We are given an array of positive integers. Find the pair in array with maximum GCD.
Examples:
Input : arr[] : { 1 2 3 4 5 }
Output : 2
Explanation : Pair {2, 4} has GCD 2 which is highest. Other pairs have a GCD of 1.
Input : arr[] : { 2 3 4 8 8 11 12 }
Output : 8
Explanation : Pair {8, 8} has GCD 8 which is highest.
Brute Force Approach:
The brute force approach to solve this problem is to generate all possible pairs of elements from the array and calculate their GCD. Then, we can find the pair with the maximum GCD among these pairs.
Below is the implementation of the above approach:
// C++ Code to find pair with // maximum GCD in an array #include <bits/stdc++.h> using namespace std;
// function to find GCD of pair with // max GCD in the array int findMaxGCD( int arr[], int n)
{ int maxGcd = 0;
for ( int i = 0; i < n; i++) {
for ( int j = i + 1; j < n; j++) {
int gcd = __gcd(arr[i], arr[j]);
maxGcd = max(maxGcd, gcd);
}
}
return maxGcd;
} // Driver code int main()
{ // Array in which pair with max GCD
// is to be found
int arr[] = { 1, 2, 4, 8, 8, 12 };
// Size of array
int n = sizeof (arr) / sizeof (arr[0]);
cout << findMaxGCD(arr,n);
return 0;
} |
import java.util.*;
public class Main {
// function to find GCD of pair with
// max GCD in the array
public static int findMaxGCD( int [] arr, int n) {
int maxGcd = 0 ;
for ( int i = 0 ; i < n; i++) {
for ( int j = i + 1 ; j < n; j++) {
int gcd = gcd(arr[i], arr[j]);
maxGcd = Math.max(maxGcd, gcd);
}
}
return maxGcd;
}
// function to calculate GCD of two numbers
public static int gcd( int a, int b) {
if (b == 0 ) {
return a;
}
return gcd(b, a % b);
}
// Driver code
public static void main(String[] args) {
// Array in which pair with max GCD
// is to be found
int [] arr = { 1 , 2 , 4 , 8 , 8 , 12 };
// Size of array
int n = arr.length;
System.out.println(findMaxGCD(arr, n));
}
} |
# Python code to find pair with # maximum GCD in an array import math
# function to find GCD of pair with # max GCD in the array def findMaxGCD(arr, n):
maxGcd = 0
for i in range (n):
for j in range (i + 1 , n):
gcd = math.gcd(arr[i], arr[j])
maxGcd = max (maxGcd, gcd)
return maxGcd
# Driver code if __name__ = = "__main__" :
# Array in which pair with max GCD
# is to be found
arr = [ 1 , 2 , 4 , 8 , 8 , 12 ]
# Size of array
n = len (arr)
print (findMaxGCD(arr,n))
|
using System;
class Program
{ // Function to find GCD of pair with
// max GCD in the array
static int FindMaxGCD( int [] arr)
{
int maxGcd = 0;
int n = arr.Length;
for ( int i = 0; i < n; i++)
{
for ( int j = i + 1; j < n; j++)
{
int gcd = GCD(arr[i], arr[j]);
maxGcd = Math.Max(maxGcd, gcd);
}
}
return maxGcd;
}
// Function to find GCD (Greatest Common Divisor)
static int GCD( int a, int b)
{
while (b != 0)
{
int temp = b;
b = a % b;
a = temp;
}
return a;
}
// Driver code
static void Main()
{
// Array in which pair with max GCD
// is to be found
int [] arr = { 1, 2, 4, 8, 8, 12 };
// Call the function to find max GCD
int maxGCD = FindMaxGCD(arr);
// Print the result
Console.WriteLine(maxGCD);
}
} |
// Function to find GCD of pair with max GCD in the array function findMaxGCD(arr) {
let maxGcd = 0;
const n = arr.length;
for (let i = 0; i < n; i++) {
for (let j = i + 1; j < n; j++) {
const gcd = findGCD(arr[i], arr[j]);
maxGcd = Math.max(maxGcd, gcd);
}
}
return maxGcd;
} // Function to find the GCD of two numbers using Euclidean algorithm function findGCD(a, b) {
if (b === 0) {
return a;
}
return findGCD(b, a % b);
} // Driver code const arr = [1, 2, 4, 8, 8, 12]; const result = findMaxGCD(arr); console.log(result); // Output the maximum GCD
|
8
Time Complexity: O(N^2)
Auxiliary Space: O(1)
Method 2 : (Efficient) In this method, we maintain a count array to store the count of divisors of every element. We will traverse the given array and for every element, we will calculate its divisors and increment at the index of count array. The process of computing divisors will take O(sqrt(arr[i])) time, where arr[i] is element in the given array at index i. After the whole traversal, we can simply traverse the count array from last index to index 1. If we found an index with a value greater than 1, then this means that it is a divisor of 2 elements and also the max GCD.
