Minimum insertions to make a Co-prime array

Last Updated : 10 Oct, 2022

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++

// 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;
}

Java

//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.*/

Python3

# 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".

C#

// 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.*/

PHP

<?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
?>

Javascript

<script>
 
// Javascript 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.
    let count = 0;
 
    for(let i = 1; i < n; i++)
        if (gcd(arr[i], arr[i - 1]) != 1)
            count++;
 
    // No.of insertions
    document.write(count + "</br>");
    document.write(arr[0] + " ");
 
    for(let i = 1; i < n; i++) 
    {
         
        // If pair is not a co-prime insert 1.
        if (gcd(arr[i], arr[i - 1]) != 1)
            document.write(1 + " ");
             
        document.write(arr[i] + " ");
    }
}
 
// Driver code
let A = [ 5, 10, 20 ];
let n = A.length;
 
printResult(A, n);
 
// This code is contributed by suresh07 
 
</script>

Output

2 5 1 10 1 20

Time Complexity: O(n log(A_i)), for using __gcd function
Auxiliary Space: O(1)

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Minimum insertions to make a Co-prime array

C++

Java

Python3

C#

PHP

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