Longest sub-sequence with maximum GCD

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

Input: arr[] = {2, 1, 2} 
Output: 2 
{2}, {2} and {2, 2} are the subsequences 
with the maximum possible GCD.
Input: arr[] = {1, 2, 3} 
Output: 1 
{3} is the required subsequence. 

Approach: The maximum possible GCD from the array will be equal to the value of the largest element in the array. Now, to maximize the length of the resulting subsequence, find the number of elements with a value equal to this largest value in the array, and the count of these elements is the required answer.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the length
// of the largest subsequence with
// maximum possible GCD
int maxLen(int* arr, int n)
{
    // Maximum value from the array
    int max_val = *max_element(arr, arr + n);

    // To store the frequency of the
    // maximum element in the array
    int freq = 0;

    for (int i = 0; i < n; i++) {

        // If current element is equal
        // to the maximum element
        if (arr[i] == max_val)
            freq++;
    }

    return freq;
}

// Driver code
int main()
{
    int arr[] = { 3, 2, 2, 3, 3, 3 };
    int n = sizeof(arr) / sizeof(int);

    cout << maxLen(arr, n);

    return 0;
}

Java

// Java implementation of the approach
import java.util.Arrays;

class GFG
{

// Function to return the length
// of the largest subsequence with
// maximum possible GCD
static int maxLen(int[] arr, int n)
{
    // Maximum value from the array
    int max_val = Arrays.stream(arr).max().getAsInt();

    // To store the frequency of the
    // maximum element in the array
    int freq = 0;

    for (int i = 0; i < n; i++)
    {

        // If current element is equal
        // to the maximum element
        if (arr[i] == max_val)
            freq++;
    }
    return freq;
}

// Driver code
public static void main(String []args)
{
    int arr[] = { 3, 2, 2, 3, 3, 3 };
    int n = arr.length;

    System.out.println(maxLen(arr, n));
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach

# Function to return the length
# of the largest subsequence with
# maximum possible GCD
def maxLen(arr, n) :

    # Maximum value from the array
    max_val = max(arr);

    # To store the frequency of the
    # maximum element in the array
    freq = 0;

    for i in range(n) :

        # If current element is equal
        # to the maximum element
        if (arr[i] == max_val) :
            freq += 1;

    return freq;

# Driver code
if __name__ == "__main__" :

    arr = [ 3, 2, 2, 3, 3, 3 ];
    n = len(arr);

    print(maxLen(arr, n));

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
using System.Linq;

class GFG
{

// Function to return the length
// of the largest subsequence with
// maximum possible GCD
static int maxLen(int[] arr, int n)
{
    // Maximum value from the array
    int max_val = arr.Max();

    // To store the frequency of the
    // maximum element in the array
    int freq = 0;

    for (int i = 0; i < n; i++)
    {

        // If current element is equal
        // to the maximum element
        if (arr[i] == max_val)
            freq++;
    }
    return freq;
}

// Driver code
public static void Main(String []args)
{
    int []arr = { 3, 2, 2, 3, 3, 3 };
    int n = arr.Length;

    Console.WriteLine(maxLen(arr, n));
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// Javascript implementation of the approach

// Function to return the length
// of the largest subsequence with
// maximum possible GCD
function maxLen(arr, n)
{
    // Maximum value from the array
    var max_val = arr.reduce((a,b) => Math.max(a,b));

    // To store the frequency of the
    // maximum element in the array
    var freq = 0;

    for (var i = 0; i < n; i++) {

        // If current element is equal
        // to the maximum element
        if (arr[i] == max_val)
            freq++;
    }

    return freq;
}

// Driver code
var arr = [3, 2, 2, 3, 3, 3];
var n = arr.length;
document.write( maxLen(arr, n));

</script>

Output:

4

Time Complexity: O(n), where n is the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.