Below is the implementation of above approach :
// C++ Code to find pair with // maximum GCD in an array #include <bits/stdc++.h> using namespace std;
// function to find GCD of pair with // max GCD in the array int findMaxGCD( int arr[], int n)
{ // Computing highest element
int high = 0;
for ( int i = 0; i < n; i++)
high = max(high, arr[i]);
// Array to store the count of divisors
// i.e. Potential GCDs
int divisors[high + 1] = { 0 };
// Iterating over every element
for ( int i = 0; i < n; i++)
{
// Calculating all the divisors
for ( int j = 1; j <= sqrt (arr[i]); j++)
{
// Divisor found
if (arr[i] % j == 0)
{
// Incrementing count for divisor
divisors[j]++;
// Element/divisor is also a divisor
// Checking if both divisors are
// not same
if (j != arr[i] / j)
divisors[arr[i] / j]++;
}
}
}
// Checking the highest potential GCD
for ( int i = high; i >= 1; i--)
// If this divisor can divide at least 2
// numbers, it is a GCD of at least 1 pair
if (divisors[i] > 1)
return i;
} // Driver code int main()
{ // Array in which pair with max GCD
// is to be found
int arr[] = { 1, 2, 4, 8, 8, 12 };
// Size of array
int n = sizeof (arr) / sizeof (arr[0]);
cout << findMaxGCD(arr,n);
return 0;
} |
// JAVA Code for Find pair with maximum GCD in an array public class GFG {
// function to find GCD of pair with
// max GCD in the array
public static int findMaxGCD( int arr[], int n)
{
// Computing highest element
int high = 0 ;
for ( int i = 0 ; i < n; i++)
high = Math.max(high, arr[i]);
// Array to store the count of divisors
// i.e. Potential GCDs
int divisors[] = new int [high + 1 ];
// Iterating over every element
for ( int i = 0 ; i < n; i++)
{
// Calculating all the divisors
for ( int j = 1 ; j <= Math.sqrt(arr[i]); j++)
{
// Divisor found
if (arr[i] % j == 0 )
{
// Incrementing count for divisor
divisors[j]++;
// Element/divisor is also a divisor
// Checking if both divisors are
// not same
if (j != arr[i] / j)
divisors[arr[i] / j]++;
}
}
}
// Checking the highest potential GCD
for ( int i = high; i >= 1 ; i--)
// If this divisor can divide at least 2
// numbers, it is a GCD of at least 1 pair
if (divisors[i] > 1 )
return i;
return 1 ;
}
/* Driver program to test above function */
public static void main(String[] args)
{
// Array in which pair with max GCD
// is to be found
int arr[] = { 1 , 2 , 4 , 8 , 8 , 12 };
// Size of array
int n = arr.length;
System.out.println(findMaxGCD(arr,n));
}
}
// This code is contributed by Arnav Kr. Mandal. |
# Python program to Find pair with # maximum GCD in an array import math
# function to find GCD of pair with # max GCD in the array def findMaxGCD(arr, n):
# Computing highest element
high = 0
i = 0
while i < n:
high = max (high, arr[i])
i = i + 1
# Array to store the count of divisors
# i.e. Potential GCDs
divisors = [ 0 ] * (high + 1 )
# Iterating over every element
i = 0
while i < n:
# Calculating all the divisors
j = 1
while j < = math.sqrt(arr[i]):
# Divisor found
if (arr[i] % j = = 0 ):
# Incrementing count for divisor
divisors[j] = divisors[j] + 1
# Element/divisor is also a divisor
# Checking if both divisors are
# not same
if (j ! = arr[i] / j):
divisors[arr[i] / j] = divisors[arr[i] / j] + 1
j = j + 1
i = i + 1
# Checking the highest potential GCD
i = high
while i > = 1 :
# If this divisor can divide at least 2
# numbers, it is a GCD of at least 1 pair
if (divisors[i] > 1 ):
return i
i = i - 1
return 1
# Driver code # Array in which pair with max GCD # is to be found arr = [ 1 , 2 , 4 , 8 , 8 , 12 ]
# Size of array n = len (arr)
print findMaxGCD(arr, n)
# This code is contributed by Nikita Tiwari. |
// C# Code for Find pair with // maximum GCD in an array using System;
class GFG {
// Function to find GCD of pair
// with max GCD in the array
public static int findMaxGCD( int []arr,
int n)
{
// Computing highest element
int high = 0;
for ( int i = 0; i < n; i++)
high = Math.Max(high, arr[i]);
// Array to store the count of
