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Leetcode #2390: Removing Stars From a String

In this guide, we solve Leetcode #2390 Removing Stars From a String in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

You are given a string s, which contains stars *. In one operation, you can: Choose a star in s.

Quick Facts

  • Difficulty: Medium
  • Premium: No
  • Tags: Stack, String, Simulation

Intuition

The problem has a natural nested or last-in-first-out structure.

A stack lets us resolve matches in the correct order as we scan.

Approach

Push items as they appear and pop when you can finalize a decision.

The stack captures the unresolved part of the input.

Steps:

  • Push elements as you scan.
  • Pop when a rule or match is satisfied.
  • Use the stack to compute results.

Example

Input: s = "leet**cod*e" Output: "lecoe" Explanation: Performing the removals from left to right: - The closest character to the 1st star is 't' in "leet**cod*e". s becomes "lee*cod*e". - The closest character to the 2nd star is 'e' in "lee*cod*e". s becomes "lecod*e". - The closest character to the 3rd star is 'd' in "lecod*e". s becomes "lecoe". There are no more stars, so we return "lecoe".

Python Solution

class Solution: def removeStars(self, s: str) -> str: ans = [] for c in s: if c == '*': ans.pop() else: ans.append(c) return ''.join(ans)

Complexity

The time complexity is O(n)O(n)O(n), where nnn is the length of the string sss. The space complexity is O(1)O(1)O(1).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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