1071. Greatest Common Divisor of Strings - Explanation

Description

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).

You are given two strings str1 and str2, return the largest string x such that x divides both str1 and str2. If there is no such string, return "".

Example 1:

Input: str1 = "ABAB", str2 = "AB"

Output: "AB"

Example 2:

Input: str1 = "NANANA", str2 = "NANA"

Output: "NA"

Constraints:

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

Topics

Math String

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Prerequisites

Before attempting this problem, you should be comfortable with:

GCD (Greatest Common Divisor) - Understanding and implementing the Euclidean algorithm
String Manipulation - Working with substrings, concatenation, and repetition
Modular Arithmetic - Using modulo operations for cyclic indexing and divisibility checks

1. Iteration

Intuition

A string t divides both str1 and str2 only if its length divides both string lengths and repeating t the appropriate number of times reconstructs each string. We search for the longest such divisor by checking all possible lengths from the minimum length down to 1.

Algorithm

For each possible length l from min(len1, len2) down to 1:
- Check if l divides both len1 and len2.
- Extract the prefix str1[:l] as the candidate divisor.
- Check if repeating this prefix len1/l times equals str1 and len2/l times equals str2.
Return the first valid prefix found.
If no valid divisor exists, return an empty string.

class Solution:
    def gcdOfStrings(self, str1: str, str2: str) -> str:
        len1, len2 = len(str1), len(str2)

        def isDivisor(l):
            if len1 % l != 0 or len2 % l != 0:
                return False
            f1, f2 = len1 // l, len2 // l
            return str1[:l] * f1 == str1 and str1[:l] * f2 == str2

        for l in range(min(len1, len2), 0, -1):
            if isDivisor(l):
                return str1[:l]

        return ""

public class Solution {
    public String gcdOfStrings(String str1, String str2) {
        int len1 = str1.length(), len2 = str2.length();

        for (int l = Math.min(len1, len2); l > 0; l--) {
            if (isDivisor(l, len1, len2, str1, str2)) {
                return str1.substring(0, l);
            }
        }

        return "";
    }

    public boolean isDivisor(int l, int len1, int len2, String str1, String str2) {
        if (len1 % l != 0 || len2 % l != 0) {
            return false;
        }
        String sub = str1.substring(0, l);
        int f1 = len1 / l, f2 = len2 / l;
        return sub.repeat(f1).equals(str1) && sub.repeat(f2).equals(str2);
    }
}

class Solution {
public:
    string gcdOfStrings(string str1, string str2) {
        int len1 = str1.size(), len2 = str2.size();

        auto isDivisor = [&](int l) {
            if (len1 % l != 0 || len2 % l != 0) {
                return false;
            }
            string sub = str1.substr(0, l);
            int f1 = len1 / l, f2 = len2 / l;
            string repeated1 = "", repeated2 = "";
            for (int i = 0; i < f1; ++i) repeated1 += sub;
            for (int i = 0; i < f2; ++i) repeated2 += sub;
            return repeated1 == str1 && repeated2 == str2;
        };

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

        return "";
    }
};

class Solution {
    /**
     * @param {string} str1
     * @param {string} str2
     * @return {string}
     */
    gcdOfStrings(str1, str2) {
        const len1 = str1.length,
            len2 = str2.length;

        const isDivisor = (l) => {
            if (len1 % l !== 0 || len2 % l !== 0) {
                return false;
            }
            const sub = str1.slice(0, l);
            const f1 = len1 / l,
                f2 = len2 / l;
            return sub.repeat(f1) === str1 && sub.repeat(f2) === str2;
        };

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

        return '';
    }
}

public class Solution {
    public string GcdOfStrings(string str1, string str2) {
        int len1 = str1.Length, len2 = str2.Length;

        bool IsDivisor(int l) {
            if (len1 % l != 0 || len2 % l != 0) return false;

            int f1 = len1 / l, f2 = len2 / l;
            string baseStr = str1.Substring(0, l);
            return new StringBuilder().Insert(0, baseStr, f1).ToString() == str1 &&
                   new StringBuilder().Insert(0, baseStr, f2).ToString() == str2;
        }

        for (int l = Math.Min(len1, len2); l >= 1; l--) {
            if (IsDivisor(l)) {
                return str1.Substring(0, l);
            }
        }

