3174 Clear Digits

The idea here is pretty straight forward - once we put every alphabet in the stack, when we come across a numeric value - pop from the stack.

Code

Python3

    def clearDigits(self, s: str) -> str:
        if not s:
            return ''
 
        st = []
 
        st.append(s[0])
 
        p = 1
        while p < len(s):
            if s[p].isnumeric():
                st.pop()
            else:
                st.append(s[p])
            p += 1
 
        return ''.join(st)

Big O Analysis

• Runtime

  The runtime complexity here is $O(N)$ since we are visiting all elements in the array only once.

• Memory

  The memory usage is $O(N)$ since we are using a stack.

— A

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