LeetCode #1071 — EASY

Greatest Common Divisor of Strings

Build confidence with an intuition-first walkthrough focused on math fundamentals.

The Problem

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: ""

Example 4:

Input: str1 = "AAAAAB", str2 = "AAA"

Output: ""

Constraints:

1 <= str1.length, str2.length <= 1000
str1 and str2 consist of English uppercase letters.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

"ABCABC"
"ABC"

Example 2

"ABABAB"
"ABAB"

Example 3

"LEET"
"CODE"

Core Insight

What unlocks the optimal approach

The greatest common divisor must be a prefix of each string, so we can try all prefixes.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1071: Greatest Common Divisor of Strings
class Solution {
    public String gcdOfStrings(String str1, String str2) {
        if (!(str1 + str2).equals(str2 + str1)) {
            return "";
        }
        int len = gcd(str1.length(), str2.length());
        return str1.substring(0, len);
    }

    private int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}

// Accepted solution for LeetCode #1071: Greatest Common Divisor of Strings
func gcdOfStrings(str1 string, str2 string) string {
	if str1+str2 != str2+str1 {
		return ""
	}
	n := gcd(len(str1), len(str2))
	return str1[:n]
}

func gcd(a, b int) int {
	if b == 0 {
		return a
	}
	return gcd(b, a%b)
}

# Accepted solution for LeetCode #1071: Greatest Common Divisor of Strings
class Solution:
    def gcdOfStrings(self, str1: str, str2: str) -> str:
        def check(a, b):
            c = ""
            while len(c) < len(b):
                c += a
            return c == b

        for i in range(min(len(str1), len(str2)), 0, -1):
            t = str1[:i]
            if check(t, str1) and check(t, str2):
                return t
        return ''

// Accepted solution for LeetCode #1071: Greatest Common Divisor of Strings
impl Solution {
    pub fn gcd_of_strings(str1: String, str2: String) -> String {
        if str1.clone() + &str2 != str2.clone() + &str1 {
            return String::from("");
        }
        fn gcd(a: usize, b: usize) -> usize {
            if b == 0 {
                return a;
            }
            gcd(b, a % b)
        }

        let (m, n) = (str1.len().max(str2.len()), str1.len().min(str2.len()));
        str1[..gcd(m, n)].to_string()
    }
}

// Accepted solution for LeetCode #1071: Greatest Common Divisor of Strings
function gcdOfStrings(str1: string, str2: string): string {
    let [len1, len2] = [str1.length, str2.length];

    function isDivisor(l: number): boolean {
        if (len1 % l || len2 % l) {
            return false;
        }

        let [f1, f2] = [Math.floor(len1 / l), Math.floor(len2 / l)];

        return (
            str1.slice(0, l).repeat(f1) == str1 &&
            str1.slice(0, l).repeat(f2) == str2
        );
    }

    for (let l = Math.min(len1, len2); l > 0; l--) {
        if (isDivisor(l)) {
            return str1.slice(0, l);
        }
    }

    return '';
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.