# Program to find Greatest Common Divisor (GCD) of N strings

Given an array of string arr[], the task is the Greatest Common Divisor of the given array of string.

In strings ‘A’ and ‘B’, we say “B divides A” if and only if A = concatenation of B more than 1 times.Find the largest string which divides both A and B.

Examples:

Input: arr[] = { “GFGGFG”, “GFGGFG”, “GFGGFGGFGGFG” }
Output: “GFGGFG”
Explanation:
“GFGGFG” is the largest string which divides the whole array elements.
Input: arr = { “Geeks”, “GFG”}
Output: “”

Approach: The idea is to use recursion. Below are the steps:

1. Create a recursive function gcd(str1, str2).
2. If the length of str2 is more than str1 then we will recur with gcd(str2, str1).
3. Now if str1 doesn’t start with str2 then return an empty string.
4. If the longer string begins with a shorter string, cut off the common prefix part of the longer string and recur or repeat until one is empty.
5. The string returned after the above steps are the gcd of the given array of string.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include``using` `namespace` `std;` `// Function that finds gcd of 2 strings``string gcd(string str1, string str2)``{``  ` `    ``// If str1 length is less than``    ``// that of str2 then recur``    ``// with gcd(str2, str1)``    ``if` `(str1.length() < str2.length()) ``    ``{``        ``return` `gcd(str2, str1);``    ``}``  ` `    ``// If str1 is not the``    ``// concatenation of str2``    ``else` `if``(str1.find(str2) != 0) ``    ``{``        ``return` `""``;``    ``}``    ``else` `if` `(str2 == ``""``) ``    ``{``      ` `        ``// GCD string is found``        ``return` `str1;``    ``}``    ``else``    ``{``      ` `        ``// Cut off the common prefix``        ``// part of str1 & then recur``        ``return` `gcd(str1.substr(str2.length()), str2);``    ``}``}` `// Function to find GCD of array of``// strings``string findGCD(string arr[], ``int` `n)``{``    ``string result = arr[0];``    ``for` `(``int` `i = 1; i < n; i++)``    ``{``        ``result = gcd(result, arr[i]);``    ``}``  ` `    ``// Return the GCD of strings``    ``return` `result;``}` `// Driver Code``int` `main()``{``  ` `    ``// Given array  of strings``    ``string arr[]={ ``"GFGGFG"``,``                         ``"GFGGFG"``,``                         ``"GFGGFGGFGGFG"` `};``    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]);``  ` `    ``// Function Call``    ``cout << findGCD(arr, n);``}` `// This code is contributed by pratham76.`

## Java

 `// Java program for the above approach` `class` `GCD {` `    ``// Function that finds gcd of 2 strings``    ``static` `String gcd(String str1, String str2)``    ``{``        ``// If str1 length is less than``        ``// that of str2 then recur``        ``// with gcd(str2, str1)``        ``if` `(str1.length() < str2.length()) {``            ``return` `gcd(str2, str1);``        ``}` `        ``// If str1 is not the``        ``// concatenation of str2``        ``else` `if` `(!str1.startsWith(str2)) {``            ``return` `""``;``        ``}` `        ``else` `if` `(str2.isEmpty()) {` `            ``// GCD string is found``            ``return` `str1;``        ``}``        ``else` `{` `            ``// Cut off the common prefix``            ``// part of str1 & then recur``            ``return` `gcd(str1.substring(str2.length()),``                       ``str2);``        ``}``    ``}` `    ``// Function to find GCD of array of``    ``// strings``    ``static` `String findGCD(String arr[], ``int` `n)``    ``{``        ``String result = arr[``0``];` `        ``for` `(``int` `i = ``1``; i < n; i++) {``            ``result = gcd(result, arr[i]);``        ``}` `        ``// Return the GCD of strings``        ``return` `result;``    ``}` `    ``// Driver Code``    ``public` `static` `void``        ``main(String[] args)``    ``{``        ``// Given array  of strings``        ``String arr[]``            ``= ``new` `String[] { ``"GFGGFG"``,``                             ``"GFGGFG"``,``                             ``"GFGGFGGFGGFG"` `};``        ``int` `n = arr.length;` `        ``// Function Call``        ``System.out.println(findGCD(arr, n));``    ``}``}`

