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Longest subsequence having maximum GCD between any pair of distinct elements
  • Difficulty Level : Medium
  • Last Updated : 08 Jun, 2021

Given an array arr[] consisting of N positive integers, the task is to find the maximum length of subsequence from the given array such that the GCD of any two distinct integers of the subsequence is maximum.

Examples:

Input: arr[] = {5, 14, 15, 20, 1}
Output: 3
Explanation: Subsequence with maximum GCD of any pair of distinct integers is {5, 15, 20}.
The GCD between any pair of distinct elements in the above sequence is 5 and it is the maximum GCD in the array.

Input: arr[] = {1, 2, 8, 5, 6}
Output: 3
Explanation:< Subsequence with maximum GCD of any pair of distinct integers is {2, 8, 6}.
The GCD between any two distinct number in the above sequence is 2 and it is the maximum GCD in the array.

Naive Approach: The simplest approach is to use recursion and generate all possible subsequence of the given array recursively and find the maximum length among those subsequences. Below are the steps:



  • Create a function to find the pair in the array with maximum GCD(say maxGCD) using the idea of Sieve Of Eratosthenes.
  • Create another function to count the maximum length of the subsequence array where GCD between any two distinct elements is maximum:
    • Create an auxiliary array arr1[] that stores the array element in sorted order.
    • Initialize the two variables as [0, 0] and recursively generate all the subsequence and return the maximum length of the subsequence having GCD as maxGCD.
  • After the above steps, print the maximum length return by the above recursive calls.

 Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
 
// Function to find GCD of pair with
// maximum GCD
int findMaxGCD(int arr[], int N)
{
     
    // Stores maximum element of arr[]
    int high = 0;
     
    // Find the maximum element
    for(int i = 0; i < N; i++)
    {
        high = max(high, arr[i]);
    }
 
    // Maintain a count array
    int count[high + 1] = {0};
     
    // Store the frequency of arr[]
    for(int i = 0; i < N; i++)
    {
        count[arr[i]] += 1;
    }
 
    // Stores the multiples of a number
    int counter = 0;
 
    // Iterate over the  range [MAX, 1]
    for(int i = high; i > 0; i--)
    {
        int j = i;
 
        // Iterate from current potential
        // GCD till it is less than MAX
        while (j <= high)
        {
             
            // A multiple found
            if (count[j] > 0)
                counter += count[j];
 
            // Increment potential GCD by
            // itself io check i, 2i, 3i...
            j += i;
 
            // If 2 multiples found max
            // GCD found
            if (counter == 2)
                return i;
        }
        counter = 0;
    }
}
     
// Function to find longest subsequence
// such that GCD of any two distinct
// integers is maximum
int maxlen(int i, int j, int arr[],
           int arr1[], int N, int maxgcd)
{
    int a = 1;
 
    // Base Cases
    if (i >= N or j >= N)
        return 0;
 
    // Compare current GCD to the
    // maximum GCD
    if (__gcd(arr[i], arr1[j]) == maxgcd &&
              arr[i] != arr1[j])
    {
         
        // If true increment and
        // move the pointer
        a = max(a, 1 + maxlen(i, j + 1,
                              arr, arr1,
                              N, maxgcd));
        return a;
    }
     
    // Return max of either subsequences
    return max(maxlen(i + 1, j, arr,
                      arr1, N, maxgcd),
               maxlen(i, j + 1, arr,
                      arr1, N, maxgcd));
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 8, 5, 6 };
    int arr1[] = { 1, 2, 8, 5, 6 };
     
    // Sorted array
    int n = sizeof(arr) / sizeof(arr[0]);
     
    sort(arr, arr + n);
    sort(arr1, arr1 + n);
     
    // Function call to calculate GCD of
    // pair with maximum GCD
    int maxgcd = findMaxGCD(arr, n);
     
    // Print the result
    cout << maxlen(0, 0, arr,
                   arr1, n, maxgcd) + 1;
}
 
