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:

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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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Output:

9

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

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