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Check if all array elements are pairwise co-prime or not
  • Last Updated : 30 Apr, 2021

Given an array A[] consisting of N positive integers, the task is to check if all the array elements are pairwise co-prime, i.e. for all pairs (Ai , Aj), such that 1<=i<j<=N, GCD(Ai, Aj) = 1.

Examples: 

Input : A[] = {2, 3, 5}
Output : Yes
Explanation : All the pairs, (2, 3), (3, 5), (2, 5) are pairwise co-prime.

Input : A[] = {5, 10}
Output : No
Explanation : GCD(5, 10)=5 so they are not co-prime.

Naive Approach: The simplest approach to solve the problem is to generate all possible pairs from a given array and for each pair, check if it is coprime or not. If any pair is found to be non-coprime, print “No“. Otherwise, print “Yes“.
Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized based on the following observation: 



If any two numbers have a common prime factor, then their GCD can never be 1.  

This can also be interpreted as: 

The LCM of the array must be equal to the product of the elements in the array.

Therefore, the solution boils down to calculating the LCM of the given array and check if it is equal to the product of all the array elements or not.

Below is the implementation of the above approach :

C++




// C++ Program for the above approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
 
// Function to calculate GCD
ll GCD(ll a, ll b)
{
    if (a == 0)
        return b;
    return GCD(b % a, a);
}
 
// Function to calculate LCM
ll LCM(ll a, ll b)
{
    return (a * b)
        / GCD(a, b);
}
 
// Function to check if all elements
// in the array are pairwise coprime
void checkPairwiseCoPrime(int A[], int n)
{
    // Initialze variables
    ll prod = 1;
    ll lcm = 1;
 
    // Itertae over the array
    for (int i = 0; i < n; i++) {
 
        // Calculate product of
        // array elements
        prod *= A[i];
 
        // Calculate LCM of
        // array elements
        lcm = LCM(A[i], lcm);
    }
 
    // If the product of array elements
    // is equal to LCM of the array
    if (prod == lcm)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
}
// Driver Code
int main()
{
    int A[] = { 2, 3, 5 };
    int n = sizeof(A) / sizeof(A[0]);
 
    // Function call
    checkPairwiseCoPrime(A, n);
}

Java




// Java program for the above approach
import java.util.*;
import java.lang.*;
 
class GFG{
 
// Function to calculate GCD
static long GCD(long a, long b)
{
    if (a == 0)
        return b;
         
    return GCD(b % a, a);
}
 
// Function to calculate LCM
static long LCM(long a, long b)
{
    return (a * b) / GCD(a, b);
}
 
// Function to check if all elements
// in the array are pairwise coprime
static void checkPairwiseCoPrime(int A[], int n)
{
     
    // Initialze variables
    long prod = 1;
    long lcm = 1;
 
    // Itertae over the array
    for(int i = 0; i < n; i++)
    {
         
        // Calculate product of
        // array elements
        prod *= A[i];
 
        // Calculate LCM of
        // array elements
        lcm = LCM(A[i], lcm);
    }
     
    // If the product of array elements
    // is equal to LCM of the array
    if (prod == lcm)
        System.out.println("Yes");
    else
        System.out.println("No");
}
 
// Driver Code
public static void main (String[] args)
{
    int A[] = { 2, 3, 5 };
    int n = A.length;
     
    // Function call
    checkPairwiseCoPrime(A, n);
}
}
 
// This code is contributed by offbeat

Python3




# Python3 program for the above approach
 
# Function to calculate GCD
def GCD(a, b):
     
    if (a == 0):
        return b
         
    return GCD(b % a, a)
 
# Function to calculate LCM
def LCM(a, b):
     
    return (a * b) // GCD(a, b)
 
# Function to check if aelements
# in the array are pairwise coprime
def checkPairwiseCoPrime(A, n):
     
    # Initialze variables
    prod = 1
    lcm = 1
 
    # Itertae over the array
    for i in range(n):
 
        # Calculate product of
        # array elements
        prod *= A[i]
 
        # Calculate LCM of
        # array elements
        lcm = LCM(A[i], lcm)
 
    # If the product of array elements
    # is equal to LCM of the array
    if (prod == lcm):
        print("Yes")
    else:
        print("No")
 
# Driver Code
if __name__ == '__main__':
     
    A = [ 2, 3, 5 ]
    n = len(A)
 
    # Function call
    checkPairwiseCoPrime(A, n)
 
# This code is contributed by mohit kumar 29

C#




// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
 
// Function to calculate GCD
static long GCD(long a,
                long b)
{
  if (a == 0)
    return b;
  return GCD(b % a, a);
}
 
// Function to calculate LCM
static long LCM(long a,
                long b)
{
  return (a * b) / GCD(a, b);
}
 
// Function to check if all elements
// in the array are pairwise coprime
static void checkPairwiseCoPrime(int []A,
                                 int n)
{    
  // Initialze variables
  long prod = 1;
  long lcm = 1;
 
  // Itertae over the array
  for(int i = 0; i < n; i++)
  {
    // Calculate product of
    // array elements
    prod *= A[i];
 
    // Calculate LCM of
    // array elements
    lcm = LCM(A[i], lcm);
  }
 
  // If the product of array elements
  // is equal to LCM of the array
  if (prod == lcm)
    Console.WriteLine("Yes");
  else
    Console.WriteLine("No");
}
 
// Driver Code
public static void Main(String[] args)
{
  int []A = {2, 3, 5};
  int n = A.Length;
 
  // Function call
  checkPairwiseCoPrime(A, n);
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
// javascript program for the above approach   
// Function to calculate GCD
    function GCD(a , b) {
        if (a == 0)
            return b;
 
        return GCD(b % a, a);
    }
 
    // Function to calculate LCM
    function LCM(a , b) {
        return (a * b) / GCD(a, b);
    }
 
    // Function to check if all elements
    // in the array are pairwise coprime
    function checkPairwiseCoPrime(A , n) {
 
        // Initialze variables
        var prod = 1;
        var lcm = 1;
 
        // Itertae over the array
        for (i = 0; i < n; i++) {
 
            // Calculate product of
            // array elements
            prod *= A[i];
 
            // Calculate LCM of
            // array elements
            lcm = LCM(A[i], lcm);
        }
 
        // If the product of array elements
        // is equal to LCM of the array
        if (prod == lcm)
            document.write("Yes");
        else
            document.write("No");
    }
 
    // Driver Code
     
        var A = [ 2, 3, 5 ];
        var n = A.length;
 
        // Function call
        checkPairwiseCoPrime(A, n);
 
// This code contributed by umadevi9616
</script>
Output: 
Yes

Time Complexity: O(N log (min(A[i])))
Auxiliary Space: O(1)

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