        return "";
    }
}

func gcdOfStrings(str1 string, str2 string) string {
    len1, len2 := len(str1), len(str2)

    isDivisor := func(l int) bool {
        if len1%l != 0 || len2%l != 0 {
            return false
        }
        f1, f2 := len1/l, len2/l
        sub := str1[:l]
        repeated1, repeated2 := "", ""
        for i := 0; i < f1; i++ {
            repeated1 += sub
        }
        for i := 0; i < f2; i++ {
            repeated2 += sub
        }
        return repeated1 == str1 && repeated2 == str2
    }

    minLen := len1
    if len2 < minLen {
        minLen = len2
    }
    for l := minLen; l > 0; l-- {
        if isDivisor(l) {
            return str1[:l]
        }
    }

    return ""
}

class Solution {
    fun gcdOfStrings(str1: String, str2: String): String {
        val len1 = str1.length
        val len2 = str2.length

        fun isDivisor(l: Int): Boolean {
            if (len1 % l != 0 || len2 % l != 0) return false
            val f1 = len1 / l
            val f2 = len2 / l
            val sub = str1.substring(0, l)
            return sub.repeat(f1) == str1 && sub.repeat(f2) == str2
        }

        for (l in minOf(len1, len2) downTo 1) {
            if (isDivisor(l)) {
                return str1.substring(0, l)
            }
        }

        return ""
    }
}

class Solution {
    func gcdOfStrings(_ str1: String, _ str2: String) -> String {
        let len1 = str1.count
        let len2 = str2.count
        let arr1 = Array(str1)
        let arr2 = Array(str2)

        func isDivisor(_ l: Int) -> Bool {
            if len1 % l != 0 || len2 % l != 0 { return false }
            let f1 = len1 / l
            let f2 = len2 / l
            let sub = String(arr1[0..<l])
            return String(repeating: sub, count: f1) == str1 &&
                   String(repeating: sub, count: f2) == str2
        }

        for l in stride(from: min(len1, len2), through: 1, by: -1) {
            if isDivisor(l) {
                return String(arr1[0..<l])
            }
        }

        return ""
    }
}

Time & Space Complexity

Time complexity: $O(min(m, n) * (m + n))$
Space complexity: $O(m + n)$

Where $m$ and $n$ are the lengths of the strings $str1$ and $str2$ respectively.

2. Iteration (Space Optimized)

Intuition

Instead of constructing repeated strings and comparing them (which uses extra space), we can validate a candidate divisor by checking character-by-character using modular indexing. Each character at position i should match the character at position i % l in the candidate prefix.

Algorithm

Ensure str1 is the longer string by swapping if necessary.
For each possible length l from n down to 1:
- Check if l divides both lengths.
- Verify that str1[i] == str2[i % l] for all positions in str1.
- Verify that str2[i] == str2[i % l] for positions from l to n.
Return the prefix str2[:l] if all checks pass.
Return an empty string if no valid divisor exists.

class Solution:
    def gcdOfStrings(self, str1: str, str2: str) -> str:
        m, n = len(str1), len(str2)
        if m < n:
            m, n = n, m
            str1, str2 = str2, str1

        for l in range(n, 0, -1):
            if m % l != 0 or n % l != 0:
                continue

            valid = True
            for i in range(m):
                if str1[i] != str2[i % l]:
                    valid = False
                    break
            if not valid: continue

            for i in range(l, n):
                if str2[i] != str2[i % l]:
                    valid = False
                    break
            if valid: return str2[:l]

        return ""

public class Solution {
    public String gcdOfStrings(String str1, String str2) {
        int m = str1.length(), n = str2.length();
        if (m < n) {
            String temp = str1;
            str1 = str2;
            str2 = temp;
            int tempLen = m;
            m = n;
            n = tempLen;
        }

        for (int l = n; l > 0; l--) {
            if (m % l != 0 || n % l != 0) {
                continue;
            }

            boolean valid = true;
            for (int i = 0; i < m; i++) {
                if (str1.charAt(i) != str2.charAt(i % l)) {
                    valid = false;
                    break;
                }
            }
            if (!valid) continue;

            for (int i = l; i < n; i++) {
                if (str2.charAt(i) != str2.charAt(i % l)) {
                    valid = false;
                    break;
                }
            }
            if (valid) {
                return str2.substring(0, l);
            }
        }

