How do you solve Find Invalid IP Addresses on LeetCode?

Find Invalid IP Addresses (LeetCode #3451) 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: Leading zeros invalidate an IP.

What is the time complexity of Find Invalid IP Addresses?

The optimal solution for Find Invalid IP Addresses runs in O(n) time and uses O(n) space. Single pass through IPs makes it linear; space for storing results.

What is the brute force solution for Find Invalid IP Addresses?

The brute force approach for Find Invalid IP Addresses is Brute Force, with O(n²) time complexity and O(1) space complexity. Check each IP address against the invalid conditions one by one. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Invalid IP Addresses?

Find Invalid IP Addresses uses the following concepts: Database. The recommended patterns to study are: String Manipulation, Regular Expressions.

What are the interview tips for Find Invalid IP Addresses?

Explain your thought process clearly. Use examples to illustrate your points. Practice coding in real-time.

What are common mistakes when solving Find Invalid IP Addresses?

Not checking for leading zeros. Counting octets incorrectly.

#3451

Find Invalid IP Addresses

Hard

DatabaseString ManipulationRegular Expressions

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Use a single pass to validate each IP address, leveraging string operations for efficiency.

⚙️

Algorithm

3 steps

1Step 1: Split the IP address by '.' and check the number of segments.
2Step 2: Validate each segment: check if it's a digit, within range, and has no leading zeros.
3Step 3: Count invalid IPs and order the results.

solution.py1 lines

1SELECT ip, COUNT(*) AS invalid_count FROM logs WHERE (LENGTH(ip) - LENGTH(REPLACE(ip, '.', '')) != 3 OR ip REGEXP '(^|\.)0[0-9]+\.' OR ip REGEXP '\\b[0-9]{3,}\\b' OR ip REGEXP '^[^0-9]|[^0-9]$') GROUP BY ip ORDER BY invalid_count DESC, ip DESC;

ℹ

Complexity note: Single pass through IPs makes it linear; space for storing results.

1Leading zeros invalidate an IP.
2Each octet must be between 0 and 255.

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