// This code is contributed by ipg2016107

Java




// Java program for the
// above approach
import java.util.*;
 
class GFG{
     
// Recursive function to
// return gcd of a and b 
static int __gcd(int a,
                 int b) 
  return b == 0 ?
         a :__gcd(b, a % b);    
}
   
// Function to find GCD of
// pair with maximum GCD
static int findMaxGCD(int arr[],
                      int N)
{   
  // Stores maximum element
  // of arr[]
  int high = 0;
 
  // Find the maximum element
  for(int i = 0; i < N; i++)
  {
    high = Math.max(high,
                    arr[i]);
  }
 
  // Maintain a count array
  int []count = new int[high + 1];
 
  // Store the frequency of arr[]
  for(int i = 0; i < N; i++)
  {
    count[arr[i]] += 1;
  }
 
  // Stores the multiples of
  // a number
  int counter = 0;
 
  // Iterate over the  range
  // [MAX, 1]
  for(int i = high; i > 0; i--)
  {
    int j = i;
 
    // Iterate from current potential
    // GCD till it is less than MAX
    while (j <= high)
    {
      // A multiple found
      if (count[j] > 0)
        counter += count[j];
 
      // Increment potential GCD by
      // itself io check i, 2i, 3i...
      j += i;
 
      // If 2 multiples found max
      // GCD found
      if (counter == 2)
        return i;
    }
    counter = 0;
  }
  return 0;
}
 
// Function to find longest
// subsequence such that GCD
// of any two distinct integers
// is maximum
static int maxlen(int i, int j,
                  int arr[],
                  int arr1[],
                  int N, int maxgcd)
{
  int a = 1;
 
  // Base Cases
  if (i >= N || j >= N)
    return 0;
 
  // Compare current GCD to
  // the maximum GCD
  if (__gcd(arr[i], arr1[j]) == maxgcd &&
      arr[i] != arr1[j])
  {
    // If true increment and
    // move the pointer
    a = Math.max(a, 1 + maxlen(i, j + 1,
                               arr, arr1,
                               N, maxgcd));
    return a;
  }
 
  // Return max of either subsequences
  return Math.max(maxlen(i + 1, j, arr,
                         arr1, N, maxgcd),
                  maxlen(i, j + 1, arr,
                         arr1, N, maxgcd));
}
 
// Driver Code
public static void main(String[] args)
{
  int arr[] = {1, 2, 8, 5, 6};
  int arr1[] = {1, 2, 8, 5, 6};
 
  // Sorted array
  int n = arr.length;
 
  Arrays.sort(arr);
 
  // Function call to calculate GCD of
  // pair with maximum GCD
  int maxgcd = findMaxGCD(arr, n);
 
  // Print the result
  System.out.print(maxlen(0, 0, arr,
                          arr1, n, maxgcd));
}
}
 
// This code is contributed by gauravrajput1

Python3




# Python3 program for the above approach
import math
 
# Function to find GCD of pair with
# maximum GCD
def findMaxGCD(arr, N):
     
    # Stores maximum element of arr[]
    high = 0
     
    # Find the maximum element
    for i in range(0, N):
        high = max(high, arr[i])
 
    # Maintain a count array
    count = [0] * (high + 1)
     
    # Store the frequency of arr[]
    for i in range(0, N):
        count[arr[i]] += 1
 
    # Stores the multiples of a number
    counter = 0
 
    # Iterate over the  range [MAX, 1]
    for i in range(high, 0, -1):
        j = i
 
        # Iterate from current potential
        # GCD till it is less than MAX
        while (j <= high):
 
            # A multiple found
            if (count[j] > 0):
                counter += count[j]
 
            # Increment potential GCD by
            # itself io check i, 2i, 3i...
            j += i
 
            # If 2 multiples found max
            # GCD found
            if (counter == 2):
                return i
 
        counter = 0
 
# Function to find longest subsequence
# such that GCD of any two distinct
# integers is maximum
def maxlen(i, j):
    a = 0
 