        return "";
    }
}

class Solution {
public:
    string gcdOfStrings(string str1, string str2) {
        int m = str1.size(), n = str2.size();
        if (m < n) {
            swap(m, n);
            swap(str1, str2);
        }

        for (int l = n; l > 0; l--) {
            if (m % l != 0 || n % l != 0) {
                continue;
            }

            bool valid = true;
            for (int i = 0; i < m; i++) {
                if (str1[i] != str2[i % l]) {
                    valid = false;
                    break;
                }
            }
            if (!valid) continue;

            for (int i = l; i < n; i++) {
                if (str2[i] != str2[i % l]) {
                    valid = false;
                    break;
                }
            }
            if (valid) {
                return str2.substr(0, l);
            }
        }

        return "";
    }
};

class Solution {
    /**
     * @param {string} str1
     * @param {string} str2
     * @return {string}
     */
    gcdOfStrings(str1, str2) {
        let m = str1.length,
            n = str2.length;
        if (m < n) {
            [m, n] = [n, m];
            [str1, str2] = [str2, str1];
        }

        for (let l = n; l > 0; l--) {
            if (m % l !== 0 || n % l !== 0) {
                continue;
            }

            let valid = true;
            for (let i = 0; i < m; i++) {
                if (str1[i] !== str2[i % l]) {
                    valid = false;
                    break;
                }
            }
            if (!valid) continue;

            for (let i = l; i < n; i++) {
                if (str2[i] !== str2[i % l]) {
                    valid = false;
                    break;
                }
            }
            if (valid) {
                return str2.slice(0, l);
            }
        }

        return '';
    }
}

public class Solution {
    public string GcdOfStrings(string str1, string str2) {
        int m = str1.Length, n = str2.Length;
        if (m < n) {
            (str1, str2) = (str2, str1);
            (m, n) = (n, m);
        }

        for (int l = n; l >= 1; l--) {
            if (m % l != 0 || n % l != 0) continue;

            bool valid = true;
            for (int i = 0; i < m; i++) {
                if (str1[i] != str2[i % l]) {
                    valid = false;
                    break;
                }
            }
            if (!valid) continue;

            for (int i = l; i < n; i++) {
                if (str2[i] != str2[i % l]) {
                    valid = false;
                    break;
                }
            }

            if (valid) return str2.Substring(0, l);
        }

        return "";
    }
}

func gcdOfStrings(str1 string, str2 string) string {
    m, n := len(str1), len(str2)
    if m < n {
        m, n = n, m
        str1, str2 = str2, str1
    }

    for l := n; l > 0; l-- {
        if m%l != 0 || n%l != 0 {
            continue
        }

        valid := true
        for i := 0; i < m; i++ {
            if str1[i] != str2[i%l] {
                valid = false
                break
            }
        }
        if !valid {
            continue
        }

        for i := l; i < n; i++ {
            if str2[i] != str2[i%l] {
                valid = false
                break
            }
        }

        if valid {
            return str2[:l]
        }
    }

    return ""
}

class Solution {
    fun gcdOfStrings(str1: String, str2: String): String {
        var s1 = str1
        var s2 = str2
        var m = s1.length
        var n = s2.length
        if (m < n) {
            s1 = str2.also { s2 = str1 }
            m = n.also { n = m }
        }

        for (l in n downTo 1) {
            if (m % l != 0 || n % l != 0) continue

            var valid = true
            for (i in 0 until m) {
                if (s1[i] != s2[i % l]) {
                    valid = false
                    break
                }
            }
            if (!valid) continue

            for (i in l until n) {
                if (s2[i] != s2[i % l]) {
                    valid = false
                    break
                }
            }

            if (valid) return s2.substring(0, l)
        }

        return ""
    }
}

class Solution {
    func gcdOfStrings(_ str1: String, _ str2: String) -> String {
        var s1 = Array(str1)
        var s2 = Array(str2)
        var m = s1.count
        var n = s2.count
        if m < n {
            swap(&s1, &s2)
            swap(&m, &n)
        }

        for l in stride(from: n, through: 1, by: -1) {
            if m % l != 0 || n % l != 0 { continue }

            var valid = true
            for i in 0..<m {
                if s1[i] != s2[i % l] {
                    valid = false
                    break
                }
            }
            if !valid { continue }

            for i in l..<n {
                if s2[i] != s2[i % l] {
                    valid = false
                    break
                }
            }

            if valid { return String(s2[0..<l]) }
        }

        return ""
    }
}

Time & Space Complexity

Time complexity: $O(min(m, n) * (m + n))$
Space complexity: $O(g)$ for the output string.

