What is the time complexity of Defanging an IP Address?

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

What data structures are used to solve Defanging an IP Address?

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

What are the key insights for solving Defanging an IP Address?

**Direct String Substitution with `replace`**: The most elegant and efficient way to solve this problem is to utilize the native `replace` function available in virtually all programming languages. This function directly addresses the problem's core requirement: finding all occurrences of a specific character ('.') and substituting them with a new string ('[.]'). **Understanding String Immutability**: In many languages (like Python and Java), strings are immutable. The `replace` method does not modify the original `address` string in place; instead, it returns a *new* string containing the defanged result. This is a crucial concept to grasp for preventing unexpected behavior when working with string operations.

Which companies ask LeetCode 1108. Defanging an IP Address in interviews?

This problem is frequently asked at Robinhood.

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Defanging an IP Address - LeetCode 1108 Solution

Defanging an IP Address - Complete Solution Guide

Defanging an IP Address is LeetCode problem 1108, 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

Given a valid (IPv4) IP address , return a defanged version of that IP address. A defanged IP address replaces every period "." with "[.]" . Example 1: Input: address = "1.1.1.1" Output: "1[.]1[.]1[.]1" Example 2: Input: address = "255.100.50.0" Output: "255[.]100[.]50[.]0" Constraints: The given address is a valid IPv4 address.

Detailed Explanation

This problem asks us to take a given IPv4 address, provided as a string, and transform it into a 'defanged' version. The rule for defanging is quite specific: every period ('.') in the original address must be replaced with the sequence '[.]'. The problem statement explicitly guarantees that the input `address` will always be a valid IPv4 address, which is a helpful constraint – it means we don't need to write any validation logic for the IP format itself; we can simply focus on the string transformation. For instance, if we're given "1.1.1.1", the expected output is "1[.]1[.]1[.]1". Similarly, "255.100.50.0" should become "255[.]100[.]50[.]0". What makes this problem a good starting point for string manipulation is its clear, unambiguous requirement and the direct applicability of fundamental string operations. While seemingly trivial, correctly handling string transformations, especially when the replacement sequence differs in length from the target character, is a common task in parsing and data processing.

Solution Approach

The most straightforward and Pythonic (or idiomatic in many languages) approach for this problem is to leverage the built-in string `replace` method. This method is specifically designed to find all occurrences of a specified substring within a string and substitute them with another specified substring. In our case, we're replacing every instance of `.` with `[.]`. Most modern languages, including Python, Java, and C++, offer equivalent high-level string functions that abstract away the manual iteration and string building process. This approach works because string `replace` functions are typically optimized to handle such transformations efficiently. Instead of manually looping through the string, checking each character, and conditionally appending to a new string builder or list, we delegate this task to a highly tuned library function. This not only results in more concise and readable code but also generally performs better as these built-in methods are often implemented in lower-level languages (like C) and can manage memory allocations and character copying more effectively than a custom loop.

Step-by-Step Algorithm

Step 1: Obtain the input IPv4 address string.
Step 2: (Built-in function approach): Use the built-in string replace function to replace all occurrences of '.' with '[.]'.
Step 3: (Manual approach): Iterate through the input string. If a character is '.', append "[.]" to the result; otherwise, append the character itself.
Step 4: (Manual approach): Return the newly created defanged string. Remember to allocate sufficient memory for the result string (crucial in C).

Key Insights

**Direct String Substitution with `replace`**: The most elegant and efficient way to solve this problem is to utilize the native `replace` function available in virtually all programming languages. This function directly addresses the problem's core requirement: finding all occurrences of a specific character ('.') and substituting them with a new string ('[.]').
**Understanding String Immutability**: In many languages (like Python and Java), strings are immutable. The `replace` method does not modify the original `address` string in place; instead, it returns a *new* string containing the defanged result. This is a crucial concept to grasp for preventing unexpected behavior when working with string operations.
**Leveraging Optimized Built-in Functions**: Relying on built-in string methods like `replace` is generally superior to manual iteration and string construction. These library functions are highly optimized for performance (often implemented in C or assembly), handling memory allocations and character manipulations much more efficiently than a custom, character-by-character loop, especially for larger strings.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String.

Companies

Asked at: Robinhood.

Defanging an IP Address | LeetCode 1108

Easy

String

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

Solution Code

Available in:

class Solution {
    public String defangIPaddr(String address) {
        return address.replace(".", "[.]");
    }
}

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Problem Info

Description

Problem 1108: Defanging an IP Address - Detailed explanation and solution approaches.

Topics

String

Companies

Robinhood

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1108

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

4 steps

1
Step 1: Obtain the input IPv4 address string.
2
Step 2: (Built-in function approach): Use the built-in string replace function to replace all occurrences of '.' with '[.]'.
3
Step 3: (Manual approach): Iterate through the input string. If a character is '.', append "[.]" to the result; otherwise, append the character itself.
4
Step 4: (Manual approach): Return the newly created defanged string. Remember to allocate sufficient memory for the result string (crucial in C).

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity comes from iterating through the string, either implicitly (built-in function) or explicitly (manual loop). The space complexity arises from creating a new string to store the result, which can be as long as the original string, plus additional characters in case of replacements in the manual approach.

📖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

Regular Expression

Use a regular expression to replace all occurrences of '.' with '[.]'. This is a concise solution but might be less efficient than the built-in replace function for very large strings.

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

Trade-offs: Regular expressions provide a powerful way to handle pattern matching but can be less efficient and harder to read than simple string replacement for this specific task.

✨

Key Insights

1**Direct String Substitution with `replace`**: The most elegant and efficient way to solve this problem is to utilize the native `replace` function available in virtually all programming languages. This function directly addresses the problem's core requirement: finding all occurrences of a specific character ('.') and substituting them with a new string ('[.]').
2**Understanding String Immutability**: In many languages (like Python and Java), strings are immutable. The `replace` method does not modify the original `address` string in place; instead, it returns a *new* string containing the defanged result. This is a crucial concept to grasp for preventing unexpected behavior when working with string operations.
3**Leveraging Optimized Built-in Functions**: Relying on built-in string methods like `replace` is generally superior to manual iteration and string construction. These library functions are highly optimized for performance (often implemented in C or assembly), handling memory allocations and character manipulations much more efficiently than a custom, character-by-character loop, especially for larger strings.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: An IP address with no periods. The solution will still work correctly and return the original string.
Edge case 2: An IP address consisting only of periods. The solution will correctly replace each period with '[.]'.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle memory allocation and deallocation correctly when implementing the manual solution in C (leading to memory leaks).

✗Common mistake 2: In languages without built-in replace functionalities, not accounting for the increased length of the string after replacing '.' with '[.]' (leading to buffer overflows).

✅

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: Data sanitization and escaping special characters in IP addresses before storing or displaying them in a context where '.' might have special meaning (e.g., URLs).

🏦Real-world scenario 2: Network programming where IP addresses need to be processed and manipulated in a consistent and safe way.

💡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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ 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: June 29, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

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

@TechLeadJul 7

❓

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.

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 "Defanging an IP Address"?

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 standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?