# Maximum GCD of all subarrays of length at least 2

Given an array arr[] of N numbers. The task is to find the maximum GCD of all subarrays of size greater than 1.
Examples:

Input: arr[] = { 3, 18, 9, 9, 5, 15, 8, 7, 6, 9 }
Output:
Explanation:
GCD of the subarray {18, 9, 9} is maximum which is 9.
Input: arr[] = { 4, 8, 12, 16, 20, 24 }
Output:
Explanation:
GCD of the subarray {4, 18, 12, 16, 20, 24} is maximum which is 4.

Naive Approach: The idea is to generate all the subarray of size greater than 1 and then find the maximum of gcd of all subarray formed.
Time complexity: O(N2)
Efficient Approach: Let GCD of two numbers be g. Now if we take gcd of g with any third number say c then, gcd will decrease or remain same, but it will never increase.
The idea is to find gcd of every consecutive pair in the arr[] and the maximum of gcd of all the pairs formed is the desired result.
Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to find GCD``int` `gcd(``int` `a, ``int` `b)``{``    ``if` `(b == 0) {``        ``return` `a;``    ``}``    ``return` `gcd(b, a % b);``}` `void` `findMaxGCD(``int` `arr[], ``int` `n)``{` `    ``// To store the maximum GCD``    ``int` `maxGCD = 0;` `    ``// Traverse the array``    ``for` `(``int` `i = 0; i < n - 1; i++) {` `        ``// Find GCD of the consecutive``        ``// element``        ``int` `val = gcd(arr[i], arr[i + 1]);` `        ``// If calculated GCD > maxGCD``        ``// then update it``        ``if` `(val > maxGCD) {``            ``maxGCD = val;``        ``}``    ``}` `    ``// Print the maximum GCD``    ``cout << maxGCD << endl;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 3, 18, 9, 9, 5,``                  ``15, 8, 7, 6, 9 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``// Function Call``    ``findMaxGCD(arr, n);``    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `class` `GFG{` `// Function to find GCD``static` `int` `gcd(``int` `a, ``int` `b)``{``    ``if` `(b == ``0``)``    ``{``        ``return` `a;``    ``}``    ``return` `gcd(b, a % b);``}` `static` `void` `findMaxGCD(``int` `arr[], ``int` `n)``{` `    ``// To store the maximum GCD``    ``int` `maxGCD = ``0``;` `    ``// Traverse the array``    ``for``(``int` `i = ``0``; i < n - ``1``; i++) ``    ``{``        ` `       ``// Find GCD of the consecutive``       ``// element``       ``int` `val = gcd(arr[i], arr[i + ``1``]);``       ` `       ``// If calculated GCD > maxGCD``       ``// then update it``       ``if` `(val > maxGCD)``       ``{``           ``maxGCD = val;``       ``}``    ``}` `    ``// Print the maximum GCD``    ``System.out.print(maxGCD + ``"\n"``);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = { ``3``, ``18``, ``9``, ``9``, ``5``,``                  ``15``, ``8``, ``7``, ``6``, ``9` `};``    ``int` `n = arr.length;` `    ``// Function call``    ``findMaxGCD(arr, n);``}``}` `// This code is contributed by amal kumar choubey`

## Python3

 `# Python3 program for the above approach` `# Function to find GCD``def` `gcd(a, b):``    ` `    ``if` `(b ``=``=` `0``):``        ``return` `a;``    ``return` `gcd(b, a ``%` `b);` `def` `findMaxGCD(arr, n):``    ` `    ``# To store the maximum GCD``    ``maxGCD ``=` `0``;` `    ``# Traverse the array``    ``for` `i ``in` `range``(``0``, n ``-` `1``):` `        ``# Find GCD of the consecutive``        ``# element``        ``val ``=` `gcd(arr[i], arr[i ``+` `1``]);` `        ``# If calculated GCD > maxGCD``        ``# then update it``        ``if` `(val > maxGCD):``            ``maxGCD ``=` `val;` `    ``# Print the maximum GCD``    ``print``(maxGCD);` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``arr ``=` `[ ``3``, ``18``, ``9``, ``9``, ``5``, ``            ``15``, ``8``, ``7``, ``6``, ``9` `];``    ``n ``=` `len``(arr);` `    ``# Function call``    ``findMaxGCD(arr, n);` `# This code is contributed by 29AjayKumar`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{` `// Function to find GCD``static` `int` `gcd(``int` `a, ``int` `b)``{``    ``if` `(b == 0)``    ``{``        ``return` `a;``    ``}``    ``return` `gcd(b, a % b);``}` `static` `void` `findMaxGCD(``int` `[]arr, ``int` `n)``{` `    ``// To store the maximum GCD``    ``int` `maxGCD = 0;` `    ``// Traverse the array``    ``for``(``int` `i = 0; i < n - 1; i++) ``    ``{``        ` `        ``// Find GCD of the consecutive``        ``// element``        ``int` `val = gcd(arr[i], arr[i + 1]);``            ` `        ``// If calculated GCD > maxGCD``        ``// then update it``        ``if` `(val > maxGCD)``        ``{``            ``maxGCD = val;``        ``}``    ``}` `    ``// Print the maximum GCD``    ``Console.Write(maxGCD + ``"\n"``);``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `[]arr = { 3, 18, 9, 9, 5,``                 ``15, 8, 7, 6, 9 };``    ``int` `n = arr.Length;` `    ``// Function call``    ``findMaxGCD(arr, n);``}``}` `// This code is contributed by Code_Mech`

## Javascript

 ``

Output:
`9`

Time Complexity: O(N), where N is the length of the array.

Auxiliary Space: O(log(max(a, b)))

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