# Longest sub-array with maximum GCD

Given an array arr[] of length N, the task is the find the length of the longest sub-array with maximum possible GCD value.

Examples:

Input: arr[] = {1, 2, 2}
Output: 2
Here all possible sub-arrays and there GCD’s are:
1) {1} -> 1
2) {2} -> 2
3) {2} -> 2
4) {1, 2} -> 1
5) {2, 2} -> 2
6) {1, 2, 3} -> 1
Here, the maximum GCD value is 2 and longest sub-array having GCD = 2 is {2, 2}.
Thus, the answer is {2, 2}.

Input: arr[] = {3, 3, 3, 3}
Output: 4

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive approach: Generate all the possible sub-arrays and find the GCD of each of them individually in order to find the longest such sub-array. This approach will take O(N3) time to solve the problem.

Better approach: The maximum GCD value will always be equal to the largest number present in the array. Let’s say that the largest number present in the array is X. Now, the task is to find the largest sub-array having all X. The same can be done using the two-pointer approach. Below is the algorithm:

• Find the largest number in the array. Let us call this number X.
• Run a loop from i = 0
• If arr[i] != X then increment i and continue.
• Else initialise j = i.
• While j < n and arr[j] = X, increment j.
• Update the answer as ans = max(ans, j – i).
• Update i as i = j.

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach #include using namespace std;    // Function to return the length of // the largest subarray with // maximum possible GCD int findLength(int* arr, int n) {     // To store the maximum number     // present in the array     int x = 0;        // Finding the maximum element     for (int i = 0; i < n; i++)         x = max(x, arr[i]);        // To store the final answer     int ans = 0;        // Two pointer     for (int i = 0; i < n; i++) {            if (arr[i] != x)             continue;            // Running a loop from j = i         int j = i;            // Condition for incrementing 'j'         while (arr[j] == x)             j++;            // Updating the answer         ans = max(ans, j - i);     }        return ans; }    // Driver code int main() {     int arr[] = { 1, 2, 2 };     int n = sizeof(arr) / sizeof(int);        cout << findLength(arr, n);        return 0; }

## Java

 // Java implementation of the approach  class GFG  {        // Function to return the length of      // the largest subarray with      // maximum possible GCD      static int findLength(int []arr, int n)      {          // To store the maximum number          // present in the array          int x = 0;                 // Finding the maximum element          for (int i = 0; i < n; i++)              x = Math.max(x, arr[i]);                 // To store the final answer          int ans = 0;                 // Two pointer          for (int i = 0; i < n; i++)          {              if (arr[i] != x)                  continue;                     // Running a loop from j = i              int j = i;                     // Condition for incrementing 'j'              while (arr[j] == x)              {                 j++;                  if (j >= n )                 break;             }                    // Updating the answer              ans = Math.max(ans, j - i);          }          return ans;      }             // Driver code      public static void main (String[] args)     {          int arr[] = { 1, 2, 2 };          int n = arr.length;                 System.out.println(findLength(arr, n));      }  }    // This code is contributed by AnkitRai01

## Python3

 # Python3 implementation of the approach     # Function to return the length of  # the largest subarray with  # maximum possible GCD  def findLength(arr, n) :        # To store the maximum number      # present in the array      x = 0;         # Finding the maximum element      for i in range(n) :          x = max(x, arr[i]);         # To store the final answer      ans = 0;         # Two pointer      for i in range(n) :            if (arr[i] != x) :             continue;             # Running a loop from j = i          j = i;             # Condition for incrementing 'j'          while (arr[j] == x) :             j += 1;                             if j >= n :                 break            # Updating the answer          ans = max(ans, j - i);         return ans;     # Driver code  if __name__ == "__main__" :         arr = [ 1, 2, 2 ];      n = len(arr);         print(findLength(arr, n));         # This code is contributed by AnkitRai01

## C#

 // C# implementation of the approach  using System;    class GFG  {        // Function to return the length of      // the largest subarray with      // maximum possible GCD      static int findLength(int []arr, int n)      {          // To store the maximum number          // present in the array          int x = 0;                 // Finding the maximum element          for (int i = 0; i < n; i++)              x = Math.Max(x, arr[i]);                 // To store the final answer          int ans = 0;                 // Two pointer          for (int i = 0; i < n; i++)          {              if (arr[i] != x)                  continue;                     // Running a loop from j = i              int j = i;                     // Condition for incrementing 'j'              while (arr[j] == x)              {                 j++;                  if (j >= n )                 break;             }                    // Updating the answer              ans = Math.Max(ans, j - i);          }          return ans;      }             // Driver code      public static void Main ()     {          int []arr = { 1, 2, 2 };          int n = arr.Length;                 Console.WriteLine(findLength(arr, n));      }  }    // This code is contributed by AnkitRai01

Output:

2

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