Minimum insertions to make a Co-prime array

Given an array of N elements, find the minimum number of insertions to convert the given array into a co-prime array. Print the resultant array also.

Co-prime Array : An array in which every pair of adjacent elements are co-primes. i.e, gcd(a, b) = 1.

Examples :



Input : A[] = {2, 7, 28}
Output : 1
Explanation : 
Here, 1st pair = {2, 7} are co-primes( gcd(2, 7) = 1).
2nd pair = {7, 28} are not co-primes, insert 9
between them. gcd(7, 9) = 1 and gcd(9, 28) = 1.


Input : A[] = {5, 10, 20}
Output : 2
Explanation : 
Here, there is no pair which are co-primes. 
Insert 7 between (5, 10) and 1 between (10, 20).

Observe that to make a pair to become co-primes we have to insert a number which makes the newly formed pairs co-primes. So, we have to check every adjacent pair for their co-primality and insert a number if required. Now, what is the number that should be inserted? Let us take two numbers a and b. If any of a or b is 1, then GCD(a, b) = 1. So, we can insert ONE(1) at every pair. For efficiency we use euler’s gcd function .


Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP program for minimum insertions to
// make a Co-prime Array.
#include <bits/stdc++.h>
using namespace std;
  
void printResult(int arr[], int n)
{
    // Counting adjacent pairs that are not
    // co-prime.
    int count = 0;
    for (int i = 1; i < n; i++)     
        if (__gcd(arr[i], arr[i - 1]) != 1)
            count++;
  
    cout << count << endl; // No.of insertions
    cout << arr[0] << " ";
    for (int i = 1; i < n; i++) 
    {
        // If pair is not a co-prime insert 1.
        if (__gcd(arr[i], arr[i - 1]) != 1)
            cout << 1 << " ";
        cout << arr[i] << " ";
    }
}
  
// Driver Function
int main()
{
    int A[] = { 5, 10, 20 };
    int n = sizeof(A) / sizeof(A[0]);
    printResult(A, n);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

//Java program for minimum insertions
// to make a Co-prime Array.
import java.io.*;
  
class GFG {
      
    // Recursive function to return 
    // gcd of a and b
    static int gcd(int a, int b)
    {
        // Everything divides 0 
        if (a == 0 || b == 0)
        return 0;
      
        // base case
        if (a == b)
            return a;
      
        // a is greater
        if (a > b)
            return gcd(a-b, b);
  
        return gcd(a, b-a);
    }
      
    static void printResult(int arr[], int n)
    {
          
        // Counting adjacent pairs that are not
        // co-prime.
        int count = 0;
  
        for (int i = 1; i < n; i++)     
            if (gcd(arr[i], arr[i - 1]) != 1)
                count++;
      
        // No.of insertions
        System.out.println(count ); 
        System.out.print (arr[0] + " ");
  
        for (int i = 1; i < n; i++) 
        {
              
            // If pair is not a co-prime insert 1.
            if (gcd(arr[i], arr[i - 1]) != 1)
                System.out.print( 1 + " ");
            System.out.print(arr[i] + " ");
        }
    }
      
    // Driver Function
    public static void main(String args[])
    {
        int A[] = { 5, 10, 20 };
        int n = A.length;
        printResult(A, n);
    }
}
  
/*This code is contributed by Nikita Tiwari.*/

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 code for minimum insertions
# to make a Co-prime Array.
from fractions import gcd
  
def printResult(arr, n):
  
    # Counting adjacent pairs that 
    # are not co-prime.
    count = 0
    for i in range(1,n):
        if (gcd(arr[i], arr[i - 1]) != 1):
            count+=1
      
    print(count)     # No.of insertions
    print( arr[0], end = " ")
    for i in range(1,n):
          
        # If pair is not a co-prime insert 1.
        if (gcd(arr[i], arr[i - 1]) != 1):
            print(1, end = " ")
        print(arr[i] , end = " ")
          
# Driver Code
A = [ 5, 10, 20 ]
n = len(A)
printResult(A, n) 
  
# This code is contributed by "Sharad_Bhardwaj".

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program for minimum insertions
// to make a Co-prime Array.
using System;
  
class GFG {
  
    // Recursive function to return
    // gcd of a and b
    static int gcd(int a, int b)
    {
        // Everything divides 0
        if (a == 0 || b == 0)
            return 0;
  
        // base case
        if (a == b)
            return a;
  
        // a is greater
        if (a > b)
            return gcd(a - b, b);
  
        return gcd(a, b - a);
    }
  
    static void printResult(int[] arr, int n)
    {
        // Counting adjacent pairs that 
        // are not co-prime.
        int count = 0;
  
        for (int i = 1; i < n; i++)
            if (gcd(arr[i], arr[i - 1]) != 1)
                count++;
  
        // No.of insertions
        Console.WriteLine(count);
        Console.Write(arr[0] + " ");
  
        for (int i = 1; i < n; i++) {
  
            // If pair is not a co-prime insert 1.
            if (gcd(arr[i], arr[i - 1]) != 1)
                Console.Write(1 + " ");
            Console.Write(arr[i] + " ");
        }
    }
  
    // Driver Function
    public static void Main()
    {
        int[] A = { 5, 10, 20 };
        int n = A.Length;
        printResult(A, n);
    }
}
  
/*This code is contributed by vt_m.*/

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program for minimum 
// insertions to make a 
// Co-prime Array. 
  
// Recursive function to 
// return gcd of a and b
function gcd($a, $b)
{
    // Everything divides 0
    if ($a == 0 || $b == 0)
        return 0;
  
    // base case
    if ($a == $b)
        return $a;
  
    // a is greater
    if ($a > $b)
        return gcd($a - $b, $b);
  
    return gcd($a, $b - $a);
}
  
function printResult($arr, $n)
{
    // Counting adjacent pairs 
    // that are not co-prime.
    $count = 0;
  
    for ($i = 1; $i < $n; $i++)
        if (gcd($arr[$i], 
                $arr[$i - 1]) != 1)
            $count++;
  
    // No.of insertions
    echo $count, "\n";
    echo $arr[0] , " ";
  
    for ($i = 1; $i < $n; $i++)
    {
  
        // If pair is not a 
        // co-prime insert 1.
        if (gcd($arr[$i], 
                $arr[$i - 1]) != 1)
            echo 1 , " ";
        echo $arr[$i] , " ";
    }
}
  
// Driver Code
$A = array(5, 10, 20);
$n = sizeof($A);
printResult($A, $n);
  
// This code is contributed
// by ajit
?>

chevron_right



Output:

2
5 1 10 1 20 

Time Complexity : O(n).



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : jit_t, nidhi_biet