Length of longest subarray of length at least 2 with maximum GCD

Given an array arr[] of length N, the task is the find the length of longest subarray of length at least 2 with maximum possible GCD value.

Examples:

Input: arr[] = {1, 2, 2}
Output: 2
Explanation:
Possible sub-arrays of size greater than 2 and there GCD’s are:
1) {1, 2} -> 1
2) {2, 2} -> 2
3) {1, 2, 3} -> 1
Here, the maximum GCD value is 2 and longest sub-array having GCD = 2 is {2, 2}.
Hence the answer is {2, 2}.

Input: arr[] = {18, 3, 6, 9}
Output: 4
Explanation:
Here, the maximum GCD value is 3 and longest sub-array having GCD = 3 is {18, 3, 6, 9}.
Hence the answer is {18, 3, 6, 9}.

Naive Approach: The idea is to generate all the possible subarray of size at least 2 and find the GCD of each subarray of them individually. Then, the length of the subarray with maximum GCD value is the required length.



Time Complexity: O(N2)

Efficient Approach:

  1. Find the maximum GCD(say g) of all the subarray with length atleast 2 by using the approach discussed in this article.
  2. Traverse the given array and count the maximum number of consecutive elements which are divisible by GCD g.

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 calculate GCD of two numbers
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
  
// Function to find maximum size subarray
// having maximum GCD
int maximumGcdSubarray(int arr[], int n)
{
    // Base Case
    if (n == 1)
        return 0;
  
    // Let the maximum GCD be 1 initially
    int k = 1;
  
    // Loop thourgh array to find maximum
    // GCD of subarray with size 2
    for (int i = 1; i < n; ++i) {
        k = max(k, gcd(arr[i], arr[i - 1]));
    }
  
    int cnt = 0;
    int maxLength = 0;
  
    // Traverse the array
    for (int i = 0; i < n; i++) {
  
        // Is a multiple of k, increase cnt
        if (arr[i] % k == 0) {
            cnt++;
        }
  
        // Else update maximum length with
        // consecutive element divisible by k
        // Set cnt to 0
        else {
            maxLength = max(maxLength, cnt);
            cnt = 0;
        }
    }
  
    // Update the maxLength
    maxLength = max(maxLength, cnt);
  
    // Return the maxLength
    return maxLength;
}
  
// Driver Code
int main()
{
    int arr[] = { 18, 3, 6, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Function Call
    cout << maximumGcdSubarray(arr, n);
    return 0;
}
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Output:
4

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

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