## Python3

 `# Python3 program for the above approach` `# Function that finds gcd of 2 strings``def` `gcd(str1, str2):` `    ``# If str1 length is less than``    ``# that of str2 then recur``    ``# with gcd(str2, str1)``    ``if``(``len``(str1) < ``len``(str2)):``        ``return` `gcd(str2, str1)``    ` `    ``# If str1 is not the``    ``# concatenation of str2``    ``elif``(``not` `str1.startswith(str2)):``        ``return` `""``    ``elif``(``len``(str2) ``=``=` `0``):``        ` `        ``# GCD string is found``        ``return` `str1``    ``else``:``        ` `        ``# Cut off the common prefix``        ``# part of str1 & then recur``        ``return` `gcd(str1[``len``(str2):], str2)` `# Function to find GCD of array of``# strings``def` `findGCD(arr, n):``    ``result ``=` `arr[``0``]` `    ``for` `i ``in` `range``(``1``, n):``        ``result ``=` `gcd(result, arr[i])``    ` `    ``# Return the GCD of strings``    ``return` `result` `# Driver Code` `# Given array  of strings``arr ``=` `[``"GFGGFG"``, ``"GFGGFG"``, ``"GFGGFGGFGGFG"` `]``n ``=` `len``(arr)` `# Function Call``print``(findGCD(arr, n))` `# This code is contributed by avanitrachhadiya2155`

## C#

 `// C# program for ``// the above approach``using` `System;``class` `GFG{` `// Function that finds gcd ``// of 2 strings``static` `String gcd(String str1, ``                  ``String str2)``{``  ``// If str1 length is less than``  ``// that of str2 then recur``  ``// with gcd(str2, str1)``  ``if` `(str1.Length < str2.Length) ``  ``{``    ``return` `gcd(str2, str1);``  ``}` `  ``// If str1 is not the``  ``// concatenation of str2``  ``else` `if` `(!str1.StartsWith(str2)) ``  ``{``    ``return` `""``;``  ``}``  ``else` `if` `(str2.Length == 0) ``  ``{``    ``// GCD string is found``    ``return` `str1;``  ``}``  ``else``  ``{``    ``// Cut off the common prefix``    ``// part of str1 & then recur``    ``return` `gcd(str1.Substring(str2.Length),``                              ``str2);``  ``}``}` `// Function to find GCD ``// of array of strings``static` `String findGCD(String []arr, ``                      ``int` `n)``{``  ``String result = arr[0];` `  ``for` `(``int` `i = 1; i < n; i++) ``  ``{``    ``result = gcd(result, arr[i]);``  ``}` `  ``// Return the GCD of strings``  ``return` `result;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``  ``// Given array  of strings``  ``String []arr = ``new` `String[] {``"GFGGFG"``,``                               ``"GFGGFG"``,``                               ``"GFGGFGGFGGFG"``};``  ``int` `n = arr.Length;` `  ``// Function Call``  ``Console.WriteLine(findGCD(arr, n));``}``}` `// This code is contributed by shikhasingrajput`

## Javascript

 ``

Output
```GFGGFG

```

Time Complexity: O(N*log(B)), where N is the number of strings, and B is the maximum length of any string in arr[].
Auxiliary Space: O(1)

#### Approach 2 (using Euclidean algorithm)

In this approach we need to first find the GCD of the individual strings in the array. Then wecan use the Euclidean algorithm to find the GCD of two strings A and B.

Algorithm:

`First find the length of strings A and B. Let them be n and m.   If m > n, swap A and B.   If A is not divisible by B, return "" (empty string) because there is no common divisor.Otherwise, repeat the following steps until A is divisible by B:    a. Let C be the remainder when A is divided by B.    b. Set A to B and B to C.Return B, which is the GCD of A and B.`

## C++

 `#include ``#include ``#include ` `using` `namespace` `std;` `// Define a function to find the GCD of two strings using the Euclidean algorithm.``string gcd_strings(string a, string b)``{``  ` `    ``// Find the lengths of the two strings.``    ``int` `n = a.length(), m = b.length();``  ` `    ``// If the second string is longer than the first, swap them.``    ``if` `(m > n) {``        ``swap(a, b);``        ``swap(n, m);``    ``}``    ``// If the first string is not divisible by the second string, there is no common divisor.``    ``if` `(a.compare(0, m, b) != 0) {``        ``return` `""``;``    ``}``    ``// Repeat the Euclidean algorithm until the second string divides the first.``    ``while` `(m > 0) {``        ``int` `tmp = m;``        ``m = n % m;``        ``n = tmp;``        ``if` `(m > 0) {``            ``a = b;``            ``b = a.substr(n - m, m);``        ``}``    ``}``    ``// Return the second string, which is the GCD of the two strings.``    ``return` `b;``}`  `// Define a function to find the GCD of an array of strings.``string gcd_array_strings(vector arr) ``{``  ` `    ``// Start with the first string in the array.``    ``string gcd = arr[0];``  ` `    ``// Iterate over the remaining strings in the array.``    ``for` `(``int` `i = 1; i < arr.size(); i++) ``    ``{``      ` `        ``// Find the GCD of the current string and the previous GCD.``        ``gcd = gcd_strings(gcd, arr[i]);``      ` `        ``// If the GCD is an empty string, there is no common divisor.``        ``if` `(gcd == ``""``) {``            ``break``;``        ``}``    ``}``    ``// Return the final GCD, which is the GCD of the entire array.``    ``return` `gcd;``}` `// Example usage``int` `main() {``    ``vector arr = {``"GFGGFG"``, ``"GFGGFG"``, ``"GFGGFGGFGGFG"``};``    ``cout << gcd_array_strings(arr) << endl; ``// output: GFGGFG``    ``return` `0;``}`

