What is the time complexity of Clear Digits?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Clear Digits?

This problem uses String, Stack, Simulation. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Clear Digits?

Insight 1: Iterative approach: The problem requires repeated removal of digits and their corresponding leftmost non-digit characters, necessitating an iterative solution. Insight 2: Two-pointer approach (or reverse iteration): Efficiently finding the closest non-digit to the left of a digit requires a search from right to left, this can be achieved using a second pointer.

Which companies ask LeetCode 3174. Clear Digits in interviews?

This problem is frequently asked at Flexera.

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Clear Digits - LeetCode 3174 Solution

Clear Digits - Complete Solution Guide

Clear Digits is LeetCode problem 3174, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a string s . Your task is to remove all digits by doing this operation repeatedly: Delete the first digit and the closest non-digit character to its left . Return the resulting string after removing all digits. Note that the operation cannot be performed on a digit that does not have any non-digit character to its left. Example 1: Input: s = "abc" Output: "abc" Explanation: There is no digit in the string. Example 2: Input: s = "cb34" Output: "" Explanation: First, we apply the ope

Detailed Explanation

The problem asks you to remove all digits from a given string. The removal isn't straightforward; you must iteratively delete the first digit encountered and the closest non-digit character to its left. If a digit doesn't have a non-digit character to its left, it's not removed. The process continues until all digits are eliminated. The input is a string containing only lowercase English letters and digits. The output is the remaining string after all digits are removed.

Solution Approach

The provided solutions utilize an iterative approach. They scan the string from left to right. When a digit is found, they search backward for the nearest non-digit character. Once found, both the digit and the non-digit character are removed. The process repeats until no more digits remain. The Python and Java solutions use in-place modification for efficiency, while the C++ solution modifies the string directly using `erase()`, which is also relatively efficient.

Step-by-Step Algorithm

Step 1: Initialize a pointer (or index) `i` to 0 to iterate through the string.
Step 2: If `s[i]` is a digit, search backward from `i-1` to find the nearest non-digit character. If found, remove both `s[i]` (the digit) and the found non-digit character.
Step 3: If no non-digit character is found to the left of the digit, proceed to the next character.
Step 4: If `s[i]` is not a digit, proceed to the next character.
Step 5: Repeat steps 2-4 until the string contains no more digits.
Step 6: Return the modified string.

Key Insights

Insight 1: Iterative approach: The problem requires repeated removal of digits and their corresponding leftmost non-digit characters, necessitating an iterative solution.
Insight 2: Two-pointer approach (or reverse iteration): Efficiently finding the closest non-digit to the left of a digit requires a search from right to left, this can be achieved using a second pointer.
Insight 3: In-place modification (or StringBuilder): To avoid unnecessary string creation, the solution should modify the string in-place (as in Python's list manipulation or Java's StringBuilder). Otherwise, creating new strings in each iteration would lead to performance degradation.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: String, Stack, Simulation.

Companies

Asked at: Flexera.

Clear Digits | LeetCode 3174

Easy

String Stack Simulation

Time: O(n^2)Space: O(1)

Solution Code

Available in:

class Solution {
    public String clearDigits(String s) {
        StringBuilder sb = new StringBuilder(s);
        while (true) {
            boolean removed = false;
            for (int i = 0; i < sb.length(); i++) {
                if (Character.isDigit(sb.charAt(i))) {
                    boolean foundLeft = false;
                    for (int j = i - 1; j >= 0; j--) {
                        if (!Character.isDigit(sb.charAt(j))) {
                            sb.deleteCharAt(i);
                            sb.deleteCharAt(j);
                            removed = true;
                            foundLeft = true;
                            break;
                        }
                    }
                    if (foundLeft) {
                        break;
                    }
                }
            }
            if (!removed) {
                break;
            }
        }
        return sb.toString();
    }
}

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Problem Info

Description

Problem 3174: Clear Digits - Detailed explanation and solution approaches.

Topics

String Stack Simulation

Companies

Flexera

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:3174

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize a pointer (or index) `i` to 0 to iterate through the string.
2
Step 2: If `s[i]` is a digit, search backward from `i-1` to find the nearest non-digit character. If found, remove both `s[i]` (the digit) and the found non-digit character.
3
Step 3: If no non-digit character is found to the left of the digit, proceed to the next character.
4
Step 4: If `s[i]` is not a digit, proceed to the next character.
5
Step 5: Repeat steps 2-4 until the string contains no more digits.
6
Step 6: Return the modified string.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The outer loop iterates through the string at most once for each character (O(n)). The inner loop, however, searches backward for the closest non-digit. In the worst-case scenario (e.g., a string with alternating digits and non-digits), the inner loop's complexity is O(n). This nested loop structure results in a time complexity of O(n^2). The space complexity is O(1) for the in-place solutions in Python, Java, and C++, because they modify the existing string directly. The C solution uses a stack with space proportional to the string length in the worst case, leading to O(n) space complexity.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Stack-based approach (C solution)

Use a stack to process the string. Push non-digits onto the stack; ignore digits, unless a non-digit is on the stack, in which case pop one element. Finally, return the stack content.

Time: O(n)Space: O(n)

Trade-offs: This approach offers a better time complexity of O(n), but at the cost of using O(n) space for the stack. It's preferable to the in-place solution if space isn't a significant constraint, or you prioritize improved time efficiency.

✨

Key Insights

1Insight 1: Iterative approach: The problem requires repeated removal of digits and their corresponding leftmost non-digit characters, necessitating an iterative solution.
2Insight 2: Two-pointer approach (or reverse iteration): Efficiently finding the closest non-digit to the left of a digit requires a search from right to left, this can be achieved using a second pointer.
3Insight 3: In-place modification (or StringBuilder): To avoid unnecessary string creation, the solution should modify the string in-place (as in Python's list manipulation or Java's StringBuilder). Otherwise, creating new strings in each iteration would lead to performance degradation.

📚

Key Terms Explained

For Students

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty string: The solution should handle an empty string gracefully. The given solutions correctly handle this case by returning an empty string.
Edge case 2: String with no digits: The solution should also handle cases where the input string contains only letters. All solutions correctly handle this, returning the original string unchanged.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect handling of edge cases: Failing to handle the empty string or a string with no digits can lead to runtime errors or incorrect output. Careful testing is crucial.

✗Common mistake 2: Inefficient searching for the closest non-digit: Using an inefficient search algorithm (like a naive linear scan) within the nested loop can drastically increase the runtime. This should be solved by going left-to-right for the digit and right-to-left for the non-digit.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Text preprocessing: This type of algorithm could be used as a preprocessing step for text analysis or natural language processing tasks where digits are considered noise and need to be removed in a specific way.

🏦Real-world scenario 2: Data cleaning: In data cleaning pipelines, such removal logic might be used to sanitize data before further analysis or processing, focusing on preserving the information around numerical data. For example, you may want to remove the `3` and `b` from `abc3def`.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 23, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

What problems are best solved with a stack?

Stacks excel at: matching problems (parentheses validation), next greater/smaller element, expression evaluation, backtracking, and undo operations. The key pattern is LIFO (Last In, First Out) - when you need to access the most recently added element.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Clear Digits"?

For this easy-level String problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether stack techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?