LeetCode 3174. Clear Digits C++ Solution & Guide

Featured Summary: To efficiently solve LeetCode 3174, "Clear Digits", a stack-based algorithm is the most optimal strategy. By iterating through the input string and using a stack to push letters and pop them whenever a digit is encountered, we effectively remove each digit alongside the closest non-digit character to its left in O(n) linear time. This technique ensures that we manage state dynamically, achieving optimal performance with minimal computational overhead.

Introduction to the Clear Digits Problem

In the rapidly evolving realm of algorithmic string manipulation, understanding how to efficiently handle dynamic removals is a critical and foundational skill for any software engineer. The "Clear Digits" problem, widely known as LeetCode 3174, presents a fascinating challenge that tests your ability to manage state, positional relationships between characters, and optimize algorithmic complexity. In this comprehensive guide, we provide a deep dive into the problem mechanics, carefully analyze potential edge cases, and present a highly optimized C++ solution that demonstrates expert-level application of the stack data structure.

Whether you are diligently preparing for technical coding interviews at top tech companies or simply looking to refine and expand your problem-solving toolkit, mastering this specific pattern will significantly enhance your overall comprehension of string processing algorithms. The problem text guarantees that for every digit present in the string, there will always be a preceding non-digit character available to delete, making it structurally sound for immediate stack resolution without requiring complex backtracking.

Detailed 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.

Example 1:

Input: s = "abc"
Output: "abc"
Explanation: There is no digit in the string, so no operations are needed.

Example 2:

Input: s = "cb34"
Output: ""
Explanation:
First, we apply the operation on s[2], and s becomes "c4".
Then we apply the operation on s[1], and s becomes "".

Constraints:

• 1 <= s.length <= 100
• s consists only of lowercase English letters and digits.
• The input is generated such that it is possible to delete all digits.

Optimized C++ Implementation Analysis

To effectively and elegantly solve this problem, we rely on a standard library stack or a string mimicking a stack. Below is the complete C++ source code utilizing this data structure. This solution safely processes every single character sequentially, successfully eliminating digits and their paired characters with O(1) time overhead per individual step.

    string clearDigits(string s) {
        stack<char> sc;
        string result = "";
        for( char c : s ){
            if( c >= 48 && c <=57 ){
                sc.pop();
            } else
                sc.push(c);
        }
        while( !sc.empty() ){
            result+= sc.top();
            sc.pop();
        }
        reverse( result.begin(), result.end());
        return result;
    }
    

As you review the code above, notice the utilization of the ASCII values directly (where 48 corresponds to '0' and 57 corresponds to '9'). While you could use built-in functions like isdigit(), explicit comparisons provide absolute clarity about the character bounds being checked. After pushing all appropriate characters, we sequentially extract them to form the resulting string. Because a stack operates strictly on a Last-In-First-Out basis, we must finally apply a string reversal to restore the chronological sequence of characters.

Algorithm Time and Space Complexity

Understanding the exact resource consumption of an algorithm is indispensable. Let us break down the complexity of our stack-based solution:

• Time Complexity: O(n), where n represents the total number of characters in the input string s. The algorithm iterates over the string precisely one time. The push and pop operations associated with the stack execute in constant O(1) time, ensuring excellent runtime efficiency even for maximum input sizes.
• Space Complexity: O(n). In the absolute worst-case scenario where there are absolutely no digits present in the entire string, the stack will store all n characters. Hence, the memory footprint scales linearly in direct proportion to the input string length.

Frequently Asked Questions (FAQ)

What is the most optimal way to solve LeetCode 3174 Clear Digits?

The optimal approach uses a stack data structure to iterate through the string. When a digit is encountered, the top character (closest non-digit to the left) is popped from the stack. Otherwise, the character is pushed onto the stack. This achieves an O(n) time complexity.

What is the time complexity of the Clear Digits solution?

The time complexity is O(n), where n is the length of the string. We process each character exactly once and stack operations (push and pop) take O(1) time.

Can we solve the Clear Digits problem without extra space?

Yes, it can be solved in O(1) auxiliary space (excluding the output string) by simulating the stack behavior in-place using a two-pointer approach, overwriting characters in the original string.

Conclusion

Mastering string manipulation problems like LeetCode 3174 Clear Digits empowers developers to write cleaner, more efficient, and highly scalable code. Understanding space and time trade-offs is an essential component of advancing your algorithmic prowess. By carefully leveraging the Last-In-First-Out (LIFO) property of a standard stack, we gracefully track and pair characters for deletion. Keep continuously practicing these core algorithmic patterns, and you will inevitably find similar problems involving parentheses matching, directory path simplification, and adjacent character removal intuitively simple and straightforward to resolve. Happy coding!