// divisors i.e. Potential GCDs
int []divisors = new int [high + 1];
// Iterating over every element
for ( int i = 0; i < n; i++)
{
// Calculating all the divisors
for ( int j = 1; j <=
Math.Sqrt(arr[i]); j++)
{
// Divisor found
if (arr[i] % j == 0)
{
// Incrementing count
// for divisor
divisors[j]++;
// Element / divisor is also
// a divisor Checking if both
// divisors are not same
if (j != arr[i] / j)
divisors[arr[i] / j]++;
}
}
}
// Checking the highest potential GCD
for ( int i = high; i >= 1; i--)
// If this divisor can divide at
// least 2 numbers, it is a
// GCD of at least 1 pair
if (divisors[i] > 1)
return i;
return 1;
}
// Driver Code
public static void Main(String []args)
{
// Array in which pair with
// max GCD is to be found
int []arr = {1, 2, 4, 8, 8, 12};
// Size of array
int n = arr.Length;
Console.WriteLine(findMaxGCD(arr,n));
}
} // This code is contributed by vt_m. |
<script> // JavaScript Code for Find pair with // maximum GCD in an array // function to find GCD of pair with // max GCD in the array function findMaxGCD(arr , n)
{ // Computing highest element
var high = 0;
for ( var i = 0; i < n; i++)
high = Math.max(high, arr[i]);
// Array to store the count of divisors
// i.e. Potential GCDs
var divisors =
Array.from({length: high + 1}, (_, i) => 0);
// Iterating over every element
for ( var i = 0; i < n; i++)
{
// Calculating all the divisors
for ( var j = 1; j <= Math.sqrt(arr[i]); j++)
{
// Divisor found
if (arr[i] % j == 0)
{
// Incrementing count for divisor
divisors[j]++;
// Element/divisor is also a divisor
// Checking if both divisors are
// not same
if (j != arr[i] / j)
divisors[arr[i] / j]++;
}
}
}
// Checking the highest potential GCD
for ( var i = high; i >= 1; i--)
// If this divisor can divide at least 2
// numbers, it is a GCD of at least 1 pair
if (divisors[i] > 1)
return i;
return 1;
} /* Driver program to test above function */ // Array in which pair with max GCD
// is to be found var arr = [ 1, 2, 4, 8, 8, 12 ];
// Size of array
var n = arr.length;
document.write(findMaxGCD(arr,n));
// This code contributed by shikhasingrajput </script> |
<?php // PHP Code for Find pair with // maximum GCD in an array // Function to find GCD // of pair with max GCD // in the array function findMaxGCD( $arr , $n )
{ // Computing highest element
$high = 0;
for ( $i = 0; $i < $n ; $i ++)
$high = max( $high , $arr [ $i ]);
// Array to store the
// count of divisors
// i.e. Potential GCDs
$divisors = array_fill (0, $high + 1, 0);
// Iterating over every element
for ( $i = 0; $i < $n ; $i ++)
{
// Calculating all
// the divisors
for ( $j = 1;
$j <= (int)(sqrt( $arr [ $i ])); $j ++)
{
// Divisor found
if ( $arr [ $i ] % $j == 0)
{
// Incrementing count
// for divisor
$divisors [ $j ]++;
// Element/divisor is also
// a divisor Checking if
// both divisors are not same
if ( $j != (int)( $arr [ $i ] / $j ))
$divisors [(int)( $arr [ $i ] / $j )]++;
}
}
}
// Checking the highest
// potential GCD
for ( $i = $high ; $i >= 1; $i --)
// If this divisor can divide
// at least 2 numbers, it is
// a GCD of at least 1 pair
if ( $divisors [ $i ] > 1)
return $i ;
} // Driver code // Array in which pair // with max GCD is to // be found $arr = array ( 1, 2, 4, 8, 8, 12 );
// Size of array $n = sizeof( $arr );
echo findMaxGCD( $arr , $n );
// This code is contributed by mits ?> |
8
Time Complexity: O(N * sqrt(arr[i]) + H) , where arr[i] denotes the element of the array and H denotes the largest number of the array.
Auxiliary Space: O(high), high is the maximum element in the array
Method 3 (Most Efficient): This approach is based on the idea of Sieve Of Eratosthenes.
First let’s solve a simpler problem, given a value X we have to tell whether a pair has a GCD equal to X. This can be done by checking that how many elements in the array are multiples of X. If the number of such multiples is greater than 1, then X will be a GCD of some pair.