## Java

 `import` `java.util.ArrayList;``import` `java.util.List;` `public` `class` `GCDStrings {``    ``// Function to find the GCD of two strings using the Euclidean algorithm``    ``public` `static` `String gcdStrings(String a, String b) {``        ``int` `n = a.length();``        ``int` `m = b.length();` `        ``// If the second string is longer than the first, swap them``        ``if` `(m > n) {``            ``String temp = a;``            ``a = b;``            ``b = temp;``            ``int` `tempLen = n;``            ``n = m;``            ``m = tempLen;``        ``}` `        ``// If the first string is not divisible by the second string, there is no common divisor``        ``if` `(!a.startsWith(b)) {``            ``return` `""``;``        ``}` `        ``// Repeat the Euclidean algorithm until the second string divides the first``        ``while` `(m > ``0``) {``            ``int` `temp = m;``            ``m = n % m;``            ``n = temp;``            ``if` `(m > ``0``) {``                ``a = b;``                ``b = a.substring(n - m, n);``            ``}``        ``}` `        ``// Return the second string, which is the GCD of the two strings``        ``return` `b;``    ``}` `    ``// Function to find the GCD of an array of strings``    ``public` `static` `String gcdArrayStrings(List arr) {``        ``// Start with the first string in the array``        ``String gcd = arr.get(``0``);` `        ``// Iterate over the remaining strings in the array``        ``for` `(``int` `i = ``1``; i < arr.size(); i++) {``            ``// Find the GCD of the current string and the previous GCD``            ``gcd = gcdStrings(gcd, arr.get(i));` `            ``// If the GCD is an empty string, there is no common divisor``            ``if` `(gcd.isEmpty()) {``                ``break``;``            ``}``        ``}` `        ``// Return the final GCD, which is the GCD of the entire array``        ``return` `gcd;``    ``}` `    ``// Example usage``    ``public` `static` `void` `main(String[] args) {``        ``List arr = ``new` `ArrayList<>();``        ``arr.add(``"GFGGFG"``);``        ``arr.add(``"GFGGFG"``);``        ``arr.add(``"GFGGFGGFGGFG"``);` `        ``System.out.println(gcdArrayStrings(arr)); ``// Output: GFGGFG``    ``}``}`

## Python

 `# Define a function to find the GCD of two strings using the Euclidean algorithm.``def` `gcd_strings(a, b):``    ``# Find the lengths of the two strings.``    ``n, m ``=` `len``(a), ``len``(b)``    ``# If the second string is longer than the first, swap them.``    ``if` `m > n:``        ``a, b ``=` `b, a``        ``n, m ``=` `m, n``    ``# If the first string is not divisible by the second string, there is no common divisor.``    ``if` `a !``=` `b ``*` `(n ``/``/` `m):``        ``return` `""``    ``# Repeat the Euclidean algorithm until the second string divides the first.``    ``while` `m > ``0``:``        ``n, m ``=` `m, n ``%` `m``        ``if` `m > ``0``:``            ``a, b ``=` `b, a[m:n] ``    ``# Return the second string, which is the GCD of the two strings.``    ``return` `b` `# Define a function to find the GCD of an array of strings.``def` `gcd_array_strings(arr):``    ``# Start with the first string in the array.``    ``gcd ``=` `arr[``0``]``    ``# Iterate over the remaining strings in the array.``    ``for` `i ``in` `range``(``1``, ``len``(arr)):``        ``# Find the GCD of the current string and the previous GCD.``        ``gcd ``=` `gcd_strings(gcd, arr[i])``        ``# If the GCD is an empty string, there is no common divisor.``        ``if` `gcd ``=``=` `"":``            ``break``    ``# Return the final GCD, which is the GCD of the entire array.``    ``return` `gcd` `# Example usage``arr ``=` `[``"GFGGFG"``, ``"GFGGFG"``, ``"GFGGFGGFGGFG"``]``print``(gcd_array_strings(arr)) ``# output: GFGGFG`

