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, 
.
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 Functionint 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 gcddef 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 CodeA = [ 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 bfunction 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 bfunction 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 codelet 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(Ai)), for using __gcd function
Auxiliary Space: O(1)
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