How do you solve Clear Digits on LeetCode?

Clear Digits (LeetCode #3174) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a stack allows for efficient removal of characters without multiple scans.

What is the brute force solution for Clear Digits?

The brute force approach for Clear Digits is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves repeatedly scanning the string to find digits and their closest non-digit characters. This is simple but inefficient as it may require multiple passes through the string. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Clear Digits?

Clear Digits uses the following concepts: String, Stack, Simulation. The recommended patterns to study are: Stack, String Manipulation.

What are the interview tips for Clear Digits?

Explain your thought process clearly as you work through the problem. Consider edge cases and how your solution handles them. Practice similar problems to become comfortable with different data structures.

What are common mistakes when solving Clear Digits?

Not considering edge cases where digits are at the start or end of the string. Overcomplicating the logic instead of using a straightforward data structure like a stack.

#3174

Clear Digits

Easy

StringStackSimulationStackString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a stack to efficiently keep track of non-digit characters. When a digit is encountered, we pop from the stack to remove the last non-digit character, achieving the desired result in a single pass.

⚙️

Algorithm

5 steps

1Step 1: Initialize an empty stack.
2Step 2: Iterate through each character in the string.
3Step 3: If the character is a digit, pop the top of the stack (if not empty).
4Step 4: If the character is not a digit, push it onto the stack.
5Step 5: After processing all characters, the stack contains the result. Convert it back to a string.

solution.py10 lines

1s = 'cb34'
2stack = []
3for char in s:
4    if char.isdigit():
5        if stack:
6            stack.pop()
7    else:
8        stack.append(char)
9result = ''.join(stack)
10print(result)

ℹ

Complexity note: This complexity is linear because we make a single pass through the string, and the stack can grow to the size of the input string in the worst case.

1Using a stack allows for efficient removal of characters without multiple scans.
2Understanding how to manipulate data structures like stacks is crucial for solving string problems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.