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# Find pair with maximum GCD in an array

• Difficulty Level : Hard
• Last Updated : 16 Dec, 2022

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

Method 1 (Brute-force): The simplest method to solve this problem is to use two loops to generate all possible pairs of elements of the array and calculate and compare the GCD at the same time. We can use the Extended Euclidean algorithm for efficiently computing GCD of two numbers.
Time Complexity: O(N^2 * log(max(a, b)))
Here, log(max(a, b)) is the time complexity to calculate GCD of a and b.
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 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

 // 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

 # 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#

 // 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.

## 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 // 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

 // 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 # 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#

 // 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.

## 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

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