What is the time complexity of Restore IP Addresses?

The optimal solution has a time complexity of O(3^n).

What data structures are used to solve Restore IP Addresses?

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

What are the key insights for solving Restore IP Addresses?

Insight 1: Backtracking is a natural fit since we need to explore different possibilities for splitting the string into four parts. Insight 2: Pruning is crucial to avoid unnecessary exploration. We can prune branches if the current part is invalid (e.g., leading zero, greater than 255) or if we've already used more than 4 parts or if the remaining string is not sufficient to form the rest of the ip address.

Which companies ask LeetCode 93. Restore IP Addresses in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Arista Networks, Autodesk and 9 more companies.

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Restore IP Addresses - LeetCode 93 Solution

Restore IP Addresses - Complete Solution Guide

Restore IP Addresses is LeetCode problem 93, 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

A valid IP address consists of exactly four integers separated by single dots. Each integer is between 0 and 255 ( inclusive ) and cannot have leading zeros. For example, "0.1.2.201" and "192.168.1.1" are valid IP addresses, but "0.011.255.245" , "192.168.1.312" and "[email protected]" are invalid IP addresses. Given a string s containing only digits, return all possible valid IP addresses that can be formed by inserting dots into s . You are not allowed to reorder or remove any digits in s . You may

Detailed Explanation

The problem asks us to take a string `s` consisting only of digits and determine all possible valid IP addresses that can be formed by inserting exactly three dots into the string. A valid IP address has four parts, each being an integer between 0 and 255 (inclusive), and no part can have leading zeros unless it is exactly zero. We are not allowed to reorder or remove any digits from the input string. The output should be a list of strings, each representing a valid IP address formed from the input string.

Solution Approach

The provided solutions use a backtracking approach to explore all possible ways to split the input string `s` into four parts. The `backtrack` function recursively explores different combinations of splitting the string by choosing the length of each part (1, 2, or 3). For each potential part, it checks if the part is a valid integer (between 0 and 255 and without leading zeros). If a valid split into four parts is found, it is added to the result list. The backtracking step involves appending the current valid part to a list representing potential valid ip address, recursively calling backtrack function with the next starting index and updated partial ip, and removing the part from the list before returning to the previous state.

Step-by-Step Algorithm

Step 1: Check if the input string length is within the valid range (4 to 12 inclusive). If not, return an empty list.
Step 2: Define a recursive `backtrack` function with arguments: `start_index` (the starting index of the current part), and `current_parts` (a list of strings representing the parts of the IP address built so far).
Step 3: In the `backtrack` function, if `current_parts` has 4 elements:
- If `start_index` has reached the end of the string, join the parts with dots and add it to the result.
- Return.
Step 4: Iterate through possible lengths (1, 2, and 3) for the next part of the IP address.
Step 5: For each length:
- Check if the current `start_index` plus the length exceeds the string length. If so, break the loop.
- Extract the substring `part` of the given length from `start_index`.
- Check for invalid parts: leading zeros (if length > 1 and `part[0]` is '0') and if the integer value of `part` exceeds 255.
- If the part is valid, add it to `current_parts`.
- Call `backtrack` recursively with the updated `start_index` and `current_parts`.
- Remove the last added part from `current_parts` (backtracking).
Step 6: Call the `backtrack` function initially with `start_index` as 0 and an empty list for `current_parts`.
Step 7: Return the result list of valid IP addresses.
Important: Free any dynamically allocated memory, especially in C, to prevent memory leaks.

Key Insights

Insight 1: Backtracking is a natural fit since we need to explore different possibilities for splitting the string into four parts.
Insight 2: Pruning is crucial to avoid unnecessary exploration. We can prune branches if the current part is invalid (e.g., leading zero, greater than 255) or if we've already used more than 4 parts or if the remaining string is not sufficient to form the rest of the ip address.
Insight 3: The length of each part of IP address can be only 1, 2 or 3. This limits the search space.

Complexity Analysis

Time Complexity: O(3^n)

Space Complexity: O(n)

Topics

This problem involves: String, Backtracking.

Companies

Asked at: Adobe, Amazon, Apple, Arista Networks, Autodesk, Cisco, Meta, Microsoft, Oracle, Palo Alto Networks, StackAdapt, TikTok, Visa, Zoho.

Restore IP Addresses | LeetCode 93

Medium

String Backtracking

Time: O(3^n)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<String> restoreIpAddresses(String s) {
        List<String> result = new ArrayList<>();
        int n = s.length();

        if (n < 4 || n > 12) {
            return result;
        }

        backtrack(0, new ArrayList<>(), s, result);
        return result;
    }

    private void backtrack(int startIndex, List<String> currentParts, String s, List<String> result) {
        if (currentParts.size() == 4) {
            if (startIndex == s.length()) {
                result.add(String.join(".", currentParts));
            }
            return;
        }

        for (int length = 1; length <= 3; length++) {
            if (startIndex + length > s.length()) {
                break;
            }

            String part = s.substring(startIndex, startIndex + length);

            if (part.length() > 1 && part.charAt(0) == '0') {
                break;
            }

            if (Integer.parseInt(part) > 255) {
                break;
            }

            currentParts.add(part);
            backtrack(startIndex + length, currentParts, s, result);
            currentParts.remove(currentParts.size() - 1);
        }
    }
}

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Problem Info

Description

Problem 93: Restore IP Addresses - Detailed explanation and solution approaches.

