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++ 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)
{ // Initialize variables
ll prod = 1;
ll lcm = 1;
// Iterate 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 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)
{ // Initialize variables
long prod = 1 ;
long lcm = 1 ;
// Iterate 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 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):
# Initialize variables
prod = 1
lcm = 1
# Iterate 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# 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)
{ // Initialize variables
long prod = 1;
long lcm = 1;
// Iterate 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 |
<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) {
// Initialize variables
var prod = 1;
var lcm = 1;
// Iterate 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> |
Yes
Time Complexity: O(N log (min(A[i])))
Auxiliary Space: O(1)
Approach#2:using nested loop
Algorithm
1. Define a function named are_elements_pairwise_coprime that takes an array arr as input.
2.Initialize a variable n to the length of the array arr.
3.Loop over all pairs of elements in the array using nested loops, i.e., for each element arr[i], loop over all elements arr[j] where j > i.
4.For each pair of elements, compute their GCD using the gcd function defined below.
5.If the GCD of any pair of elements is not equal to 1, return “No”.
6.If all pairs of elements have a GCD equal to 1, return “Yes”.
#include <iostream> using namespace std;
// Function to calculate the GCD of two numbers using Euclidean algorithm int gcd( int a, int b) {
// Base case: GCD is the non-zero value
if (b == 0) {
return a;
} else {
// Recursive call to find GCD
return gcd(b, a % b);
}
} // Function to check if elements in the array are pairwise coprime // Returns "Yes" if all pairs are coprime, "No" otherwise string areElementsPairwiseCoprime( int arr[], int n) {
// Iterate through all pairs of elements in the array
for ( int i = 0; i < n; ++i) {
for ( int j = i + 1; j < n; ++j) {
// Check if the GCD (Greatest Common Divisor) of the pair is not 1
if (gcd(arr[i], arr[j]) != 1) {
return "No" ;
}
}
}
return "Yes" ;
} int main() {
// Example array
int arr[] = {2, 3, 5};
int n = sizeof (arr) / sizeof (arr[0]);
// Check if elements in the array are pairwise coprime
cout << areElementsPairwiseCoprime(arr, n) << endl;
return 0;
} |
import java.util.Arrays;
public class Main {
// Function to calculate the GCD of two numbers using Euclidean algorithm
static int gcd( int a, int b) {
// Base case: GCD is the non-zero value
if (b == 0 ) {
return a;
} else {
// Recursive call to find GCD
return gcd(b, a % b);
}
}
// Function to check if elements in the array are pairwise coprime
// Returns "Yes" if all pairs are coprime, "No" otherwise
static String areElementsPairwiseCoprime( int [] arr) {
int n = arr.length;
// Iterate through all pairs of elements in the array
for ( int i = 0 ; i < n; ++i) {
for ( int j = i + 1 ; j < n; ++j) {
// Check if the GCD (Greatest Common Divisor) of the pair is not 1
if (gcd(arr[i], arr[j]) != 1 ) {
return "No" ;
}
}
}
return "Yes" ;
}
public static void main(String[] args) {
// Example array
int [] arr = { 2 , 3 , 5 };
// Check if elements in the array are pairwise coprime
System.out.println(areElementsPairwiseCoprime(arr));
}
} |
def are_elements_pairwise_coprime(arr):
n = len (arr)
for i in range (n):
for j in range (i + 1 , n):
if gcd(arr[i], arr[j]) ! = 1 :
return "No"
return "Yes"
def gcd(a, b):
if b = = 0 :
return a
else :
return gcd(b, a % b)
arr = [ 2 , 3 , 5 ]
print (are_elements_pairwise_coprime(arr))
|
using System;
class Program {
// Function to calculate the GCD of two numbers using
// Euclidean algorithm
static int GCD( int a, int b)
{
// Base case: GCD is the non-zero value
if (b == 0) {
return a;
}
else {
// Recursive call to find GCD
return GCD(b, a % b);
}
}
// Function to check if elements in the array are
// pairwise coprime Returns "Yes" if all pairs are
// coprime, "No" otherwise
static string AreElementsPairwiseCoprime( int [] arr,
int n)
{
// Iterate through all pairs of elements in the
// array
for ( int i = 0; i < n; ++i) {
for ( int j = i + 1; j < n; ++j) {
// Check if the GCD (Greatest Common
// Divisor) of the pair is not 1
if (GCD(arr[i], arr[j]) != 1) {
return "No" ;
}
}
}
return "Yes" ;
}
static void Main()
{
// Example array
int [] arr = { 2, 3, 5 };
int n = arr.Length;
// Check if elements in the array are pairwise
// coprime
Console.WriteLine(
AreElementsPairwiseCoprime(arr, n));
}
} |
// Function to calculate the GCD of two numbers using Euclidean algorithm function gcd(a, b) {
// Base case: GCD is the non-zero value
if (b === 0) {
return a;
} else {
// Recursive call to find GCD
return gcd(b, a % b);
}
} // Function to check if elements in the array are pairwise coprime // Returns "Yes" if all pairs are coprime, "No" otherwise function areElementsPairwiseCoprime(arr) {
// Iterate through all pairs of elements in the array
for (let i = 0; i < arr.length; ++i) {
for (let j = i + 1; j < arr.length; ++j) {
// Check if the GCD (Greatest Common Divisor) of the pair is not 1
if (gcd(arr[i], arr[j]) !== 1) {
return "No" ;
}
}
}
return "Yes" ;
} // Example array const arr = [2, 3, 5]; // Check if elements in the array are pairwise coprime console.log(areElementsPairwiseCoprime(arr)); |
Yes
The time complexity of this algorithm is O(n^2), where n is the length of the input array, since we are checking all possible pairs of elements.
The space complexity is O(1), since we are not using any additional data structures.