What is the time complexity of Greatest Common Divisor of Strings?

The optimal solution has a time complexity of O(log(min(n, m))).

What data structures are used to solve Greatest Common Divisor of Strings?

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

What are the key insights for solving Greatest Common Divisor of Strings?

**The Concatenation Commutativity Test:** A fundamental insight is that if two strings `str1` and `str2` share a common string divisor `x`, then `str1 + str2` must be identical to `str2 + str1`. This provides an efficient O(L1+L2) check to confirm the *existence* of a common divisor string before any further calculations. **Reduction to Numerical GCD:** Once the existence of a common divisor string is established, the problem of finding its *length* transforms into a standard numerical GCD problem. The length of the greatest common divisor string will always be `gcd(len(str1), len(str2))`, allowing us to leverage efficient algorithms like Euclidean algorithm.

Which companies ask LeetCode 1071. Greatest Common Divisor of Strings in interviews?

This problem is frequently asked at Datadog, Infosys, Nvidia.

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Greatest Common Divisor of Strings - LeetCode 1071 Solution

Q: How to solve Greatest Common Divisor of Strings?

The provided solution is remarkably elegant, hinging on two key insights. The first is a crucial pre-condition check: `if str1 + str2 != str2 + str1: return ""`. This condition is both necessary and sufficient for two strings to share a common repeating unit. If `str1` is `x` repeated `n` times and `str2` is `x` repeated `m` times, then `str1 + str2` will be `x` repeated `n+m` times, and `str2 + str1` will be `x` repeated `m+n` times. Since `n+m` always equals `m+n`, these concatenated strings *must* be identical. If they're not, it immediately tells us no such common `x` exists, streamlining the process significantly by handling all non-divisible cases (e.g., "ABC" and "ABA"). Once this initial check confirms that a common divisor string *can* exist, the problem simplifies. If `x` divides both `str1` and `str2`, then the length of `x` must divide both `len(str1)` and `len(str2)`. To find the *largest* such `x`, we need to find the *largest possible length* for `x`. This is precisely the greatest common divisor of the numerical lengths: `gcd(len(str1), len(str2))`. The solution leverages the standard Euclidean algorithm for this. Finally, knowing this maximum possible length, the actual common divisor string `x` is simply the prefix of `str1` (or `str2`) up to that calculated GCD length. For instance, with `str1 = "ABABAB"` and `str2 = "ABAB"`, `gcd(6, 4)` is 2, and the answer is `str1[:2]` which is "AB". This works because if `x` is the repeating unit, it logically forms the beginning of both `str1` and `str2`.

Greatest Common Divisor of Strings - Complete Solution Guide

Greatest Common Divisor of Strings is LeetCode problem 1071, 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

For two strings s and t , we say " t divides s " if and only if s = t + t + t + ... + t + t (i.e., t is concatenated with itself one or more times). Given two strings str1 and str2 , return the largest string x such that x divides both str1 and str2 . Example 1: Input: str1 = "ABCABC", str2 = "ABC" Output: "ABC" Example 2: Input: str1 = "ABABAB", str2 = "ABAB" Output: "AB" Example 3: Input: str1 = "LEET", str2 = "CODE" Output: "" Constraints: 1 <= str1.length, str2.length <= 1000 str1 and str2 c

Detailed Explanation

This problem introduces a unique definition of string division, asking us to find the "greatest common divisor" (GCD) of two strings, `str1` and `str2`. A string `t` is said to "divide" `s` if `s` can be formed by concatenating `t` with itself one or more times. Our task is to locate the *largest* string `x` that satisfies this division property for both `str1` and `str2`. This isn't about traditional substring matching; it's about identifying a common, repeating elemental string that builds up both given strings.

Solution Approach

The provided solution is remarkably elegant, hinging on two key insights. The first is a crucial pre-condition check: `if str1 + str2 != str2 + str1: return ""`. This condition is both necessary and sufficient for two strings to share a common repeating unit. If `str1` is `x` repeated `n` times and `str2` is `x` repeated `m` times, then `str1 + str2` will be `x` repeated `n+m` times, and `str2 + str1` will be `x` repeated `m+n` times. Since `n+m` always equals `m+n`, these concatenated strings *must* be identical. If they're not, it immediately tells us no such common `x` exists, streamlining the process significantly by handling all non-divisible cases (e.g., "ABC" and "ABA"). Once this initial check confirms that a common divisor string *can* exist, the problem simplifies. If `x` divides both `str1` and `str2`, then the length of `x` must divide both `len(str1)` and `len(str2)`. To find the *largest* such `x`, we need to find the *largest possible length* for `x`. This is precisely the greatest common divisor of the numerical lengths: `gcd(len(str1), len(str2))`. The solution leverages the standard Euclidean algorithm for this. Finally, knowing this maximum possible length, the actual common divisor string `x` is simply the prefix of `str1` (or `str2`) up to that calculated GCD length. For instance, with `str1 = "ABABAB"` and `str2 = "ABAB"`, `gcd(6, 4)` is 2, and the answer is `str1[:2]` which is "AB". This works because if `x` is the repeating unit, it logically forms the beginning of both `str1` and `str2`.