    # Base Cases
    if i >= N or j >= N:
        return 0
 
    # Compare current GCD to the
    # maximum GCD
    if math.gcd(arr[i], arr1[j]) == maxgcd and arr[i] != arr1[j]:
 
        # If true increment and
        # move the pointer
        a = max(a, 1 + maxlen(i, j + 1))
        return a
 
    # Return max of either subsequences
    return max(maxlen(i + 1, j), maxlen(i, j + 1))
 
 
# Drivers Code
arr = [1, 2, 8, 5, 6]
 
# Sorted array
arr1 = sorted(arr)
 
# Length of the array
N = len(arr)
 
# Function call to calculate GCD of
# pair with maximum GCD
maxgcd = findMaxGCD(arr, N)
 
# Print the result
print(maxlen(0, 0))

C#




// C# program for the
// above approach
using System;
 
class GFG{
     
// Recursive function to
// return gcd of a and b 
static int __gcd(int a, int b) 
{
  return b == 0 ?
         a :__gcd(b, a % b);    
}
   
// Function to find GCD of
// pair with maximum GCD
static int findMaxGCD(int []arr,
                      int N)
{   
   
  // Stores maximum element
  // of []arr
  int high = 0;
 
  // Find the maximum element
  for(int i = 0; i < N; i++)
  {
    high = Math.Max(high,
                    arr[i]);
  }
 
  // Maintain a count array
  int []count = new int[high + 1];
 
  // Store the frequency of []arr
  for(int i = 0; i < N; i++)
  {
    count[arr[i]] += 1;
  }
 
  // Stores the multiples of
  // a number
  int counter = 0;
 
  // Iterate over the  range
  // [MAX, 1]
  for(int i = high; i > 0; i--)
  {
    int j = i;
 
    // Iterate from current potential
    // GCD till it is less than MAX
    while (j <= high)
    {
       
      // A multiple found
      if (count[j] > 0)
        counter += count[j];
 
      // Increment potential GCD by
      // itself io check i, 2i, 3i...
      j += i;
 
      // If 2 multiples found max
      // GCD found
      if (counter == 2)
        return i;
    }
    counter = 0;
  }
  return 0;
}
 
// Function to find longest
// subsequence such that GCD
// of any two distinct integers
// is maximum
static int maxlen(int i, int j,
                  int []arr,
                  int []arr1,
                  int N, int maxgcd)
{
  int a = 1;
 
  // Base Cases
  if (i >= N || j >= N)
    return 0;
 
  // Compare current GCD to
  // the maximum GCD
  if (__gcd(arr[i], arr1[j]) == maxgcd &&
                      arr[i] != arr1[j])
  {
     
    // If true increment and
    // move the pointer
    a = Math.Max(a, 1 + maxlen(i, j + 1,
                               arr, arr1,
                               N, maxgcd));
    return a;
  }
 
  // Return max of either subsequences
  return Math.Max(maxlen(i + 1, j, arr,
                         arr1, N, maxgcd),
                  maxlen(i, j + 1, arr,
                         arr1, N, maxgcd));
}
 
// Driver Code
public static void Main(String[] args)
{
  int []arr = { 1, 2, 8, 5, 6 };
  int []arr1 = { 1, 2, 8, 5, 6 };
 
  // Sorted array
  int n = arr.Length;
 
  Array.Sort(arr);
 
  // Function call to calculate GCD of
  // pair with maximum GCD
  int maxgcd = findMaxGCD(arr, n);
 
  // Print the result
  Console.Write(maxlen(0, 0, arr,
                       arr1, n, maxgcd));
}
}
 
// This code is contributed by Princi Singh

Javascript




<script>
// javascript program for the
// above approach
 
    // Recursive function to
    // return gcd of a and b
    function __gcd(a , b) {
        return b == 0 ? a : __gcd(b, a % b);
    }
 
    // Function to find GCD of
    // pair with maximum GCD
    function findMaxGCD(arr, N)
    {
     
        // Stores maximum element
        // of arr
        var high = 0;
 