Now for pair with maximum GCD, we maintain a count array of the original array. Our method is based on the above problem with Sieve-like approach for loop. Below is the step by step algorithm of this approach:
- Iterate ‘i’ from MAX (maximum array element) to 1.
- Iterate ‘j’ from ‘i’ to MAX. We will check if the count array is 1 at index ‘j’.
-
Increment the index ‘j’ everytime with ‘i’. This way, we can check for
i, 2i, 3i, and so on. - If we get 1 two times at count array that means 2 multiples of i exists. This makes it the highest GCD.
Below is the implementation of above approach :
// C++ Code to // Find pair with // maximum GCD in // an array #include <bits/stdc++.h> using namespace std;
// function to find // GCD of pair with // max GCD in the // array int findMaxGCD( int arr[], int n)
{ // Calculating MAX in array
int high = 0;
for ( int i = 0; i < n; i++)
high = max(high, arr[i]);
// Maintaining count array
int count[high + 1] = {0};
for ( int i = 0; i < n; i++)
count[arr[i]]++;
// Variable to store the
// multiples of a number
int counter = 0;
// Iterating from MAX to 1
// GCD is always between
// MAX and 1. The first
// GCD found will be the
// highest as we are
// decrementing the potential
// GCD
for ( int i = high; i >= 1; i--)
{
int j = i;
counter = 0;
// Iterating from current
// potential GCD
// till it is less than
// MAX
while (j <= high)
{
// A multiple found
if (count[j] >=2)
return j;
else if (count[j] == 1)
counter++;
// Incrementing potential
// GCD by itself
// To check i, 2i, 3i....
j += i;
// 2 multiples found,
// max GCD found
if (counter == 2)
return i;
}
}
} // Driver code int main()
{ // Array in which pair
// with max GCD is to
// be found
int arr[] = { 1, 2, 4, 8, 8, 12 };
// Size of array
int n = sizeof (arr) / sizeof (arr[0]);
cout << findMaxGCD(arr, n);
return 0;
} |
// Java Code to // Find pair with // maximum GCD in // an array class GFG {
// function to find
// GCD of pair with
// max GCD in the
// array
public static int findMaxGCD( int arr[], int n)
{
// Calculating MAX in
// array
int high = 0 ;
for ( int i = 0 ; i < n; i++)
high = Math.max(high, arr[i]);
// Maintaining count array
int count[]= new int [high + 1 ];
for ( int i = 0 ; i < n; i++)
count[arr[i]]++;
// Variable to store
// the multiples of
// a number
int counter = 0 ;
// Iterating from MAX
// to 1 GCD is always
// between MAX and 1
// The first GCD found
// will be the highest
// as we are decrementing
// the potential GCD
for ( int i = high; i >= 1 ; i--)
{
int j = i;
// Iterating from current
// potential GCD till it
// is less than MAX
while (j <= high)
{
// A multiple found
if (count[j]> 0 )
counter+=count[j];
// Incrementing potential
// GCD by itself
// To check i, 2i, 3i....
j += i;
// 2 multiples found,
// max GCD found
if (counter == 2 )
return i;
}
counter= 0 ;
}
return 1 ;
}
/* Driver program to test above function */
public static void main(String[] args)
{
// Array in which pair
// with max GCD is to
// be found
int arr[] = { 1 , 2 , 4 , 8 , 8 , 12 };
// Size of array
int n = arr.length;
System.out.println(findMaxGCD(arr,n));
}
} // This code is contributed by Arnav Kr. Mandal. |
# Python3 Code to # Find pair with # maximum GCD in # an array # function to find # GCD of pair with # max GCD in the # array def findMaxGCD(arr, n) :
# Calculating MAX in
# array
high = 0
for i in range ( 0 , n) :
high = max (high, arr[i])
# Maintaining count array
count = [ 0 ] * (high + 1 )
for i in range ( 0 , n) :
count[arr[i]] + = 1
# Variable to store the
# multiples of a number
counter = 0
# Iterating from MAX
# to 1 GCD is always
# between MAX and 1
# The first GCD found
# will be the highest
# as we are decrementing
# the potential GCD
for i in range (high, 0 , - 1 ) :
j = i
# Iterating from current
# potential GCD till it
# is less than MAX
while (j < = high) :
# A multiple found
if (count[j] > 0 ) :
counter + = count[j]
