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

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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<bits/stdc++.h>
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

<script>
    // Javascript program for the above approach
     
    // Function that finds gcd
    // of 2 strings
    function gcd(str1, 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.substr(str2.length), str2);
      }
    }
 
    // Function to find GCD
    // of array of strings
    function findGCD(arr, n)
    {
      let result = arr[0];
 
      for (let i = 1; i < n; i++)
      {
        result = gcd(result, arr[i]);
      }
 
      // Return the GCD of strings
      return result;
    }
     
    // Given array  of strings
    let arr = ["GFGGFG", "GFGGFG", "GFGGFGGFGGFG"];
    let n = arr.length;
 
    // Function Call
    document.write(findGCD(arr, n));
   
  // This code is contributed by decode2207.
</script>

                    

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 <iostream>
#include <string>
#include <vector>
 
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<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.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<string> 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<String> 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<String> 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) 



Last Updated : 12 Oct, 2023
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