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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

9

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

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