Length of the Smallest Subarray that must be removed in order to Maximise the GCD

Given an array arr[] of N elements, the task is to find the length of the smallest subarray such that when this subarray is removed from the array, the GCD of the resultant array is maximum. 
Note: The resulting array should be non-empty.
Examples: 
 

Input: N = 4, arr[] = {3, 6, 1, 2} 
Output: 2 
Explanation: 
If we remove the subarray {1, 2} then the resulting subarray will be {3, 6} and the GCD of this is 3 which is the maximum possible value.
Input: N = 3, arr[] = {4, 8, 4} 
Output: 0 
Explanation: 
Here we don't need to remove any subarray and the maximum GCD possible is 4. 
 

Approach: It is known that GCD is a non-increasing function. That is, if we add elements in the array, then the gcd will either be decreasing or remain constant. Therefore, the idea is to use this concept to solve this problem: 
 

• Now we have to notice that after removal of a subarray the resulting subarray should have either the first or the last element or both the elements. This is because we need to make sure that the resultant array after removing the subarray should be non-empty. So, we cannot remove all the elements.
• So the maximum GCD possible will be max(A[0], A[N - 1]).
• Now we have to minimize the length of the subarray which we need to remove to obtain this answer.
• To do that, we will use two pointers technique, pointing to the first and the last elements respectively.
• Now we will increase the first pointer if the element is divisible by that gcd and decrease the last pointer if the element is divisible by gcd as it will not affect our answer.
• So, at last, the number of elements between the two pointers will be the length of the subarray which we need to remove.

Below is the implementation of the above approach: 
 

// C++ program to find the length of
// the smallest subarray that must be
// removed in order to maximise the GCD

#include <bits/stdc++.h>
using namespace std;

// Function to find the length of
// the smallest subarray that must be
// removed in order to maximise the GCD
int GetMinSubarrayLength(int a[], int n)
{

    // Store the maximum possible
    // GCD of the resulting subarray
    int ans = max(a[0], a[n - 1]);

    // Two pointers initially pointing
    // to the first and last element
    // respectively
    int lo = 0, hi = n - 1;

    // Moving the left pointer to the
    // right if the elements are
    // divisible by the maximum GCD
    while (lo < n and a[lo] % ans == 0)
        lo++;

    // Moving the right pointer to the
    // left if the elements are
    // divisible by the maximum GCD
    while (hi > lo and a[hi] % ans == 0)
        hi--;

    // Return the length of
    // the subarray
    return (hi - lo + 1);
}

// Driver code
int main()
{

    int arr[] = { 4, 8, 2, 1, 4 };
    int N = sizeof(arr) / sizeof(arr[0]);

    int length = GetMinSubarrayLength(arr, N);

    cout << length << "\n";

    return 0;
}

// Java program to find the length of
// the smallest subarray that must be
// removed in order to maximise the GCD
class GFG {
    
    // Function to find the length of
    // the smallest subarray that must be
    // removed in order to maximise the GCD
    static int GetMinSubarrayLength(int a[], int n)
    {
    
        // Store the maximum possible
        // GCD of the resulting subarray
        int ans = Math.max(a[0], a[n - 1]);
    
        // Two pointers initially pointing
        // to the first and last element
        // respectively
        int lo = 0, hi = n - 1;
    
        // Moving the left pointer to the
        // right if the elements are
        // divisible by the maximum GCD
        while (lo < n && a[lo] % ans == 0)
            lo++;
    
        // Moving the right pointer to the
        // left if the elements are
        // divisible by the maximum GCD
        while (hi > lo && a[hi] % ans == 0)
            hi--;
    
        // Return the length of
        // the subarray
        return (hi - lo + 1);
    }
    
    // Driver code
    public static void main (String[] args) 
    {
    
        int arr[] = { 4, 8, 2, 1, 4 };
        int N = arr.length;
    
        int Length = GetMinSubarrayLength(arr, N);
    
        System.out.println(Length);
    
    }
}

// This code is contributed by Yash_R

# Python3 program to find the length of
# the smallest subarray that must be
# removed in order to maximise the GCD

# Function to find the length of
# the smallest subarray that must be
# removed in order to maximise the GCD
def GetMinSubarrayLength(a, n):

    # Store the maximum possible
    # GCD of the resulting subarray
    ans = max(a[0], a[n - 1])

    # Two pointers initially pointing
    # to the first and last element
    # respectively
    lo = 0
    hi = n - 1

    # Moving the left pointer to the
    # right if the elements are
    # divisible by the maximum GCD
    while (lo < n and a[lo] % ans == 0):
        lo += 1

    # Moving the right pointer to the
    # left if the elements are
    # divisible by the maximum GCD
    while (hi > lo and a[hi] % ans == 0):
        hi -= 1

    # Return the length of
    # the subarray
    return (hi - lo + 1)

# Driver code
if __name__ == '__main__':

    arr = [4, 8, 2, 1, 4]
    N = len(arr)

    length = GetMinSubarrayLength(arr, N)

    print(length)

# This code is contributed by mohit kumar 29

// C# program to find the length of
// the smallest subarray that must be
// removed in order to maximise the GCD
using System;

class GFG {
    
    // Function to find the length of
    // the smallest subarray that must be
    // removed in order to maximise the GCD
    static int GetMinSubarrayLength(int []a, int n)
    {
    
        // Store the maximum possible
        // GCD of the resulting subarray
        int ans = Math.Max(a[0], a[n - 1]);
    
        // Two pointers initially pointing
        // to the first and last element
        // respectively
        int lo = 0, hi = n - 1;
    
        // Moving the left pointer to the
        // right if the elements are
        // divisible by the maximum GCD
        while (lo < n && a[lo] % ans == 0)
            lo++;
    
        // Moving the right pointer to the
        // left if the elements are
        // divisible by the maximum GCD
        while (hi > lo && a[hi] % ans == 0)
            hi--;
    
        // Return the length of
        // the subarray
        return (hi - lo + 1);
    }
    
    // Driver code
    public static void Main (string[] args) 
    {
    
        int []arr = { 4, 8, 2, 1, 4 };
        int N = arr.Length;
    
        int Length = GetMinSubarrayLength(arr, N);
    
        Console.WriteLine(Length);
    
    }
}

// This code is contributed by Yash_R

<script>

// Javascript program to find the length of
// the smallest subarray that must be
// removed in order to maximise the GCD

// Function to find the length of
// the smallest subarray that must be
// removed in order to maximise the GCD
function GetMinSubarrayLength(a, n)
{

    // Store the maximum possible
    // GCD of the resulting subarray
    var ans = Math.max(a[0], a[n - 1]);

    // Two pointers initially pointing
    // to the first and last element
    // respectively
    var lo = 0, hi = n - 1;

    // Moving the left pointer to the
    // right if the elements are
    // divisible by the maximum GCD
    while (lo < n && a[lo] % ans == 0)
        lo++;

    // Moving the right pointer to the
    // left if the elements are
    // divisible by the maximum GCD
    while (hi > lo && a[hi] % ans == 0)
        hi--;

    // Return the length of
    // the subarray
    return (hi - lo + 1);
}

// Driver code
var arr = [4, 8, 2, 1, 4 ];
var N = arr.length;
var length = GetMinSubarrayLength(arr, N);
document.write( length );

// This code is contributed by itsok.
</script> 

2

Time Complexity: O(N)

Space Complexity: O(1)