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:2Explanation:

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:4Explanation:

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(N^{2})

**Efficient Approach:**

- Find the maximum GCD(say
**g**) of all the subarray with length atleast 2 by using the approach discussed in this article. - 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:

`// 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; `
`} ` |

*chevron_right*

*filter_none*

**Output:**

4

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

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