# Incrementing potential
# GCD by itself
# To check i, 2i, 3i....
j + = i
# 2 multiples found,
# max GCD found
if (counter = = 2 ) :
return i
counter = 0
# Driver code # Array in which pair # with max GCD is to # be found arr = [ 1 , 2 , 4 , 8 , 8 , 12 ]
# Size of array n = len (arr)
print (findMaxGCD(arr, n))
#This code is contributed by Nikita Tiwari. |
// C# Code to find pair with // maximum GCD in an array using System;
class GFG {
// function to find GCD
// of pair with max
// max GCD in the array
public static int findMaxGCD( int []arr,
int n)
{
// Calculating Max
// in array
int high = 0;
for ( int i = 0; i < n; i++)
high = Math.Max(high, arr[i]);
// Maintaining count array
int []count= new int [high + 1];
for ( int i = 0; i < n; i++)
count[arr[i]]++;
// Variable to store
// the multiples of
// a number
int counter = 0;
// Iterating from MAX
// to 1 GCD is always
// between MAX and 1
// The first GCD found
// will be the highest
// as we are decrementing
// the potential GCD
for ( int i = high; i >= 1; i--)
{
int j = i;
// Iterating from current
// potential GCD till it
// is less than MAX
while (j <= high)
{
// A multiple found
if (count[j]>0)
counter+=count[j];
// Incrementing potential
// GCD by itself
// To check i, 2i, 3i....
j += i;
// 2 multiples found,
// max GCD found
if (counter == 2)
return i;
}
counter=0;
}
return 1;
}
// Driver Code
public static void Main(String []args)
{
// Array in which pair
// with max GCD is to
// be found
int []arr = {1, 2, 4, 8, 8, 12};
// Size of array
int n = arr.Length;
Console.WriteLine(findMaxGCD(arr,n));
}
} // This code is contributed by vt_m. |
<script> // javascript Code to // Find pair with // maximum GCD in // an array // function to find
// GCD of pair with
// max GCD in the
// array
function findMaxGCD(arr , n)
{
// Calculating MAX in
// array
var high = 0;
for (let i = 0; i < n; i++)
high = Math.max(high, arr[i]);
// Maintaining count array
var count = Array(high + 1).fill(0);
for (let i = 0; i < n; i++)
count[arr[i]]++;
// Variable to store
// the multiples of
// a number
var counter = 0;
// Iterating from MAX
// to 1 GCD is always
// between MAX and 1
// The first GCD found
// will be the highest
// as we are decrementing
// the potential GCD
for (let i = high; i >= 1; i--)
{
var j = i;
// Iterating from current
// potential GCD till it
// is less than MAX
while (j <= high)
{
// A multiple found
if (count[j] > 0)
counter += count[j];
// Incrementing potential
// GCD by itself
// To check i, 2i, 3i....
j += i;
// 2 multiples found,
// max GCD found
if (counter == 2)
return i;
}
counter = 0;
}
return 1;
}
/* Driver program to test above function */
// Array in which pair
// with max GCD is to
// be found
var arr = [ 1, 2, 4, 8, 8, 12 ];
// Size of array
var n = arr.length;
document.write(findMaxGCD(arr, n));
// This code is contributed by aashish1995 </script> |
<?php // PHP Code to Find pair with maximum // GCD in an array // function to find GCD of pair with // max GCD in the array function findMaxGCD( $arr , $n )
{ // Calculating MAX in array
$high = 0;
for ( $i = 0; $i < $n ; $i ++)
$high = max( $high , $arr [ $i ]);
// Maintaining count array
$count = array_fill (0, $high + 1, 0);
for ( $i = 0; $i < $n ; $i ++)
$count [ $arr [ $i ]]++;
// Variable to store the multiples
// of a number
$counter = 0;
// Iterating from MAX to 1 GCD is always
// between MAX and 1. The first GCD found
// will be the highest as we are decrementing
// the potential GCD
for ( $i = $high ; $i >= 1; $i --)
{
$j = $i ;
$counter = 0;
// Iterating from current potential GCD
// till it is less than MAX
while ( $j <= $high )
{
// A multiple found
if ( $count [ $j ] >= 2)
return $j ;
else if ( $count [ $j ] == 1)
$counter ++;
// Incrementing potential GCD by itself
// To check i, 2i, 3i....
$j += $i ;
// 2 multiples found, max GCD found
if ( $counter == 2)
return $i ;
}
}
} // Driver code // Array in which pair with max GCD // is to be found $arr = array ( 1, 2, 4, 8, 8, 12 );
// Size of array $n = count ( $arr );
print (findMaxGCD( $arr , $n ));
// This code is contributed by mits ?> |
8
Time Complexity: The time complexity of this approach is till an open problem known as the Dirichlet divisor problem.
Time Complexity: O(high2) , high is the maximum element in the array
Auxiliary Space: O(high), high is the maximum element in the array