What is the time complexity of Remove All Occurrences of a Substring?

The optimal solution has a time complexity of O(n*m).

What data structures are used to solve Remove All Occurrences of a Substring?

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

What are the key insights for solving Remove All Occurrences of a Substring?

Insight 1: The core idea is to repeatedly find and remove the leftmost occurrence of `part` from `s` until no more occurrences exist. Insight 2: The problem can be efficiently solved using string manipulation functions like `replaceFirst` (Java), `replace` (Python & C++), or `strstr` and manual string manipulation (C).

Which companies ask LeetCode 1910. Remove All Occurrences of a Substring in interviews?

This problem is frequently asked at Arista Networks, X, Zoho.

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Remove All Occurrences of a Substring - LeetCode 1910 Solution

Remove All Occurrences of a Substring - Complete Solution Guide

Remove All Occurrences of a Substring is LeetCode problem 1910, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given two strings s and part , perform the following operation on s until all occurrences of the substring part are removed: Find the leftmost occurrence of the substring part and remove it from s . Return s after removing all occurrences of part . A substring is a contiguous sequence of characters in a string. Example 1: Input: s = "daabcbaabcbc", part = "abc" Output: "dab" Explanation : The following operations are done: - s = "da abc baabcbc", remove "abc" starting at index 2, so s = "dabaabc

Detailed Explanation

The problem requires removing all occurrences of a substring `part` from a given string `s`. The removal should be done repeatedly, always removing the leftmost occurrence of `part` in `s`, until `part` no longer exists as a substring within `s`. The inputs are the string `s` and the string `part`. The output is the modified string `s` after all occurrences of `part` have been removed. The constraints specify the lengths of `s` and `part` can be up to 1000, and they contain only lowercase English letters.

Solution Approach

The provided solutions iteratively search for and remove the leftmost occurrence of the substring `part` from the string `s`. The `while` loop continues as long as `part` is found within `s`. Inside the loop, the leftmost occurrence of `part` is located and removed using language-specific string manipulation functions. Once the loop completes (i.e., `part` is no longer a substring of `s`), the modified `s` is returned.

Step-by-Step Algorithm

Step 1: Check if the substring `part` exists in the string `s`.
Step 2: If `part` exists in `s`, find the leftmost occurrence of `part`.
Step 3: Remove the leftmost occurrence of `part` from `s`.
Step 4: Repeat steps 1-3 until `part` no longer exists in `s`.
Step 5: Return the modified string `s`.

Key Insights

Insight 1: The core idea is to repeatedly find and remove the leftmost occurrence of `part` from `s` until no more occurrences exist.
Insight 2: The problem can be efficiently solved using string manipulation functions like `replaceFirst` (Java), `replace` (Python & C++), or `strstr` and manual string manipulation (C).
Insight 3: Consider the edge case where `part` might not exist in `s` initially, in which case the original `s` should be returned directly.

Complexity Analysis

Time Complexity: O(n*m)

Space Complexity: O(1)

Topics

This problem involves: String, Stack, Simulation.

Companies

Asked at: Arista Networks, X, Zoho.

Remove All Occurrences of a Substring | LeetCode 1910

Medium

String Stack Simulation

Time: O(n*m)Space: O(1)

Solution Code

Available in:

class Solution {
    public String removeOccurrences(String s, String part) {
        while (s.indexOf(part) >= 0) {
            s = s.replaceFirst(part, "");
        }
        return s;
    }
}

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Problem Info

Description

Problem 1910: Remove All Occurrences of a Substring - Detailed explanation and solution approaches.

Topics

String Stack Simulation

Companies

Arista NetworksXZoho

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:1910

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

5 steps

1
Step 1: Check if the substring `part` exists in the string `s`.
2
Step 2: If `part` exists in `s`, find the leftmost occurrence of `part`.
3
Step 3: Remove the leftmost occurrence of `part` from `s`.
4
Step 4: Repeat steps 1-3 until `part` no longer exists in `s`.
5
Step 5: Return the modified string `s`.

💡

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*m)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity arises from the nested iterations of searching for `part` within `s` and then removing it. The space complexity is primarily determined by whether the string manipulation is done in-place (or close to in-place) or if new strings are created during the process. In C, we allocate a new string `new_s` in each iteration. While overall it won't exceed `s` length, space complexity within the loop is still `O(n)`.

📖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

Use a stack to store the characters of `s`. When the top of the stack matches the end of `part`, pop `part.length()` characters from the stack. After processing all characters of `s`, the characters remaining in the stack form the result.

Time: O(n), where n is the length of `s`.Space: O(n), where n is the length of `s` (for the stack).

Trade-offs: The stack-based approach can achieve O(n) time complexity, which is better than the O(n*m) complexity of the original solution. However, it requires extra space for the stack. This approach is more complex to implement but provides better performance for large inputs.

✨

Key Insights

1Insight 1: The core idea is to repeatedly find and remove the leftmost occurrence of `part` from `s` until no more occurrences exist.
2Insight 2: The problem can be efficiently solved using string manipulation functions like `replaceFirst` (Java), `replace` (Python & C++), or `strstr` and manual string manipulation (C).
3Insight 3: Consider the edge case where `part` might not exist in `s` initially, in which case the original `s` should be returned directly.

📚

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: `part` is an empty string. The Python/Java/C++ solutions will result in infinite loops, the C solution leads to memory allocation issues, though constraints disallow empty string for `part`.
Edge case 2: `part` is not a substring of `s`. The algorithm will correctly terminate and return the original `s` without any modifications.
Edge case 3: `s` is an empty string. The algorithm will terminate without modifications, and the empty string `s` will be returned.
Edge case 4: `s` and `part` are the same. The algorithm will remove the single occurrence of `part` and return an empty string.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not removing only the *leftmost* occurrence of `part`. Using `replace(part, '')` in Python (without the `1` argument) removes all occurrences at once, which is incorrect.

✗Common mistake 2: Infinite loop if the `part` is derived from `s` such that removing the part also re-inserts it during replacement. Consider the case where s = 'aaaaa' and part = 'aa'. Removing 'aa' always produces an 'aa' substring.

✗Common mistake 3: Forgetting to handle memory allocation correctly in C. Failing to allocate enough memory for the new string or not freeing the old string leads to memory leaks or segmentation faults.

✅

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 editing software where you want to automatically remove specific phrases or keywords from a document.

🏦Real-world scenario 2: Data cleaning processes where you need to remove unwanted patterns or substrings from data entries before further analysis.

🎮Real-world scenario 3: Spam filtering systems that remove known spam keywords from email content.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 19, 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

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

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 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.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Remove All Occurrences of a Substring"?

For this medium-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.

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