Minimum insertions to make a Coprime array (original) (raw)

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).

Try It Yourself

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

JavaScript

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