Convert the array such that the GCD of the array becomes 1

Given an array of positive elements and a positive integer k, the task is to convert the GCD of the array to 1. To make it possible, only one operation is allowed any number of times i.e. choose any element of the array & divide with a number d where d <= k.

Examples:

Input : arr = {10, 15, 30}, k = 6
Output : Yes
Divide all the elements of array by 5 as it is
the divisor of all elements and also less than k i.e. 6.
It gives as the sequence like {2, 3, 6}. Now, divide
2 by 2, 3 by 3 and 6 by 2. Then, the sequence
becomes {1, 1, 3}. Now, the gcd of the array is 1.

Input : arr = {5, 10, 20}, k = 4
Output : No
Here, divide 10 with 2 the sequence becomes
{5, 5, 20}. Then, divide 20 by 2 and again by 2.
Finally, the sequence becomes {5, 5, 5}. To make
the gcd of this sequence 1, divide each element
by 5. But 5 > 4 so here it is impossible to make the
gcd 1.

Approach:

If there exists a positive prime number greater than k that divides each element of the array then the answer is No.
If the largest prime factor of the GCD of the array is less than or equal to k then the answer is Yes.
First, find the GCD of the array then check if there exists a prime factor of the GCD that is greater than k.
For this, calculate largest prime factor of GCD.

C++

// C++ program to check if it is 
// possible to convert the gcd of 
// the array to 1 by applying the 
// given operation 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to get gcd of the array. 
int getGcd(int* arr, int n) 
{ 
    int gcd = arr[0]; 
    for (int i = 1; i < n; i++) 
        gcd = __gcd(arr[i], gcd); 
  
    return gcd; 
} 
  
// Function to check if it is possible. 
bool convertGcd(int* arr, int n, int k) 
{ 
    // Getting the gcd of array 
    int gcd = getGcd(arr, n); 
  
    // Initially taking max_prime factor is 1. 
    int max_prime = 1; 
  
    // find maximum of all the prime factors 
    // till sqrt(gcd). 
    for (int i = 2; i <= sqrt(gcd); i++) { 
        while (gcd % i == 0) { 
            gcd /= i; 
            max_prime = max(max_prime, i); 
        } 
    } 
  
    // either GCD is reduced to 1 or a prime factor 
    // greater than sqrt(gcd) 
    max_prime = max(max_prime, gcd); 
  
    return (max_prime <= k); 
} 
  
// Drivers code 
int main() 
{ 
    int arr[] = { 10, 15, 30 }; 
    int k = 6; 
    int n = sizeof(arr) / sizeof(arr[0]); 
  
    if (convertGcd(arr, n, k) == true) 
       cout << "Yes"; 
    else
       cout << "No"; 
  
    return 0; 
} 

Java

// Java program to check if it is 
// possible to convert the gcd of 
// the array to 1 by applying the 
// given operation 
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); 
    } 
      
    // Function to get gcd of the array. 
    static int getGcd(int arr[], int n) 
    { 
        int gcd = arr[0]; 
        for (int i = 1; i < n; i++) 
            gcd = __gcd(arr[i], gcd); 
      
        return gcd; 
    } 
  
    // Function to check if it is possible. 
    static boolean convertGcd(int []arr,  
                             int n, int k) 
    { 
        // Getting the gcd of array 
        int gcd = getGcd(arr, n); 
      
        // Initially taking max_prime 
        // factor is 1. 
        int max_prime = 1; 
      
        // find maximum of all the prime  
        // factors till sqrt(gcd). 
        for (int i = 2; i <= Math.sqrt(gcd); 
                                        i++) 
        { 
            while (gcd % i == 0) { 
                gcd /= i; 
                max_prime =  
                     Math.max(max_prime, i); 
            } 
        } 
      
        // either GCD is reduced to 1 or a 
        // prime factor greater than sqrt(gcd) 
        max_prime = Math.max(max_prime, gcd); 
      
        return (max_prime <= k); 
    } 
  
    // Drivers code 
    public static void main (String[] args) 
    { 
        int []arr = { 10, 15, 30 }; 
        int k = 6; 
        int n = arr.length; 
      
        if (convertGcd(arr, n, k) == true) 
            System.out.println( "Yes"); 
        else
            System.out.println( "No"); 
    } 
} 
  
// This code is contributed by anuj_67. 

