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NeetCode #368LC-3174EasyStack
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#368 · #3174 · Clear Digits(清除数字)

📌 Problem Statement & Constraints

You are given a string s. Repeatedly remove every digit and the closest non-digit character to its left, until no digits remain. Return the resulting string. Constraints: 1 <= s.length <= 100; the input guarantees the operation is always possible.

💡 Core Algorithmic Approaches

  1. This is Removing Stars From a String with digits in place of stars.
  2. Push non-digit characters; on a digit, pop the top.
  3. The guarantee that the operation is always possible means the stack is never empty when a digit arrives.
  4. The result is the stack joined.

💻 Benchmark Python3 Implementation

class Solution:
    def clearDigits(self, s: str) -> str:
        st = []
        for ch in s:
            if ch.isdigit():
                st.pop()                   # remove the closest non-digit to the left
            else:
                st.append(ch)
        return "".join(st)

⚡ Complexity Deep Dive

⏱️ Time Complexity
O(n): each character is pushed or popped once.
💾 Space Complexity
O(n) for the stack.

⚠️ Interview Pitfalls & Follow-ups

  • Using ch in '0123456789': isdigit() is clearer and handles any digit-like character.
  • Forgetting that the digit itself is also removed: it is never pushed, so this is automatic.
  • Scanning backwards: also valid, but the forward stack is the natural model.
  • Assuming the stack cannot be empty: the problem guarantees safety, but a defensive check is cheap.