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: 
 



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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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// 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
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# 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
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// 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
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Output: 
4

 

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

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