GFG App
Open App
Browser
Continue

# Pair with maximum GCD from two arrays

Given two arrays of n integers with values of the array being small (values never exceed a small number say 100). Find the pair(x, y) which has maximum gcd. x and y cannot be of the same array. If multiple pairs have the same gcd, then consider the pair which has the maximum sum.
Examples:

Input : a[] = {3, 1, 4, 2, 8}
b[] = {5, 2, 12, 8, 3}
Output : 8 8
Explanation: The maximum gcd is 8 which is
of pair(8, 8).

Input: a[] = {2, 3, 5}
b[] = {7, 11, 13}
Output: 5 13
Explanation: Every pair has a gcd of 1.
The maximum sum pair with GCD 1 is (5, 13)

A naive approach will be to iterate for every pair in both the arrays and find out the maximum gcd possible. Below is the code for naive approach.

## C++

 #include using namespace std;   int gcd(int a, int b) {     if (b == 0) return a;     return gcd(b, a % b); }   int main() {     int a[] = {12, 18, 24};     int b[] = {36, 8, 72};     int m = sizeof(a)/sizeof(a[0]);     int n = sizeof(b)/sizeof(b[0]);     int max_gcd = INT_MIN;     int num1, num2;     for (int i=0; i max_gcd) {                 max_gcd = g;                 num1 = a[i];                 num2 = b[j];             }         }     }     cout << "Pair with greatest GCD: (" << num1 << ", " << num2 << ") with GCD: " << max_gcd << endl;     return 0; }

## Java

 import java.lang.Math;   public class GCD {       // Function to find the gcd of two numbers using     // Euclidean algorithm     public static int gcd(int a, int b)     {         // If b is 0, return a as gcd         if (b == 0) {             return a;         }         // Recursive call to find the gcd of b and a%b         return gcd(b, a % b);     }       public static void main(String[] args)     {         // Initialize two arrays a and b         int[] a = { 12, 18, 24 };         int[] b = { 36, 8, 72 };           // Set max_gcd to negative infinity, and initialize         // num1 and num2 to None         int max_gcd = Integer.MIN_VALUE;         Integer num1 = null;         Integer num2 = null;           // Loop through elements of both arrays and find the         // maximum gcd         for (int i = 0; i < a.length; i++) {             for (int j = 0; j < b.length; j++) {                 // Call gcd function to find the gcd of a[i]                 // and b[j]                 int g = gcd(a[i], b[j]);                 // If current gcd is greater than the                 // maximum, update max_gcd, num1 and num2                 if (g > max_gcd) {                     max_gcd = g;                     num1 = a[i];                     num2 = b[j];                 }             }         }         // Print the pair with the greatest GCD         System.out.println("Pair with greatest GCD: ("                            + num1 + ", " + num2                            + ") with GCD: " + max_gcd);     } }

## Python3

 import math   # Function to find the gcd of two numbers using Euclidean algorithm def gcd(a, b):     # If b is 0, return a as gcd     if b == 0:         return a     # Recursive call to find the gcd of b and a%b     return gcd(b, a % b)   # Initialize two arrays a and b a = [12, 18, 24] b = [36, 8, 72]   # Set max_gcd to negative infinity, and initialize num1 and num2 to None max_gcd = float("-inf") num1, num2 = None, None   # Loop through elements of both arrays and find the maximum gcd for i in range(len(a)):     for j in range(len(b)):         # Call gcd function to find the gcd of a[i] and b[j]         g = gcd(a[i], b[j])         # If current gcd is greater than the maximum, update max_gcd, num1 and num2         if g > max_gcd:             max_gcd = g             num1 = a[i]             num2 = b[j]   # Print the pair with the greatest GCD print(f"Pair with greatest GCD: ({num1}, {num2}) with GCD: {max_gcd}")

## C#

 using System;   class MainClass {     // Define a function to compute the greatest common divisor of two integers     static int GCD(int a, int b) {         if (b == 0) return a;         return GCD(b, a % b);     }       static void Main() {         // Define two arrays of integers         int[] a = new int[] {12, 18, 24};         int[] b = new int[] {36, 8, 72};         int m = a.Length; // Get the size of the first array         int n = b.Length; // Get the size of the second array         int max_gcd = int.MinValue; // Initialize the maximum GCD to the smallest possible integer value         int num1 = 0, num2 = 0; // Initialize variables to store the two numbers with the greatest GCD         for (int i=0; i max_gcd) { // If the GCD is greater than the current maximum GCD                     max_gcd = g; // Update the maximum GCD                     num1 = a[i]; // Update the first number with the greatest GCD                     num2 = b[j]; // Update the second number with the greatest GCD                 }             }         }         // Output the two numbers with the greatest GCD and their GCD         Console.WriteLine("Pair with greatest GCD: ({0}, {1}) with GCD: {2}", num1, num2, max_gcd);     } } //This code is contributed by chinmaya121221

