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

  • Last Updated : 04 Mar, 2021

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++




// 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;
}

Java




// 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




# 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#




// 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

Javascript




<script>
 
// Javascript program for the above approach
 
// Function to calculate GCD of two numbers
function gcd(a, b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}
 
// Function to find maximum size subarray
// having maximum GCD
function maximumGcdSubarray(arr, n)
{
    // Base Case
    if (n == 1)
        return 0;
 
    // Let the maximum GCD be 1 initially
    let k = 1;
 
    // Loop thourgh array to find maximum
    // GCD of subarray with size 2
    for (let i = 1; i < n; ++i) {
        k = Math.max(k, gcd(arr[i], arr[i - 1]));
    }
 
    let cnt = 0;
    let maxLength = 0;
 
    // Traverse the array
    for (let 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
 
    let arr = [ 18, 3, 6, 9 ];
    let n = arr.length;
 
    // Function Call
    document.write(maximumGcdSubarray(arr, n));
 
// This code is contributed by Mayank Tyagi
 
</script>
Output: 
4

 

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

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