Topics

String Backtracking

Companies

AdobeAmazonAppleArista NetworksAutodeskCiscoMetaMicrosoft

Stats

Difficulty:Medium

Languages:4

Companies:14

Problem #:93

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

16 steps

1
Step 1: Check if the input string length is within the valid range (4 to 12 inclusive). If not, return an empty list.
2
Step 2: Define a recursive `backtrack` function with arguments: `start_index` (the starting index of the current part), and `current_parts` (a list of strings representing the parts of the IP address built so far).
3
Step 3: In the `backtrack` function, if `current_parts` has 4 elements:
4
- If `start_index` has reached the end of the string, join the parts with dots and add it to the result.
5
- Return.
6
Step 4: Iterate through possible lengths (1, 2, and 3) for the next part of the IP address.
7
Step 5: For each length:
8
- Check if the current `start_index` plus the length exceeds the string length. If so, break the loop.
9
- Extract the substring `part` of the given length from `start_index`.
10
- Check for invalid parts: leading zeros (if length > 1 and `part[0]` is '0') and if the integer value of `part` exceeds 255.
11
- If the part is valid, add it to `current_parts`.
12
- Call `backtrack` recursively with the updated `start_index` and `current_parts`.
13
- Remove the last added part from `current_parts` (backtracking).
14
Step 6: Call the `backtrack` function initially with `start_index` as 0 and an empty list for `current_parts`.
15
Step 7: Return the result list of valid IP addresses.
16
Important: Free any dynamically allocated memory, especially in C, to prevent memory leaks.

💡

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(3^n)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is exponential because we are exploring all possible combinations of splitting the string into four parts. The space complexity is primarily determined by the recursion depth, which is at most 4. In the Python version the current_parts is also stored, resulting to 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

Iterative Approach

Instead of recursion, we can use nested loops to iterate through all possible combinations of the lengths of the four parts. This will make the solution iterative rather than recursive. Though it is still a kind of brute-force way, the loops can avoid recursive calls.

Time: O(1). Essentially constant time because the loops will run a maximum of 3*3*3 times. Space: O(n) because there can be up to a linear number of potential solutions stored in the result array.

Trade-offs: Iterative approach can be slightly more efficient in terms of stack usage, but can be harder to read, especially for more complex backtracking problems.

✨

Key Insights

1Insight 1: Backtracking is a natural fit since we need to explore different possibilities for splitting the string into four parts.
2Insight 2: Pruning is crucial to avoid unnecessary exploration. We can prune branches if the current part is invalid (e.g., leading zero, greater than 255) or if we've already used more than 4 parts or if the remaining string is not sufficient to form the rest of the ip address.
3Insight 3: The length of each part of IP address can be only 1, 2 or 3. This limits the search space.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: Input string is too short (less than 4 digits) or too long (more than 12 digits). The solution correctly returns an empty list.
Edge case 2: Leading zeros in a part of the IP address. The solution correctly handles this by checking if `len(part) > 1 and part[0] == '0'`, and breaks the loop when detecting it.
Edge case 3: A part of the IP address is greater than 255. The solution correctly handles it by checking if `int(part) > 255` and breaks if detected.
Edge case 4: Input like "0000" produces ["0.0.0.0"].
Edge case 5: Input string that contains only 0, such as "010010", and various placements of numbers. All should still pass the validation for each split.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check for leading zeros. This can lead to invalid IP addresses being included in the result.

✗Common mistake 2: Not handling the case where the input string is too short or too long. This can cause index out of bounds errors or incorrect results.

✗Common mistake 3: Not correctly backtracking - forgetting to remove the last added part from `current_parts` after the recursive call. This can lead to incorrect combinations being explored.

✗Common mistake 4: Incorrectly converting the string part to an integer. Always make sure the conversion is handled correctly, and that potential exceptions are handled.

✅

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: Validating user input for IP address fields in web forms or configuration files. This algorithm can be used to ensure that the entered IP address is in a valid format.

🏦Real-world scenario 2: Analyzing network traffic logs. IP addresses in logs need to be parsed and validated, and this algorithm can be used as a component of the parsing process.

🎮Real-world scenario 3: Network configuration tools, that help visualize all the possible IP configuration for a given subnet.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: September 2, 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

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

@DevExpert92Jun 30

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How do I avoid duplicate solutions in backtracking?

Sort the input first, then skip consecutive duplicates at the same recursion level: if (i > start && nums[i] == nums[i-1]) continue. For permutations, use a visited array or swap-based approach. Always consider if order matters (permutations vs combinations).

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Restore IP Addresses"?

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

🤖 Need help with this problem?