What is the time complexity of Rotate String?

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

What data structures are used to solve Rotate String?

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

What are the key insights for solving Rotate String?

Insight 1: If the lengths of `s` and `goal` are different, `goal` cannot be a rotation of `s`. This is a necessary initial check. Insight 2: If `goal` is a rotation of `s`, then `goal` must be a substring of `s + s`. For instance, if `s = 'abcde'` and `goal = 'cdeab'`, then `s + s = 'abcdeabcde'`, and `goal` is a substring of it.

Which companies ask LeetCode 796. Rotate String in interviews?

This problem is frequently asked at Amdocs, Rivian, SAP, Visa, Wells Fargo and 1 more companies.

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Rotate String - LeetCode 796 Solution

Q: How to solve Rotate String?

The solution checks if the lengths of the strings `s` and `goal` are equal. If not, it returns `false`. Otherwise, it concatenates `s` with itself (i.e., creates `s + s`). It then checks if `goal` is a substring of the concatenated string. If it is, then `goal` is a rotation of `s`, and the function returns `true`. Otherwise, it returns `false`.

Rotate String - Complete Solution Guide

Rotate String is LeetCode problem 796, 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 two strings s and goal , return true if and only if s can become goal after some number of shifts on s . A shift on s consists of moving the leftmost character of s to the rightmost position. For example, if s = "abcde" , then it will be "bcdea" after one shift. Example 1: Input: s = "abcde", goal = "cdeab" Output: true Example 2: Input: s = "abcde", goal = "abced" Output: false Constraints: 1 <= s.length, goal.length <= 100 s and goal consist of lowercase English letters.

Detailed Explanation

The problem asks us to determine if a string `goal` can be obtained by rotating another string `s` a certain number of times. A single rotation involves moving the leftmost character of `s` to the rightmost position. The input consists of two strings, `s` and `goal`, and the output should be `true` if `goal` can be obtained by rotating `s`, and `false` otherwise. The length of both strings is between 1 and 100 characters.

Solution Approach

The solution checks if the lengths of the strings `s` and `goal` are equal. If not, it returns `false`. Otherwise, it concatenates `s` with itself (i.e., creates `s + s`). It then checks if `goal` is a substring of the concatenated string. If it is, then `goal` is a rotation of `s`, and the function returns `true`. Otherwise, it returns `false`.

Step-by-Step Algorithm

Step 1: Check if the lengths of `s` and `goal` are equal. If not, return `false`.
Step 2: Concatenate `s` with itself to create a new string `ss = s + s`.
Step 3: Check if `goal` is a substring of `ss` using the `contains` or `find` method. If it is, return `true`.
Step 4: If `goal` is not a substring of `ss`, return `false`.

Key Insights

Insight 1: If the lengths of `s` and `goal` are different, `goal` cannot be a rotation of `s`. This is a necessary initial check.
Insight 2: If `goal` is a rotation of `s`, then `goal` must be a substring of `s + s`. For instance, if `s = 'abcde'` and `goal = 'cdeab'`, then `s + s = 'abcdeabcde'`, and `goal` is a substring of it.
Insight 3: Check for edge cases like empty strings or null strings.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, String Matching.

Companies

Asked at: Amdocs, Rivian, SAP, Visa, Wells Fargo, tcs.

Rotate String | LeetCode 796

Easy

String String Matching

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

Solution Code

Available in:

class Solution {
    public boolean rotateString(String s, String goal) {
        if (s.length() != goal.length()) return false;
        return (s + s).contains(goal);
    }
}

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Problem Info

Description

Problem 796: Rotate String - Detailed explanation and solution approaches.

Topics

String String Matching

Companies

AmdocsRivianSAPVisaWells Fargotcs

Stats

Difficulty:Easy

Languages:4

Companies:6

Problem #:796

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: Check if the lengths of `s` and `goal` are equal. If not, return `false`.
2
Step 2: Concatenate `s` with itself to create a new string `ss = s + s`.
3
Step 3: Check if `goal` is a substring of `ss` using the `contains` or `find` method. If it is, return `true`.
4
Step 4: If `goal` is not a substring of `ss`, return `false`.

💡

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 is dominated by the string concatenation `s + s` and the substring search `(s + s).contains(goal)`. Both operations take O(n) time. The space complexity is O(n) because a new string `s + s` of length 2n is created.

📖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 Rotation

Iteratively rotate string `s` and compare it with `goal` after each rotation.

Time: O(n^2), where n is the length of the strings.Space: O(n) or O(1) depending on implementation details.

Trade-offs: This approach has a higher time complexity than the main solution (O(n^2) vs O(n)). However, it might use less space (O(1) if the rotation is done in-place) if in-place rotation is possible, at the expense of making the code harder to read.

KMP Algorithm

Use the Knuth-Morris-Pratt (KMP) algorithm for substring search. Concatenate `s` with itself as before. Then use KMP to search for `goal` within `s+s`.

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

Trade-offs: The KMP algorithm guarantees O(n) time complexity for substring search. This is not necessarily better than the standard substring search implemented in many languages. The code is more complex.

✨

Key Insights

1Insight 1: If the lengths of `s` and `goal` are different, `goal` cannot be a rotation of `s`. This is a necessary initial check.
2Insight 2: If `goal` is a rotation of `s`, then `goal` must be a substring of `s + s`. For instance, if `s = 'abcde'` and `goal = 'cdeab'`, then `s + s = 'abcdeabcde'`, and `goal` is a substring of it.
3Insight 3: Check for edge cases like empty strings or null strings.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: `s` and `goal` are empty strings. The code should handle this case gracefully (most languages do by default with the `contains` method) and return `true` if both are empty and `false` if one is empty and the other is not.
Edge case 2: `s` and `goal` are equal but not empty. The concatenation and substring check will correctly identify that goal is a rotation of s.
Edge case 3: When `s` and `goal` are very large, the concatenation `s + s` might cause memory issues. However, given the problem constraints (length <= 100), this is unlikely to be a problem.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking if the lengths of `s` and `goal` are equal before proceeding. This can lead to incorrect results and unnecessary computation.

✗Common mistake 2: Failing to consider the edge case where either string is empty or null.

✗Common mistake 3: Overlooking the fact that `s + s` contains all possible rotations of `s`.

✅

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 editors or IDEs might use string rotation detection to implement features like code completion or refactoring. For example, identifying rotated function names or variable names.

🏦Real-world scenario 2: Data encryption or encoding schemes might use string rotation as a simple transformation. Though it's not a strong encryption method, it demonstrates a basic rotation concept that is useful in cryptography.

🎮Real-world scenario 3: Circular buffer implementations. String rotation conceptually mirrors how data wraps around in a circular buffer or queue. The ability to identify rotations is useful in reasoning about the state of data within such structures.

💡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: August 16, 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

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

@AlgoMasterJun 5

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProMay 29

❓

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 "Rotate String"?

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

🤖 Need help with this problem?