        // Find the maximum element
        for (i = 0; i < N; i++)
        {
            high = Math.max(high, arr[i]);
        }
 
        // Maintain a count array
        var count = Array(high + 1).fill(0);
 
        // Store the frequency of arr
        for (i = 0; i < N; i++)
        {
            count[arr[i]] += 1;
        }
 
        // Stores the multiples of
        // a number
        var counter = 0;
 
        // Iterate over the range
        // [MAX, 1]
        for (i = high; i > 0; i--)
        {
            var j = i;
 
            // Iterate from current potential
            // GCD till it is less than MAX
            while (j <= high)
            {
             
                // A multiple found
                if (count[j] > 0)
                    counter += count[j];
 
                // Increment potential GCD by
                // itself io check i, 2i, 3i...
                j += i;
 
                // If 2 multiples found max
                // GCD found
                if (counter == 2)
                    return i;
            }
            counter = 0;
        }
        return 0;
    }
 
    // Function to find longest
    // subsequence such that GCD
    // of any two distinct integers
    // is maximum
    function maxlen(i , j , arr , arr1 , N , maxgcd)
    {
        var a = 1;
 
        // Base Cases
        if (i >= N || j >= N)
            return 0;
 
        // Compare current GCD to
        // the maximum GCD
        if (__gcd(arr[i], arr1[j]) == maxgcd && arr[i] != arr1[j])
        {
         
            // If true increment and
            // move the pointer
            a = Math.max(a, 1 + maxlen(i, j + 1, arr, arr1, N, maxgcd));
            return a;
        }
 
        // Return max of either subsequences
        return Math.max(maxlen(i + 1, j, arr, arr1, N, maxgcd), maxlen(i, j + 1, arr, arr1, N, maxgcd));
    }
 
    // Driver Code
     
        var arr = [ 1, 2, 8, 5, 6 ];
        var arr1 = [ 1, 2, 8, 5, 6 ];
 
        // Sorted array
        var n = arr.length;
 
        arr.sort();
 
        // Function call to calculate GCD of
        // pair with maximum GCD
        var maxgcd = findMaxGCD(arr, n);
 
        // Prvar the result
        document.write(maxlen(0, 0, arr, arr1, n, maxgcd));
 
// This code is contributed by gauravrajput1.
</script>
Output: 
3

 

Time Complexity: O(2N)
Auxiliary Space: O(N)

Efficient Approach: To optimize the above approach, the idea is to use Dynamic Programming as there are many overlapping subproblems in the above solution which are solved again and again. Therefore, the recomputation of the same subproblems can be avoided by either using Memoization or Tabulation. Use a dictionary to memorize the maximum length of the subsequence between two indexes and use those values in the next recursion calls. Below are the steps:

  1. Perform all the above-explained steps.
  2. Use a dictionary dp to store the maximum length of the subsequence between two indexes.
  3. While having the recursive call just check in the table dp[][] whether the value is previously computed or not. If found to be true then use this value. Otherwise, call the function for other index range.
  4. Print the maximum length of subsequence calculated after the above steps.

 Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
 
map<pair<int, int>, int>dp;
int maxgcd;
vector<int>arr, arr1;
int N;
 
// Function to find GCD of pair with
// maximum GCD
int findMaxGCD(vector<int>arr, int  N)
{
     
    // Stores maximum element of arr[]
    int high = 0;
     
    // Find the maximum element
    for(int i = 0; i < N; i++)
        high = max(high, arr[i]);
         
    // Maintain a count array
    int count[high + 1] = {0};
 
    // Store the frequency of arr[]
    for(int i = 0; i < N; i++)
        count[arr[i]] += 1;
         
    // Stores the multiples of a number
    int counter = 0;
     
    // Iterate over the  range [MAX, 1]
    for(int i = high; i > 0; i--)
    {
        int j = i;
         
        // Iterate from current potential
        // GCD till it is less than MAX
        while (j <= high)
        {
             
            // A multiple found
            if (count[j] > 0)
                counter += count[j];
                 