Python3

# Python 3 program to check if it is 
# possible to convert the gcd of 
# the array to 1 by applying the 
# given operation 
from math import gcd as __gcd, sqrt 
  
# Function to get gcd of the array. 
def getGcd(arr, n): 
    gcd = arr[0]; 
    for i in range(1, n, 1): 
        gcd = __gcd(arr[i], gcd) 
  
    return gcd 
  
# Function to check if it is possible. 
def convertGcd(arr, n, k): 
      
    # Getting the gcd of array 
    gcd = getGcd(arr, n) 
  
    # Initially taking max_prime  
    # factor is 1. 
    max_prime = 1
  
    # find maximum of all the  
    # prime factors till sqrt(gcd) 
    p = int(sqrt(gcd)) + 1
    for i in range(2, p, 1): 
        while (gcd % i == 0): 
            gcd = int(gcd / i) 
            max_prime = max(max_prime, i) 
  
    # either GCD is reduced to 1 or a  
    # prime factor greater than sqrt(gcd) 
    max_prime = max(max_prime, gcd) 
  
    return (max_prime <= k) 
  
# Drivers code 
if __name__ == '__main__': 
    arr = [10, 15, 30] 
    k = 6
    n = len(arr) 
  
    if (convertGcd(arr, n, k) == True): 
        print("Yes") 
    else: 
        print("No") 
  
# This code is contributed by 
# Sahil_Shelangia 

C#

// C# program to check if it is 
// possible to convert the gcd of 
// the array to 1 by applying the 
// given operation 
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); 
    } 
      
    // Function to get gcd of the array. 
    static int getGcd(int []arr, int n) 
    { 
        int gcd = arr[0]; 
        for (int i = 1; i < n; i++) 
            gcd = __gcd(arr[i], gcd); 
      
        return gcd; 
    } 
  
    // Function to check if it is possible. 
    static bool convertGcd(int []arr,  
                            int n, int k) 
    { 
        // Getting the gcd of array 
        int gcd = getGcd(arr, n); 
      
        // Initially taking max_prime 
        // factor is 1. 
        int max_prime = 1; 
      
        // find maximum of all the prime  
        // factors till sqrt(gcd). 
        for (int i = 2; i <= Math.Sqrt(gcd); 
                                        i++) 
        { 
            while (gcd % i == 0) { 
                gcd /= i; 
                max_prime =  
                    Math.Max(max_prime, i); 
            } 
        } 
      
        // either GCD is reduced to 1 or a 
        // prime factor greater than sqrt(gcd) 
        max_prime = Math.Max(max_prime, gcd); 
      
        return (max_prime <= k); 
    } 
  
    // Drivers code 
    public static void Main () 
    { 
        int []arr = { 10, 15, 30 }; 
        int k = 6; 
        int n = arr.Length; 
      
        if (convertGcd(arr, n, k) == true) 
            Console.WriteLine( "Yes"); 
        else
            Console.WriteLine( "No"); 
    } 
} 
  
// This code is contributed by anuj_67. 

PHP

<?php 
// PHP program to check if it is 
// possible to convert the gcd of 
// the array to 1 by applying the 
// given operation 
  
// Recursive function to  
// return gcd of a and b 
function __gcd($a, $b) 
{ 
      
    // Everything divides 0 
    if($a == 0 or $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 to get gcd of the array. 
function getGcd($arr, $n) 
{ 
    $gcd = $arr[0]; 
    for ($i = 1; $i < $n; $i++) 
        $gcd = __gcd($arr[$i], $gcd); 
  
    return $gcd; 
} 
  
// Function to check if it is possible. 
function convertGcd( $arr, $n, $k) 
{ 
      
    // Getting the gcd of array 
    $gcd = getGcd($arr, $n); 
  
    // Initially taking max_prime 
    // factor is 1. 
    $max_prime = 1; 
  
    // find maximum of all the prime 
    // factors till sqrt(gcd). 
    for($i = 2; $i <= sqrt($gcd); $i++) 
    { 
        while ($gcd % $i == 0) 
        { 
            $gcd /= $i; 
            $max_prime = max($max_prime, $i); 
        } 
    } 
  
    // either GCD is reduced  
    // to 1 or a prime factor 
    // greater than sqrt(gcd) 
    $max_prime = max($max_prime, $gcd); 
  
    return ($max_prime <= $k); 
} 
  
    // Driver Code 
    $arr = array(10, 15, 30); 
    $k = 6; 
    $n = count($arr); 
  
    if (convertGcd($arr, $n, $k) == true) 
        echo "Yes"; 
    else
        echo "No"; 
  
// This code is contributed by anuj_67. 
?> 

Output:

Yes

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