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: 9 
Explanation:
GCD of the subarray {18, 9, 9} is maximum which is 9.
Input: arr[] = { 4, 8, 12, 16, 20, 24 } 
Output: 4 
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 <bits/stdc++.h>
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

<script>

// Javascript program for the above approach

// Function to find GCD
function gcd(a, b)
{
    if (b == 0) {
        return a;
    }
    return gcd(b, a % b);
}

function findMaxGCD(arr, n)
{

    // To store the maximum GCD
    let maxGCD = 0;

    // Traverse the array
    for (let i = 0; i < n - 1; i++) {

        // Find GCD of the consecutive
        // element
        let val = gcd(arr[i], arr[i + 1]);

        // If calculated GCD > maxGCD
        // then update it
        if (val > maxGCD) {
            maxGCD = val;
        }
    }

    // Print the maximum GCD
    document.write(maxGCD + "<br>");
}

// Driver Code

    let arr = [ 3, 18, 9, 9, 5,
                15, 8, 7, 6, 9 ];

    let n = arr.length;

    // Function Call
    findMaxGCD(arr, n);

// This code is contributed by Mayank Tyagi

</script>

Output:

9

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

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