            // Increment potential GCD by
            // itself io check i, 2i, 3i...
            j += i;
             
            // If 2 multiples found max
            // GCD found
            if (counter == 2)
                return i;
        }
        counter = 0;
    }
    return 1;
}
     
// Function to find longest subsequence
// such that GCD of any two distinct
// integers is maximum
int maxlen(int i, int j)
{
    int  a = 0;
 
    // Create the key
    // Base Case
    if (i >= N or j >= N)
        return 0;
         
    // Check if the current state is
    // already calculated
    if (dp[{i, j}])
        return dp[{i, j}];
       
    // Compare current GCD to the
    // maximum GCD
    if (__gcd(arr[i], arr1[j]) ==
        maxgcd && arr[i] != arr1[j])
    {
         
        // If true increment and
        // move the pointer
        dp[{i, j}] = 1 + maxlen(i, j + 1);
        return dp[{i, j}];
    }
     
    // Return max of either subsequences
    return (max(maxlen(i + 1, j),
                maxlen(i, j + 1)));
}
   
// Driver Code
int main()
{
    arr = { 1, 2, 8, 5, 6 };
    arr1 = arr;
     
    // Sorted array
    sort(arr1.begin(), arr1.end());
     
    // Length of the array
    N = arr.size();
     
    // Function Call
    maxgcd = findMaxGCD(arr, N);
     
    // Print the result
    cout << (maxlen(0, 0));
}
 
// This code is contributed by Stream_Cipher

Java




// Java program for the above approach
import java.util.*;
import java.awt.Point;
public class Main
{
  static HashMap<Point, Integer> dp = new HashMap<>();
  static int maxgcd, N;
  static Vector<Integer> arr;
  static Vector<Integer> arr1;
 
  static int __gcd(int a, int b)
  {
    if (b == 0)
      return a;
 
    return __gcd(b, a % b);
  }
 
  // Function to find GCD of pair with
  // maximum GCD
  static int findMaxGCD(Vector<Integer> arr, int N)
  {
 
    // Stores maximum element of arr[]
    int high = 0;
 
    // Find the maximum element
    for(int i = 0; i < N; i++)
      high = Math.max(high, arr.get(i));
 
    // Maintain a count array
    int[] count = new int[high + 1];
 
    // Store the frequency of arr[]
    for(int i = 0; i < N; i++)
      count[arr.get(i)] += 1;
 
    // Stores the multiples of a number
    int counter = 0;
 
    // Iterate over the  range [MAX, 1]
    for(int i = high; i > 0; i--)
    {
      int j = i;
 
      // Iterate from current potential
      // GCD till it is less than MAX
      while (j <= high)
      {
 
        // A multiple found
        if (count[j] > 0)
          counter += count[j];
 
        // Increment potential GCD by
        // itself io check i, 2i, 3i...
        j += i;
 
        // If 2 multiples found max
        // GCD found
        if (counter == 2)
          return i;
      }
      counter = 0;
    }
    return 1;
  }
 
  // Function to find longest subsequence
  // such that GCD of any two distinct
  // integers is maximum
  static int maxlen(int i, int j)
  {
 
    // Create the key
    // Base Case
    if (i >= N || j >= N)
      return 0;
 
    // Check if the current state is
    // already calculated
    if (dp.containsKey(new Point(i, j)))
      return dp.get(new Point(i, j));
 
    // Compare current GCD to the
    // maximum GCD
    if (__gcd(arr.get(i), arr1.get(j)) ==
        maxgcd && arr.get(i) != arr1.get(j))
    {
 
      // If true increment and
      // move the pointer
      dp.put(new Point(i, j), 1 + maxlen(i, j + 1));
      return dp.get(new Point(i, j));
    }
 
    // Return max of either subsequences
    return (Math.max(maxlen(i + 1, j), maxlen(i, j + 1)));
  }
 
  public static void main(String[] args) {
    arr = new Vector<Integer>();
    arr.add(1);
    arr.add(2);
    arr.add(8);
    arr.add(5);
    arr.add(6);
    arr1 = arr;
 
    // Sorted array
    Collections.sort(arr1);
 
    // Length of the array
    N = arr.size();
 
    // Function Call
    maxgcd = findMaxGCD(arr, N);
 
    // Print the result
    System.out.println(maxlen(0, 0) + 1);
  }
}
 
// This code is contributed by divyeshrabadiya07.

