Find the length of the Largest subset such that all elements are Pairwise Coprime

Given an array A of size N, our task is to find the length of the largest subset such that all elements in the subset are pairwise coprime that is for any two elements x and y where x and y are not the same, the gcd(x, y) is equal to 1.
Note: All array elements are <= 50.

Examples:

Input: A = [2, 5, 2, 5, 2] 
Output: 2 
Explanation:
The largest subset satisfying the condition is: {2, 5} 

Input: A = [2, 3, 13, 5, 14, 6, 7, 11] 
Output: 6 

Naive Approach:
To solve the problem mentioned above we have to generate all subsets, and for each subset check whether the given condition holds or not. But this method takes O(N² * 2^N) time and can be optimized further.

Below is the implementation of the above approach:

C++

// C++ implementation to Find the length of the Largest
// subset such that all elements are Pairwise Coprime
#include <bits/stdc++.h>
using namespace std;

// Function to find the largest subset possible
int largestSubset(int a[], int n)
{
    int answer = 0;

    // Iterate through all the subsets
    for (int i = 1; i < (1 << n); i++) {
        vector<int> subset;

        /* Check if jth bit in the counter is set */
        for (int j = 0; j < n; j++) {
            if (i & (1 << j))
                subset.push_back(a[j]);
        }

        bool flag = true;

        for (int j = 0; j < subset.size(); j++) {
            for (int k = j + 1; k < subset.size(); k++) {
                // Check if the gcd is not equal to 1
                if (__gcd(subset[j], subset[k]) != 1)
                    flag = false;
            }
        }

        if (flag == true)
            // Update the answer with maximum value
            answer = max(answer, (int)subset.size());
    }

    // Return the final result
    return answer;
}

// Driver code
int main()
{

    int A[] = { 2, 3, 13, 5, 14, 6, 7, 11 };

    int N = sizeof(A) / sizeof(A[0]);

    cout << largestSubset(A, N);

    return 0;
}

Java

// Java implementation to find the length
// of the largest subset such that all
// elements are Pairwise Coprime
import java.util.*;

class GFG{

static int gcd(int a, int b)
{

    // Everything divides 0
    if (a == 0)
        return b;
    if (b == 0)
        return a;

    // Base case
    if (a == b)
        return a;

    // a is greater
    if (a > b)
        return gcd(a - b, b);

    return gcd(a, b - a);
}

// Function to find the largest subset possible
static int largestSubset(int a[], int n)
{
    int answer = 0;

    // Iterate through all the subsets
    for(int i = 1; i < (1 << n); i++)
    {
        Vector<Integer> subset = new Vector<Integer>();

        // Check if jth bit in the counter is set
        for(int j = 0; j < n; j++)
        {
            if ((i & (1 << j)) != 0)
                subset.add(a[j]);
        }

        boolean flag = true;

        for(int j = 0; j < subset.size(); j++)
        {
            for(int k = j + 1; k < subset.size(); k++)
            {

                // Check if the gcd is not equal to 1
                if (gcd((int)subset.get(j),
                       (int) subset.get(k)) != 1)
                    flag = false;
            }
        }

        if (flag == true)

            // Update the answer with maximum value
            answer = Math.max(answer,
                             (int)subset.size());
    }

    // Return the final result
    return answer;
}

// Driver code
public static void main(String args[])
{
    int A[] = { 2, 3, 13, 5, 14, 6, 7, 11 };

    int N = A.length;

    System.out.println(largestSubset(A, N));
}
}

// This code is contributed by Stream_Cipher

Python3

# Python3 implementation to Find
# the length of the Largest subset
# such that all elements are Pairwise Coprime
import math

# Function to find the largest subset possible
def largestSubset(a, n):
    answer = 0

    # Iterate through all the subsets
    for i in range(1, (1 << n)):
        subset = []

        # Check if jth bit in the counter is set
        for j in range(0, n):
            if (i & (1 << j)):
                subset.insert(j, a[j])

        flag = True

        for j in range(0, len(subset)):
            for k in range(j + 1, len(subset)):

