How do you solve Defanging an IP Address on LeetCode?

Defanging an IP Address (LeetCode #1108) 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: String manipulation can often be optimized with built-in methods.

What is the brute force solution for Defanging an IP Address?

The brute force approach for Defanging an IP Address is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves iterating through each character of the IP address and building a new string. Whenever we encounter a period, we replace it with '[.]'. This is straightforward but not the most efficient way to handle string manipulation. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Defanging an IP Address?

Defanging an IP Address uses the following concepts: String. The recommended patterns to study are: String Manipulation, Regular Expressions.

What are the interview tips for Defanging an IP Address?

Always discuss your thought process and approach before coding. Consider edge cases, such as the minimum and maximum length of the input. Practice writing clean and efficient code to improve readability.

What are common mistakes when solving Defanging an IP Address?

Not considering string immutability in some languages, leading to inefficient concatenation. Overcomplicating the problem by trying to handle each character individually.

#1108

Defanging an IP Address

Easy

StringString ManipulationRegular Expressions

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal approach uses built-in string manipulation methods to replace all occurrences of '.' in a single pass. This is efficient and leverages the language's capabilities to handle string replacement directly.

⚙️

Algorithm

2 steps

1Step 1: Use the string replace method to replace all '.' with '[.]'.
2Step 2: Return the modified string.

solution.py2 lines

1def defangIPaddr(address):
2    return address.replace('.', '[.]')

ℹ

Complexity note: The time complexity is O(n) because we traverse the string once to replace the characters. The space complexity is O(n) since we create a new string for the result.

1String manipulation can often be optimized with built-in methods.
2Understanding time complexity is crucial for evaluating performance.

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