Python3




# Python3 program for the above approach
import math
 
# Function to find GCD of pair with
# maximum GCD
def findMaxGCD(arr, N):
     
    # Stores maximum element of arr[]
    high = 0
     
    # Find the maximum element
    for i in range(0, N):
        high = max(high, arr[i])
 
    # Maintain a count array
    count = [0] * (high + 1)
     
    # Store the frequency of arr[]
    for i in range(0, N):
        count[arr[i]] += 1
 
    # Stores the multiples of a number
    counter = 0
 
    # Iterate over the  range [MAX, 1]
    for i in range(high, 0, -1):
        j = i
 
        # Iterate from current potential
        # GCD till it is less than MAX
        while (j <= high):
 
            # A multiple found
            if (count[j] > 0):
                counter += count[j]
 
            # Increment potential GCD by
            # itself io check i, 2i, 3i...
            j += i
 
            # If 2 multiples found max
            # GCD found
            if (counter == 2):
                return i
 
        counter = 0
 
# Function to find longest subsequence
# such that GCD of any two distinct
# integers is maximum
def maxlen(i, j):
   
    a = 0
     
    # Create the key
    key = (i, j)
 
    # Base Case
    if i >= N or j >= N:
        return 0
     
    # Check if the current state is
    # already calculated
    if key in dp:
        return dp[key]
       
    # Comapare current GCD to the
    # maximum GCD
    if math.gcd(arr[i], arr1[j]) == maxgcd and arr[i] != arr1[j]:
 
        # If true increment and
        # move the pointer
        dp[key] = 1 + maxlen(i, j + 1)
 
        return dp[key]
 
    # Return max of either subsequences
    return max(maxlen(i + 1, j), maxlen(i, j + 1))
 
 
# Drivers code
arr = [1, 2, 8, 5, 6]
 
# Empty dictionary
dp = dict()
 
# Sorted array
arr1 = sorted(arr)
 
# Length of the array
N = len(arr)
 
# Function Call
maxgcd = findMaxGCD(arr, N)
 
# Print the result
print(maxlen(0, 0))

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
static Dictionary<Tuple<int,
                        int>, int> dp = new Dictionary<Tuple<int,
                                                             int>, int>();
static int maxgcd, N;
static List<int> arr;
static List<int> arr1;
 
static int __gcd(int a, int b)
{
    if (b == 0)
        return a;
         
    return __gcd(b, a % b);
}
 
// Function to find GCD of pair with
// maximum GCD
static int findMaxGCD(List<int> arr, int N)
{
     
    // Stores maximum element of arr[]
    int high = 0;
      
    // Find the maximum element
    for(int i = 0; i < N; i++)
        high = Math.Max(high, arr[i]);
          
    // Maintain a count array
    int[] count = new int[high + 1];
  
    // Store the frequency of arr[]
    for(int i = 0; i < N; i++)
        count[arr[i]] += 1;
          
    // Stores the multiples of a number
    int counter = 0;
      
    // Iterate over the  range [MAX, 1]
    for(int i = high; i > 0; i--)
    {
        int j = i;
          
        // Iterate from current potential
        // GCD till it is less than MAX
        while (j <= high)
        {
             
            // A multiple found
            if (count[j] > 0)
                counter += count[j];
                  
            // Increment potential GCD by
            // itself io check i, 2i, 3i...
            j += i;
              
            // If 2 multiples found max
            // GCD found
            if (counter == 2)
                return i;
        }
        counter = 0;
    }
    return 1;
}
      