Where $m$ is the length of the string $str1$ , $n$ is the length of the string $str2$ , and $g$ is the length of the output string.

3. Greatest Common Divisor

Intuition

If a common divisor string exists, then concatenating the two strings in either order must produce the same result: str1 + str2 == str2 + str1. This is because both strings are built from the same repeating pattern. Once verified, the GCD of the two string lengths gives us the length of the largest common divisor.

Algorithm

Check if str1 + str2 == str2 + str1. If not, return an empty string.
Compute g = gcd(len(str1), len(str2)).
Return the prefix str1[:g].

class Solution:
    def gcdOfStrings(self, str1: str, str2: str) -> str:
        if str1 + str2 != str2 + str1:
            return ""

        g = gcd(len(str1), len(str2))
        return str1[:g]

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

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

class Solution {
public:
    string gcdOfStrings(string str1, string str2) {
        if (str1 + str2 != str2 + str1) {
            return "";
        }
        int g = __gcd((int)str1.size(), (int)str2.size());
        return str1.substr(0, g);
    }
};

class Solution {
    /**
     * @param {string} str1
     * @param {string} str2
     * @return {string}
     */
    gcdOfStrings(str1, str2) {
        if (str1 + str2 !== str2 + str1) {
            return '';
        }
        const gcd = (a, b) => (b === 0 ? a : gcd(b, a % b));
        const g = gcd(str1.length, str2.length);
        return str1.slice(0, g);
    }
}

public class Solution {
    public string GcdOfStrings(string str1, string str2) {
        if (str1 + str2 != str2 + str1) {
            return "";
        }

        int g = GCD(str1.Length, str2.Length);
        return str1.Substring(0, g);
    }

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

func gcdOfStrings(str1 string, str2 string) string {
    if str1+str2 != str2+str1 {
        return ""
    }
    var gcd func(a, b int) int
    gcd = func(a, b int) int {
        if b == 0 {
            return a
        }
        return gcd(b, a%b)
    }
    return str1[:gcd(len(str1), len(str2))]
}

class Solution {
    fun gcdOfStrings(str1: String, str2: String): String {
        if (str1 + str2 != str2 + str1) {
            return ""
        }
        fun gcd(a: Int, b: Int): Int = if (b == 0) a else gcd(b, a % b)
        return str1.substring(0, gcd(str1.length, str2.length))
    }
}

class Solution {
    func gcdOfStrings(_ str1: String, _ str2: String) -> String {
        if str1 + str2 != str2 + str1 {
            return ""
        }
        func gcd(_ a: Int, _ b: Int) -> Int {
            return b == 0 ? a : gcd(b, a % b)
        }
        let g = gcd(str1.count, str2.count)
        return String(str1.prefix(g))
    }
}

Time & Space Complexity

Time complexity: $O(m + n)$
Space complexity: $O(m + n)$ .

Where $m$ and $n$ are the lengths of the strings $str1$ and $str2$ respectively.

4. Greatest Common Divisor (Space Optimized)

Intuition

We can avoid the concatenation check (which requires O(m+n) space) by directly validating the divisor property. Compute the GCD of the lengths, then verify that every character in both strings matches the corresponding character in the first g characters, using modular indexing.

Algorithm

Compute g = gcd(len(str1), len(str2)).
For each index i in str1, verify str1[i] == str1[i % g].
For each index i in str2, verify str2[i] == str1[i % g].
If all checks pass, return str1[:g].
Otherwise, return an empty string.

