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# Find pair with maximum GCD in an array

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.
Recommended Practice

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++

 // C++ Code to find pair with// maximum GCD in an array#include  using namespace std; // function to find GCD of pair with// max GCD in the arrayint 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 codeint 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;}

## Python3

 # Python code to find pair with# maximum GCD in an array import math # function to find GCD of pair with# max GCD in the arraydef 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 codeif __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))

## Java

 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));    }}

Output

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++

 // C++ Code to find pair with// maximum GCD in an array#include  using namespace std; // function to find GCD of pair with// max GCD in the arrayint 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 codeint 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

 // JAVA Code for Find pair with maximum GCD in an arraypublic 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

 # Python program to Find pair with# maximum GCD in an arrayimport 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 foundarr = [1, 2, 4, 8, 8, 12] # Size of arrayn = len(arr) print findMaxGCD(arr, n) # This code is contributed by Nikita Tiwari.

## C#

 // C# Code for Find pair with// maximum GCD in an arrayusing 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.

## PHP

 = 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?>

## Javascript



Output

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:

1. Iterate ‘i’ from MAX (maximum array element) to 1.
2. Iterate ‘j’ from ‘i’ to MAX. We will check if the count array is 1 at index ‘j’.
3. Increment the index ‘j’ everytime with ‘i’. This way, we can check for
i, 2i, 3i, and so on.
4. 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++

 // C++ Code to// Find pair with// maximum GCD in// an array#include using namespace std; // function to find// GCD of pair with// max GCD in the// arrayint 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 codeint 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

 // 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

 # Python3 Code to# Find pair with# maximum GCD in# an array # function to find# GCD of pair with# max GCD in the# arraydef 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 foundarr = [1, 2, 4, 8, 8, 12]# Size of arrayn = len(arr)print(findMaxGCD(arr, n)) #This code is contributed by Nikita Tiwari.

## C#

 // C# Code to find pair with// maximum GCD in an arrayusing 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.

## PHP

 = 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?>

## Javascript



Output

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

This article is contributed by Aarti_Rathi and Rohit Thapliyal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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