                # Check if the gcd is not equal to 1
                if (math.gcd(subset[j], subset[k]) != 1) :
                    flag = False

        if (flag == True):

            # Update the answer with maximum value
            answer = max(answer, len(subset))

    # Return the final result
    return answer

# Driver code
A = [ 2, 3, 13, 5, 14, 6, 7, 11 ]
N = len(A)
print(largestSubset(A, N))

# This code is contributed by Sanjit_Prasad

C#

// C# implementation to Find the length
// of the largest subset such that all
// elements are Pairwise Coprime
using System;
using System.Collections.Generic;

class GFG{

static int gcd(int a, int b)
{

    // Everything divides 0
    if (a == 0)
        return b;
    if (b == 0)
        return a;

    // base case
    if (a == b)
        return a;

    // a is greater
    if (a > b)
        return gcd(a - b, b);

    return gcd(a, b - a);
}

// Function to find the largest subset possible
static int largestSubset(int []a, int n)
{
    int answer = 0;

    // Iterate through all the subsets
    for(int i = 1; i < (1 << n); i++)
    {
        List<int> subset = new List<int>();

        // Check if jth bit in the counter is set
        for(int j = 0; j < n; j++)
        {
            if ((i & (1 << j)) != 0)
                subset.Add(a[j]);
        }

        int flag = 1;

        for(int j = 0; j < subset.Count; j++)
        {
            for(int k = j + 1; k < subset.Count; k++)
            {
                // Check if the gcd is not equal to 1
                if (gcd((int)subset[j],
                       (int) subset[k]) != 1)
                    flag = 0;
            }
        }

        if (flag == 1)

            // Update the answer with maximum value
            answer = Math.Max(answer,
                             (int)subset.Count);
    }

    // Return the final result
    return answer;
}

// Driver code
public static void Main()
{
    int []A = { 2, 3, 13, 5, 14, 6, 7, 11 };

    int N = A.Length;

    Console.WriteLine(largestSubset(A, N));
}
}

// This code is contributed by Stream_Cipher

Javascript

<script>
    // Javascript implementation to Find the length
    // of the largest subset such that all
    // elements are Pairwise Coprime

    function gcd(a, b)
    {

        // Everything divides 0
        if (a == 0)
            return b;
        if (b == 0)
            return a;

        // base case
        if (a == b)
            return a;

        // a is greater
        if (a > b)
            return gcd(a - b, b);

        return gcd(a, b - a);
    }

    // Function to find the largest subset possible
    function largestSubset(a, n)
    {
        let answer = 0;

        // Iterate through all the subsets
        for(let i = 1; i < (1 << n); i++)
        {
            let subset = [];

            // Check if jth bit in the counter is set
            for(let j = 0; j < n; j++)
            {
                if ((i & (1 << j)) != 0)
                    subset.push(a[j]);
            }

            let flag = 1;

            for(let j = 0; j < subset.length; j++)
            {
                for(let k = j + 1; k < subset.length; k++)
                {
                    // Check if the gcd is not equal to 1
                    if (gcd(subset[j], subset[k]) != 1)
                        flag = 0;
                }
            }

            if (flag == 1)

                // Update the answer with maximum value
                answer = Math.max(answer, subset.length);
        }

        // Return the final result
        return answer;
    }

    let A = [ 2, 3, 13, 5, 14, 6, 7, 11 ];

    let N = A.length;

    document.write(largestSubset(A, N));

</script>

Output:

6

Efficient Approach:
The above method can be optimized and the approach depends on the fact that there are only 15 prime numbers in the first 50 natural numbers. So all the numbers in array will have prime factors among these 15 numbers only. We will use Bitmasking and Dynamic Programming to optimize the problem.