Step-by-Step Algorithm

Step 1: Check if `str1 + str2` is equal to `str2 + str1`. If not, return an empty string because no common divisor string exists.
Step 2: Calculate the greatest common divisor (GCD) of the lengths of `str1` and `str2` using Euclid's algorithm.
Step 3: Extract the substring of `str1` (or `str2`) from index 0 up to the GCD length calculated in Step 2.
Step 4: Return the extracted substring as the result.

Key Insights

**The Concatenation Commutativity Test:** A fundamental insight is that if two strings `str1` and `str2` share a common string divisor `x`, then `str1 + str2` must be identical to `str2 + str1`. This provides an efficient O(L1+L2) check to confirm the *existence* of a common divisor string before any further calculations.
**Reduction to Numerical GCD:** Once the existence of a common divisor string is established, the problem of finding its *length* transforms into a standard numerical GCD problem. The length of the greatest common divisor string will always be `gcd(len(str1), len(str2))`, allowing us to leverage efficient algorithms like Euclidean algorithm.
**Prefix as the Divisor String:** With the existence confirmed and the length determined via numerical GCD, the final step is straightforward: the actual greatest common divisor string is simply the prefix of `str1` (or `str2`) of that calculated GCD length. This holds true because the repeating unit forming both strings must inherently be present at their respective beginnings.

Complexity Analysis

Time Complexity: O(log(min(n, m)))

Space Complexity: O(1)

Topics

This problem involves: Math, String.

Companies

Asked at: Datadog, Infosys, Nvidia.

Greatest Common Divisor of Strings | LeetCode 1071

Easy

Math String

Time: O(log(min(n, m)))Space: O(1)

Solution Code

Available in:

class Solution {
    public String gcdOfStrings(String str1, String str2) {
        if (!(str1 + str2).equals(str2 + str1)) {
            return "";
        }
        int gcdLength = gcd(str1.length(), str2.length());
        return str1.substring(0, gcdLength);
    }

    private int gcd(int a, int b) {
        while (b != 0) {
            int temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }
}

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Problem Info

Description

Problem 1071: Greatest Common Divisor of Strings - Detailed explanation and solution approaches.

Topics

Math String

Companies

DatadogInfosysNvidia

Stats

Difficulty:Easy

Languages:4

Companies:3

Problem #:1071

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 `str1 + str2` is equal to `str2 + str1`. If not, return an empty string because no common divisor string exists.
2
Step 2: Calculate the greatest common divisor (GCD) of the lengths of `str1` and `str2` using Euclid's algorithm.
3
Step 3: Extract the substring of `str1` (or `str2`) from index 0 up to the GCD length calculated in Step 2.
4
Step 4: Return the extracted substring as the result.

💡

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(log(min(n, m)))

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The dominant operation is the GCD calculation, which takes O(log(min(m, n))) time using Euclid's algorithm. String concatenation and substring extraction take linear time in the length of the GCD which is at most min(m,n), but this is dominated by the logarithmic time of GCD. The space used is constant because only a few integer variables are used. The C solution allocates memory, but this is still considered O(1) as the allocated memory is always a small, constant size related to the GCD length and not dependent on the input string length.

📖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

Iterate through all possible substring lengths from the minimum length down to 1. Check if the substring of that length divides both input strings.

Time: O(m*n)Space: O(min(m,n))

Trade-offs: Less efficient than the GCD approach. The GCD approach is significantly faster and more optimized for this specific problem.

✨

Key Insights

1**The Concatenation Commutativity Test:** A fundamental insight is that if two strings `str1` and `str2` share a common string divisor `x`, then `str1 + str2` must be identical to `str2 + str1`. This provides an efficient O(L1+L2) check to confirm the *existence* of a common divisor string before any further calculations.
2**Reduction to Numerical GCD:** Once the existence of a common divisor string is established, the problem of finding its *length* transforms into a standard numerical GCD problem. The length of the greatest common divisor string will always be `gcd(len(str1), len(str2))`, allowing us to leverage efficient algorithms like Euclidean algorithm.
3**Prefix as the Divisor String:** With the existence confirmed and the length determined via numerical GCD, the final step is straightforward: the actual greatest common divisor string is simply the prefix of `str1` (or `str2`) of that calculated GCD length. This holds true because the repeating unit forming both strings must inherently be present at their respective beginnings.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: When `str1` and `str2` have no common divisor string. The code correctly handles this by returning an empty string when `str1 + str2 != str2 + str1`.
Edge case 2: When one string is a prefix of the other. The solution correctly handles this, returning the shorter string as the common divisor.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check if `str1 + str2 == str2 + str1` before calculating the GCD. This check is crucial to determine if a common divisor string exists.

✗Common mistake 2: Incorrectly calculating or using the GCD. Using a naive GCD calculation or misinterpreting the GCD's relation to the substring length will lead to incorrect results.

✅

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 compression and pattern recognition. Finding repeating patterns in data streams for efficient compression techniques.

🏦Real-world scenario 2: Bioinformatics. Analyzing DNA or protein sequences to identify repeated motifs or common subsequences.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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: November 19, 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.

@AlgoMasterJul 14

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

@InterviewProJul 7

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

@CodeNinjaJun 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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Greatest Common Divisor of Strings"?

For this easy-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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