class Solution:
    def gcdOfStrings(self, str1: str, str2: str) -> str:
        g = gcd(len(str1), len(str2))

        if all(str1[i] == str1[i % g] for i in range(len(str1))) and \
           all(str2[i] == str1[i % g] for i in range(len(str2))):
            return str1[:g]
        return ""

public class Solution {
    public String gcdOfStrings(String str1, String str2) {
        int g = gcd(str1.length(), str2.length());

        for (int i = 0; i < str1.length(); i++) {
            if (str1.charAt(i) != str1.charAt(i % g)) {
                return "";
            }
        }

        for (int i = 0; i < str2.length(); i++) {
            if (str2.charAt(i) != str1.charAt(i % g)) {
                return "";
            }
        }

        return str1.substring(0, g);
    }

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

class Solution {
public:
    string gcdOfStrings(string str1, string str2) {
        int g = __gcd((int)str1.size(), (int)str2.size());

        for (int i = 0; i < str1.size(); i++) {
            if (str1[i] != str1[i % g]) {
                return "";
            }
        }

        for (int i = 0; i < str2.size(); i++) {
            if (str2[i] != str1[i % g]) {
                return "";
            }
        }

        return str1.substr(0, g);
    }
};

class Solution {
    /**
     * @param {string} str1
     * @param {string} str2
     * @return {string}
     */
    gcdOfStrings(str1, str2) {
        const gcd = (a, b) => (b === 0 ? a : gcd(b, a % b));
        const g = gcd(str1.length, str2.length);

        for (let i = 0; i < str1.length; i++) {
            if (str1[i] !== str1[i % g]) {
                return '';
            }
        }

        for (let i = 0; i < str2.length; i++) {
            if (str2[i] !== str1[i % g]) {
                return '';
            }
        }

        return str1.slice(0, g);
    }
}

public class Solution {
    public string GcdOfStrings(string str1, string str2) {
        int g = GCD(str1.Length, str2.Length);

        for (int i = 0; i < str1.Length; i++) {
            if (str1[i] != str1[i % g]) {
                return "";
            }
        }

        for (int i = 0; i < str2.Length; i++) {
            if (str2[i] != str1[i % g]) {
                return "";
            }
        }

        return str1.Substring(0, g);
    }

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

func gcdOfStrings(str1 string, str2 string) string {
    var gcd func(a, b int) int
    gcd = func(a, b int) int {
        if b == 0 {
            return a
        }
        return gcd(b, a%b)
    }
    g := gcd(len(str1), len(str2))

    for i := 0; i < len(str1); i++ {
        if str1[i] != str1[i%g] {
            return ""
        }
    }

    for i := 0; i < len(str2); i++ {
        if str2[i] != str1[i%g] {
            return ""
        }
    }

    return str1[:g]
}

class Solution {
    fun gcdOfStrings(str1: String, str2: String): String {
        fun gcd(a: Int, b: Int): Int = if (b == 0) a else gcd(b, a % b)
        val g = gcd(str1.length, str2.length)

        for (i in str1.indices) {
            if (str1[i] != str1[i % g]) {
                return ""
            }
        }

        for (i in str2.indices) {
            if (str2[i] != str1[i % g]) {
                return ""
            }
        }

        return str1.substring(0, g)
    }
}

class Solution {
    func gcdOfStrings(_ str1: String, _ str2: String) -> String {
        func gcd(_ a: Int, _ b: Int) -> Int {
            return b == 0 ? a : gcd(b, a % b)
        }
        let g = gcd(str1.count, str2.count)
        let s1 = Array(str1)
        let s2 = Array(str2)

        for i in 0..<s1.count {
            if s1[i] != s1[i % g] {
                return ""
            }
        }

        for i in 0..<s2.count {
            if s2[i] != s1[i % g] {
                return ""
            }
        }

        return String(s1[0..<g])
    }
}

Time & Space Complexity

Time complexity: $O(m + n)$
Space complexity: $O(g)$ for the output string.

Where $m$ is the length of the string $str1$ , $n$ is the length of the string $str2$ , and $g$ is the GCD of $m$ and $n$ .

Common Pitfalls

Skipping the Concatenation Check

The key insight is that if a common divisor string exists, then str1 + str2 must equal str2 + str1. Skipping this check and directly returning str1[:gcd(len1, len2)] will produce wrong answers for cases where no valid divisor exists, such as str1 = "LEET" and str2 = "CODE".

Confusing GCD of Lengths with GCD of Strings

The GCD of the string lengths tells you the candidate divisor length, but it does not guarantee that the prefix of that length actually divides both strings. You must verify that the candidate pattern repeats correctly to form both original strings, either through the concatenation check or by explicitly validating the pattern.