Find the GCD of an array made up of numeric strings

Last Updated : 25 Aug, 2021

Given an array arr[] consisting of numeric strings, the task is to calculate Greatest Common Divisor of the given array.

Considering strings ‘A’ and ‘B’, “B divides A” if and only if A is a concatenation of B more than once. Find the largest string which divides both A and B.

Examples:

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

Input: arr = { “Geeks”, “GFG”}
Output: “”

Approach: Follow the steps below to solve the problem:

Calculate GCD of the length of all the strings of the given array, say GCD.
Check if all the strings of the given array can be made by concatenating the substring {arr[0][0], .., arr[0][GCD – 1]} any number of times or not. If found to be true, then print {arr[0][0], .., arr[0][GCD – 1]}.
Otherwise, print an empty string.

Below is the implementation of the above approach:

C++

// CPP program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Recursive function
// to return gcd of A and B
int GCD(int lena, int lenb)
{
 
  if (lena == 0)
    return lenb;
 
  if (lenb == 0)
    return lena;
 
  // Base case
  if (lena == lenb)
    return lena;
 
  // Length of A is greater
  if (lena > lenb)
    return GCD(lena - lenb, lenb);
 
  return GCD(lena, lenb - lena);
}
 
// Calculate GCD
string StringGCD(string a, string b)
{
 
  // Store the GCD of the
  // length of the strings
  int gcd = GCD(a.size(), b.size());
  if (a.substr(0, gcd) == b.substr(0, gcd))
  {
    int x = ((int)b.size()/gcd);
    int y = ((int)a.size()/gcd);
    string r="",s="";
 
    while (x--) s += a;
    while (y--) r += b;
 
    if (s == r)
      return a.substr(0, gcd);
  }
  return "-1";
}
 
// Driver Code
int main()
{
  string a = "geeksgeeks";
  string b = "geeks";
 
  // Function call
  cout<<(StringGCD(a, b));
}
 
// This code is contributed by mohit kumar 29

Java

// JAVA program for the above approach
import java.util.*;
class GFG
{
 
// Recursive function
// to return gcd of A and B
static int GCD(int lena, int lenb)
{
  if (lena == 0)
    return lenb;
  if (lenb == 0)
    return lena;
 
  // Base case
  if (lena == lenb)
    return lena;
 
  // Length of A is greater
  if (lena > lenb)
    return GCD(lena - lenb, lenb);
  return GCD(lena, lenb - lena);
}
 
// Calculate GCD
static String StringGCD(String a, String b)
{
 
  // Store the GCD of the
  // length of the Strings
  int gcd = GCD(a.length(), b.length());
  if (a.substring(0, gcd).equals(b.substring(0, gcd)))
  {
    int x = ((int)b.length()/gcd);
    int y = ((int)a.length()/gcd);
    String r="",s="";
 
    while (x-- >0) s += a;
    while (y-- >0) r += b;
 
    if (s.equals(r))
      return a.substring(0, gcd);
  }
  return "-1";
}
 
// Driver Code
public static void main(String[] args)
{
  String a = "geeksgeeks";
  String b = "geeks";
 
  // Function call
  System.out.print(StringGCD(a, b));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python implementation of the above approach
 
# Recursive function
# to return gcd of A and B
def GCD(lena, lenb):    
 
  if (lena == 0):
    return lenb
 
  if (lenb == 0):
    return lena
   
  # Base case
  if (lena == lenb):
    return lena
   
  # Length of A is greater
  if (lena > lenb):
    return GCD(lena-lenb, lenb)
  return GCD(lena, lenb-lena)
 
# Calculate GCD
def StringGCD(a, b):
   
  # Store the GCD of the
  # length of the strings
  gcd = GCD(len(a), len(b))
  if a[:gcd] == b[:gcd]:
 
    if a*(len(b)//gcd) == b*(len(a)//gcd):
      return a[:gcd]
 
  return -1
 
# Driver Code
a = 'geeksgeeks'
b = 'geeks'
 
# Function call
print(StringGCD(a, b))

C#

// C# program for the above approach
using System;
class GFG
{
 
// Recursive function
// to return gcd of A and B
static int GCD(int lena, int lenb)
{
  if (lena == 0)
    return lenb;
  if (lenb == 0)
    return lena;
 
  // Base case
  if (lena == lenb)
    return lena;
 
  // Length of A is greater
  if (lena > lenb)
    return GCD(lena - lenb, lenb);
  return GCD(lena, lenb - lena);
}
 
// Calculate GCD
static String StringGCD(String a, String b)
{
 
  // Store the GCD of the
  // length of the Strings
  int gcd = GCD(a.Length, b.Length);
  if (a.Substring(0, gcd).Equals(b.Substring(0, gcd)))
  {
    int x = ((int)b.Length/gcd);
    int y = ((int)a.Length/gcd);
    String r="", s="";
 
    while (x-- >0) s += a;
    while (y-- >0) r += b;
    if (s.Equals(r))
      return a.Substring(0, gcd);
  }
  return "-1";
}
 
// Driver Code
public static void Main(String[] args)
{
  String a = "geeksgeeks";
  String b = "geeks";
 
  // Function call
  Console.Write(StringGCD(a, b));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
// JAVASCRIPT program for the above approach
 
// Recursive function
// to return gcd of A and B
function GCD(lena,lenb)
{
    if (lena == 0)
        return lenb;
      if (lenb == 0)
        return lena;
  
      // Base case
      if (lena == lenb)
        return lena;
  
      // Length of A is greater
      if (lena > lenb)
        return GCD(lena - lenb, lenb);
      return GCD(lena, lenb - lena);
}
 
// Calculate GCD
function StringGCD(a,b)
{
    // Store the GCD of the
      // length of the Strings
      let gcd = GCD(a.length, b.length);
      if (a.substring(0, gcd) == (b.substring(0, gcd)))
      {
        let x = Math.floor(b.length/gcd);
        let y = Math.floor(a.length/gcd);
        let r="",s="";
  
        while (x-- >0) 
            s += a;
        while (y-- >0) 
            r += b;
  
        if (s == (r))
              return a.substring(0, gcd);
      }
      return "-1";
}
 
// Driver Code
let a = "geeksgeeks";
let b = "geeks";
 
// Function call
document.write(StringGCD(a, b));
 
 
// This code is contributed by patel2127
</script>

Output:

geeks

Time Complexity: O(N * M), where M is the length of strings
Auxiliary Space: O(1)

Flip all K-bits of a given number

rohitsingh07052

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Article Tags :

Practice Tags :

Find the GCD of an array made up of numeric strings

C++

Java

Python3

C#

Javascript

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