## Javascript

 function gcd(a, b) {     if (b == 0) return a;     return gcd(b, a % b); }   let a = [12, 18, 24]; let b = [36, 8, 72]; let m = a.length; let n = b.length; let max_gcd = Number.MIN_SAFE_INTEGER; let num1, num2;   for (let i=0; i max_gcd) {             max_gcd = g;             num1 = a[i];             num2 = b[j];         }     } }   console.log("Pair with greatest GCD: (" + num1 + ", " + num2 + ") with GCD: " + max_gcd);

Output

Pair with greatest GCD: (24, 72) with GCD: 24

Time complexity : O(M*N)  where M , N are the sizes of the array

Auxiliary space : O(1)

An efficient (only when elements are small) is to apply the sieve property and for that, we need to pre-calculate the following things.

1. A cnt array to mark the presence of array elements.
2. We check for all the numbers from 1 to N and for each multiple, we check that if the number exists then the max of the pre-existing number or the present existing multiple is stored.
3. Step 1 and 2 is repeated for the other array also.
4. At the end we check for the maximum multiple which is common in both first and second array to get the maximum GCD, and in the position of is stored the element, in first the element of an array is stored, and in second the element of b array is stored, so we print the pair.

Below is the implementation of the above approach :

## C++

 // CPP program to find maximum GCD pair // from two arrays #include using namespace std;   // Find the maximum GCD pair with maximum // sum void gcdMax(int a[], int b[], int n, int N) {     // array to keep a count of existing elements     int cnt[N] = { 0 };       // first[i] and second[i] are going to store     // maximum multiples of i in a[] and b[]     // respectively.     int first[N] = { 0 }, second[N] = { 0 };       // traverse through the first array to     // mark the elements in cnt     for (int i = 0; i < n; ++i)         cnt[a[i]] = 1;       // Find maximum multiple of every number     // in first array     for (int i = 1; i < N; ++i)         for (int j = i; j < N; j += i)             if (cnt[j])                 first[i] = max(first[i], j);       // Find maximum multiple of every number     // in second array     // We re-initialise cnt[] and traverse     // through the second array to mark the     // elements in cnt     memset(cnt, 0, sizeof(cnt));     for (int i = 0; i < n; ++i)         cnt[b[i]] = true;     for (int i = 1; i < N; ++i)         for (int j = i; j < N; j += i)               // if the multiple is present in the             // second array then store the  max             // of number or the  pre-existing             // element             if (cnt[j])                 second[i] = max(second[i], j);       // traverse for every elements and checks     // the maximum N that is present in both     // the arrays     int i;     for (i = N - 1; i >= 0; i--)         if (first[i] && second[i])             break;       cout << "Maximum GCD pair with maximum "             "sum is " << first[i] << " "          << second[i] << endl; }   // driver program to test the above function int main() {     int a[] = { 3, 1, 4, 2, 8 };     int b[] = { 5, 2, 12, 8, 3 };     int n = sizeof(a) / sizeof(a[0]);       // Maximum possible value of elements     // in both arrays.     int N = 20;       gcdMax(a, b, n, N);     return 0; }

## Java

 // Java program to find maximum // GCD pair from two arrays import java.io.*; public class GFG {       // Find the maximum GCD // pair with maximum sum static void gcdMax(int[] a, int[] b,                    int n, int N) {     // array to keep a count     // of existing elements     int[] cnt = new int[N];       // first[i] and second[i]     // are going to store     // maximum multiples of     // i in a[] and b[]     // respectively.     int[] first = new int[N];     int[] second = new int[N];       // traverse through the     // first array to mark     // the elements in cnt     for (int i = 0; i < n; ++i)         cnt[a[i]] = 1;       // Find maximum multiple     // of every number in     // first array     for (int i = 1; i < N; ++i)         for (int j = i; j < N; j += i)             if (cnt[j] > 0)                 first[i] = Math.max(first[i], j);       // Find maximum multiple     // of every number in second     // array. We re-initialise     // cnt[] and traverse through     // the second array to mark     // the elements in cnt     cnt = new int[N];     for (int i = 0; i < n; ++i)         cnt[b[i]] = 1;     for (int i = 1; i < N; ++i)         for (int j = i; j < N; j += i)               // if the multiple is present             // in the second array then             // store the max of number or             // the pre-existing element             if (cnt[j] > 0)                 second[i] = Math.max(second[i], j);       // traverse for every     // elements and checks     // the maximum N that     // is present in both     // the arrays     int x;     for (x = N - 1; x >= 0; x--)         if (first[x] > 0 &&             second[x] > 0)             break;       System.out.println(first[x] + " " +                             second[x]); }   // Driver Code public static void main(String[] args) {     int[] a = { 3, 1, 4, 2, 8 };     int[] b = { 5, 2, 12, 8, 3 };     int n = a.length;       // Maximum possible     // value of elements     // in both arrays.     int N = 20;       gcdMax(a, b, n, N); } }   // This code is contributed // by mits

