearDigits(self, s: str) -> str:
        st = []
        for ch in s:
            if '0' <= ch and ch <= '9':
                st.pop()
            else:
                st.append(ch)
        
        return ''.join(st)