// Function to find longest subsequence
// such that GCD of any two distinct
// integers is maximum
static int maxlen(int i, int j)
{
     
    // Create the key
    // Base Case
    if (i >= N || j >= N)
        return 0;
          
    // Check if the current state is
    // already calculated
    if (dp.ContainsKey(new Tuple<int, int>(i, j)))
        return dp[new Tuple<int, int>(i, j)];
        
    // Compare current GCD to the
    // maximum GCD
    if (__gcd(arr[i], arr1[j]) ==
             maxgcd && arr[i] != arr1[j])
    {
         
        // If true increment and
        // move the pointer
        dp[new Tuple<int, int>(i, j)] = 1 + maxlen(
                                     i, j + 1);
        return dp[new Tuple<int, int>(i, j)];
    }
      
    // Return max of either subsequences
    return (Math.Max(maxlen(i + 1, j),
                  maxlen(i, j + 1)));
}
 
// Driver code
static void Main()
{
    arr = new List<int>(new int[]{ 1, 2, 8, 5, 6 });
    arr1 = arr;
      
    // Sorted array
    arr1.Sort();
      
    // Length of the array
    N = arr.Count;
      
    // Function Call
    maxgcd = findMaxGCD(arr, N);
      
    // Print the result
    Console.Write(maxlen(0, 0) + 1);
}
}
 
// This code is contributed by divyesh072019

Javascript




<script>
 
// JavaScript program for the above approach
 
var dp = new Map();
var maxgcd;
var arr = [], arr1 = [];
var N;
 
// Function to find GCD of pair with
// maximum GCD
function findMaxGCD(arr, N)
{
     
    // Stores maximum element of arr[]
    var high = 0;
     
    // Find the maximum element
    for(var i = 0; i < N; i++)
        high = Math.max(high, arr[i]);
         
    // Maintain a count array
    var count = Array(high+1).fill(0);
 
    // Store the frequency of arr[]
    for(var i = 0; i < N; i++)
        count[arr[i]] += 1;
         
    // Stores the multiples of a number
    var counter = 0;
     
    // Iterate over the  range [MAX, 1]
    for(var i = high; i > 0; i--)
    {
        var j = i;
         
        // Iterate from current potential
        // GCD till it is less than MAX
        while (j <= high)
        {
             
            // A multiple found
            if (count[j] > 0)
                counter += count[j];
                 
            // Increment potential GCD by
            // itself io check i, 2i, 3i...
            j += i;
             
            // If 2 multiples found max
            // GCD found
            if (counter == 2)
                return i;
        }
        counter = 0;
    }
    return 1;
}
 
function __gcd(a, b)
{
    if (b == 0)
        return a;
         
    return __gcd(b, a % b);
}
     
// Function to find longest subsequence
// such that GCD of any two distinct
// integers is maximum
function maxlen(i, j)
{
    var a = 0;
 
    // Create the key
    // Base Case
    if (i >= N ||  j >= N)
        return 0;
         
    // Check if the current state is
    // already calculated
    if (dp.has([i, j].toString()))
        return dp.get([i, j].toString());
       
    // Compare current GCD to the
    // maximum GCD
    if (__gcd(arr[i], arr1[j]) ==
        maxgcd && arr[i] != arr1[j])
    {
         
        // If true increment and
        // move the pointer
        dp.set([i, j].toString(), 1 + maxlen(i, j + 1));
        return dp.get([i, j].toString());
    }
     
    // Return max of either subsequences
    return (Math.max(maxlen(i + 1, j),
                maxlen(i, j + 1)));
}
   
// Driver Code
arr = [1, 2, 8, 5, 6];
arr1 = JSON.parse(JSON.stringify(arr));
 
// Sorted array
arr1.sort((a,b)=>a-b);
 
// Length of the array
N = arr.length;
 
// Function Call
maxgcd = findMaxGCD(arr, N);
 
// Print the result
document.write(maxlen(0, 0));
 
 
</script>
Output: 
3

 

Time Complexity: O(N2)
Auxiliary Space: O(N)

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