## Python3

 # Python 3 program to find maximum GCD pair # from two arrays   # Find the maximum GCD pair with maximum # sum def gcdMax(a, b, n, N):       # array to keep a count of existing elements     cnt = [0]*N       # first[i] and second[i] are going to store     # maximum multiples of i in a[] and b[]     # respectively.     first = [0]*N     second = [0]*N       # traverse through the first array to     # mark the elements in cnt     for i in range(n):         cnt[a[i]] = 1       # Find maximum multiple of every number     # in first array     for i in range(1,N):         for j in range(i,N,i):             if (cnt[j]):                 first[i] = max(first[i], j)       # Find maximum multiple of every number     # in second array     # We re-initialise cnt[] and traverse     # through the second array to mark the     # elements in cnt     cnt = [0]*N     for i in range(n):         cnt[b[i]] = 1     for i in range(1,N):         for j in range(i,N,i):               # if the multiple is present in the             # second array then store the max             # of number or the pre-existing             # element             if (cnt[j]>0):                 second[i] = max(second[i], j)                   # traverse for every elements and checks     # the maximum N that is present in both     # the arrays           i = N-1     while i>= 0:         if (first[i]>0 and second[i]>0):             break         i -= 1           print( str(first[i]) + " " + str(second[i]))   # driver program to test the above function if __name__ == "__main__":     a = [ 3, 1, 4, 2, 8 ]     b = [ 5, 2, 12, 8, 3 ]     n = len(a)       # Maximum possible value of elements     # in both arrays.     N = 20     gcdMax(a, b, n, N)   # this code is contributed by ChitraNayal

## C#

 // C# program to find // maximum GCD pair // from two arrays using System;   class GFG { // Find the maximum GCD // pair with maximum sum static void gcdMax(int[] a, int[] b,                    int n, int N) {     // array to keep a count     // of existing elements     int[] cnt = new int[N];       // first[i] and second[i]     // are going to store     // maximum multiples of     // i in a[] and b[]     // respectively.     int[] first = new int[N];     int[] second = new int[N];       // traverse through the     // first array to mark     // the elements in cnt     for (int i = 0; i < n; ++i)         cnt[a[i]] = 1;       // Find maximum multiple     // of every number in     // first array     for (int i = 1; i < N; ++i)         for (int j = i; j < N; j += i)             if (cnt[j] > 0)                 first[i] = Math.Max(first[i], j);       // Find maximum multiple     // of every number in second     // array. We re-initialise     // cnt[] and traverse through     // the second array to mark     // the elements in cnt     cnt = new int[N];     for (int i = 0; i < n; ++i)         cnt[b[i]] = 1;     for (int i = 1; i < N; ++i)         for (int j = i; j < N; j += i)               // if the multiple is present             // in the second array then             // store the max of number or             // the pre-existing element             if (cnt[j] > 0)                 second[i] = Math.Max(second[i], j);       // traverse for every     // elements and checks     // the maximum N that     // is present in both     // the arrays     int x;     for (x = N - 1; x >= 0; x--)         if (first[x] > 0 &&             second[x] > 0)             break;       Console.WriteLine(first[x] +                       " " + second[x]); }   // Driver Code static int Main() {     int[] a = { 3, 1, 4, 2, 8 };     int[] b = { 5, 2, 12, 8, 3 };     int n = a.Length;       // Maximum possible     // value of elements     // in both arrays.     int N = 20;       gcdMax(a, b, n, N);     return 0; } }   // This code is contributed // by mits

## PHP

 = 0; \$x--)         if (\$first[\$x] && \$second[\$x])             break;           echo \$first[\$x] . " " .              \$second[\$x] . "\n"; }   // Driver code \$a = array(3, 1, 4, 2, 8); \$b = array(5, 2, 12, 8, 3); \$n = sizeof(\$a);   // Maximum possible value // of elements in both arrays. \$N = 20;   gcdMax(\$a, \$b, \$n, \$N);   // This code is contributed // by mits ?>

## Javascript



Output

Maximum GCD pair with maximum sum is 8 8

Time complexity : O(N Log N + N), as we are using nested loops where outer loop traverses N times and inner loop traverses logN times (as N + (N/2) + (N/3) + ….. + 1 = N log N) and we are using extra Loop to traverse N times.
Auxiliary Space : O(N), as we are using extra space for cnt array.

This article is contributed by Raj. 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.