## C#

 `using` `System;``using` `System.Collections.Generic;` `namespace` `GCDStringCalculator``{``    ``class` `Program``    ``{``        ``// Define a function to find the GCD of two strings using the Euclidean algorithm.``        ``static` `string` `GcdStrings(``string` `a, ``string` `b)``        ``{``            ``// Find the lengths of the two strings.``            ``int` `n = a.Length, m = b.Length;` `            ``// If the second string is longer than the first, swap them.``            ``if` `(m > n)``            ``{``                ``string` `temp = a;``                ``a = b;``                ``b = temp;``                ``int` `tempLength = n;``                ``n = m;``                ``m = tempLength;``            ``}` `            ``// If the first string is not divisible by the second string, there is no common divisor.``            ``if` `(a.Substring(0, m) != b)``            ``{``                ``return` `""``;``            ``}` `            ``// Repeat the Euclidean algorithm until the second string divides the first.``            ``while` `(m > 0)``            ``{``                ``int` `tmp = m;``                ``m = n % m;``                ``n = tmp;``                ``if` `(m > 0)``                ``{``                    ``a = b;``                    ``b = a.Substring(n - m, m);``                ``}``            ``}` `            ``// Return the second string, which is the GCD of the two strings.``            ``return` `b;``        ``}` `        ``// Define a function to find the GCD of an array of strings.``        ``static` `string` `GcdArrayStrings(List<``string``> arr)``        ``{``            ``// Start with the first string in the array.``            ``string` `gcd = arr[0];` `            ``// Iterate over the remaining strings in the array.``            ``for` `(``int` `i = 1; i < arr.Count; i++)``            ``{``                ``// Find the GCD of the current string and the previous GCD.``                ``gcd = GcdStrings(gcd, arr[i]);` `                ``// If the GCD is an empty string, there is no common divisor.``                ``if` `(gcd == ``""``)``                ``{``                    ``break``;``                ``}``            ``}` `            ``// Return the final GCD, which is the GCD of the entire array.``            ``return` `gcd;``        ``}` `        ``// Example usage``        ``static` `void` `Main(``string``[] args)``        ``{``            ``List<``string``> arr = ``new` `List<``string``> { ``"GFGGFG"``, ``"GFGGFG"``, ``"GFGGFGGFGGFG"` `};``            ``Console.WriteLine(GcdArrayStrings(arr)); ``        ``}``    ``}``}`

## Javascript

 `// Define a function to find the GCD of two strings using the Euclidean algorithm.``function` `gcdStrings(a, b) {``    ``// If the second string is longer than the first, swap them.``    ``if` `(b.length > a.length) {``        ``[a, b] = [b, a];``    ``}``    ` `    ``// If the first string is not divisible by the second string, there is no common divisor.``    ``if` `(!a.startsWith(b)) {``        ``return` `""``;``    ``}``    ` `    ``// Repeat the Euclidean algorithm until the second string divides the first.``    ``while` `(b.length > 0) {``        ``const tmp = b;``        ``b = a % b;``        ``a = tmp;``        ` `        ``if` `(b.length > 0) {``            ``a = b;``            ``b = tmp.substring(tmp.length - b.length);``        ``}``    ``}``    ` `    ``// Return the second string, which is the GCD of the two strings.``    ``return` `a;``}` `// Define a function to find the GCD of an array of strings.``function` `gcdArrayStrings(arr) {``    ``// Start with the first string in the array.``    ``let gcd = arr[0];``    ` `    ``// Iterate over the remaining strings in the array.``    ``for` `(let i = 1; i < arr.length; i++) {``        ``// Find the GCD of the current string and the previous GCD.``        ``gcd = gcdStrings(gcd, arr[i]);``        ` `        ``// If the GCD is an empty string, there is no common divisor.``        ``if` `(gcd === ``""``) {``            ``break``;``        ``}``    ``}``    ` `    ``// Return the final GCD, which is the GCD of the entire array.``    ``return` `gcd;``}` `// Example usage``const arr = [``"GFGGFG"``, ``"GFGGFG"``, ``"GFGGFGGFGGFG"``];``console.log(gcdArrayStrings(arr)); ``// Output: GFGGFG`

Output
```GFGGFG

```

Time complexity ;O(mn), where m is the length of the longest string in the vector and n is the length of the shortest string in the vector.
Space complexity   O(1)

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