# 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:
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:
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:

 // C++ program for the above approach     #include   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;  }

 // Java program for the above approach class GFG{        // Function to calculate GCD of  // two numbers static 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  static int maximumGcdSubarray(int arr[], int n) {            // Base case     if (n == 1)         return 0;                // Let the maximum GCD be 1 initially     int k = 1;                // Loop through array to find maximum      // GCD of subarray with size 2     for(int i = 1; i < n; i++)     {        k = Math.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 = Math.max(maxLength, cnt);            cnt = 0;        }     }            // Update the maxLength     maxLength = Math.max(maxLength, cnt);                // Return the maxLength     return maxLength; }        // Driver Code public static void main(String args[]) {     int arr[] = { 18, 3, 6, 9 };     int n = arr.length;                // Function call     System.out.println(maximumGcdSubarray(arr, n)); } }    // This code is contributed by stutipathak31jan

 # Python3 program for the above approach     # Function to calculate GCD of  # two numbers def gcd(a, b):            if b == 0:         return a     return gcd(b, a % b)    # Function to find maximum size  # subarray having maximum GCD  def maxGcdSubarray(arr, n):            # Base case     if n == 1:         return 0            # Let the maximum GCD be 1 initially     k = 1            # Loop through array to find maximum      # GCD of subarray with size 2     for i in range(1, n):         k = max(k, gcd(arr[i], arr[i - 1]))                cnt = 0     maxLength = 0            # Traverse the array     for i in range(n):                    # Is a multiple of k, increase cnt         if arr[i] % k == 0:             cnt += 1                    # 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 arr = [ 18, 3, 6, 9 ] n = len(arr)        # Function call print(maxGcdSubarray(arr, n))    # This code is contributed by stutipathak31jan

 // C# program for the above approach using System; using System.Collections.Generic;    class GFG{        // Function to calculate GCD of  // two numbers  static 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  static int maximumGcdSubarray(int[] arr, int n)  {             // Base case      if (n == 1)          return 0;                 // Let the maximum GCD be 1 initially      int k = 1;                 // Loop through array to find maximum      // GCD of subarray with size 2      for(int i = 1; i < n; i++)      {          k = Math.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 = Math.Max(maxLength, cnt);              cnt = 0;          }      }             // Update the maxLength      maxLength = Math.Max(maxLength, cnt);                 // Return the maxLength      return maxLength;  }     // Driver code     static void Main() {     int[] arr = { 18, 3, 6, 9 };      int n = arr.Length;                 // Function call      Console.WriteLine(maximumGcdSubarray(arr, n)); } }    // This code is contributed by divyeshrabadiya07

Output:
4

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

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