• Since there are 15 primes only, consider a 15-bit representation of every number where each bit is 1 if that index of prime is a factor of that number. We will index prime numbers by 0 indexing, which means 2 at 0th position 3 at index 1 and so on.
• An integer variable ‘mask‘ indicates the prime factors which have already occurred in the subset. If i’th bit is set in the mask, then i’th prime factor has occurred, otherwise not.
• At each step of recurrence relation, the element can either be included in the subset or cannot be included. If the element is not included in the subarray, then simply move to the next index. If it is included, change the mask by setting all the bits corresponding to the current element’s prime factors, ON in the mask. The current element can only be included if all of its prime factors have not occurred previously.
• This condition will be satisfied only if the bits corresponding to the current element’s digits in the mask are OFF.

If we draw the complete recursion tree, we can observe that many subproblems are being solved which were occurring again and again. So we use a table dp[][] such that for every index dp[i][j], i is the position of the element in the array, and j is the mask.

Below is the implementation of the above approach:

C++

// C++ implementation to Find the length of the Largest
// subset such that all elements are Pairwise Coprime
#include <bits/stdc++.h>
using namespace std;

// Dynamic programming table
int dp[5000][(1 << 10) + 5];

// Function to obtain the mask for any integer
int getmask(int val)
{
    int mask = 0;

    // List of prime numbers till 50
    int prime[15] = { 2, 3, 5, 7, 11, 13, 17, 19,
                      23, 29, 31, 37, 41, 43, 47 };

    // Iterate through all prime numbers to obtain the mask
    for (int i = 0; i < 15; i++) {
        if (val % prime[i] == 0) {
            // Set this prime's bit ON in the mask
            mask = mask | (1 << i);
        }
    }

    // Return the mask value
    return mask;
}

// Function to count the number of ways
int calculate(int pos, int mask,
              int a[], int n)
{
    if (pos == n || mask == (1 << n - 1))
        return 0;

    // Check if subproblem has been solved
    if (dp[pos][mask] != -1)
        return dp[pos][mask];

    int size = 0;

    // Excluding current element in the subset
    size = max(size, calculate(pos + 1, mask, a, n));

    // Check if there are no common prime factors
    // then only this element can be included
    if ((getmask(a[pos]) & mask) == 0) {

        // Calculate the new mask if this element is included
        int new_mask = (mask | (getmask(a[pos])));

        size = max(size, 1 + calculate(pos + 1, new_mask, a, n));
    }

    // Store and return the answer
    return dp[pos][mask] = size;
}

// Function to find the count of
// subarray with all digits unique
int largestSubset(int a[], int n)
{
    // Initializing dp
    memset(dp, -1, sizeof(dp));

    return calculate(0, 0, a, n);
}

// Driver code
int main()
{

    int A[] = { 2, 3, 13, 5, 14, 6, 7, 11 };

    int N = sizeof(A) / sizeof(A[0]);

    cout << largestSubset(A, N);

    return 0;
}

Java

// Java implementation to find the length
// of the largest subset such that all
// elements are Pairwise Coprime
import java.util.*;

class GFG{

// Dynamic programming table
static int dp[][] = new int [5000][1029];

// Function to obtain the mask for any integer
static int getmask(int val)
{
    int mask = 0;

    // List of prime numbers till 50
    int prime[] = { 2, 3, 5, 7, 11, 13, 17, 19,
                    23, 29, 31, 37, 41, 43, 47 };

    // Iterate through all prime numbers
    // to obtain the mask
    for(int i = 0; i < 15; i++)
    {
        if (val % prime[i] == 0)
        {

            // Set this prime's bit ON in the mask
            mask = mask | (1 << i);
        }
    }

    // Return the mask value
    return mask;
}

// Function to count the number of ways
static int calculate(int pos, int mask,
                     int a[], int n)
{
    if (pos == n ||
       mask == (int)(1 << n - 1))
        return 0;

    // Check if subproblem has been solved
    if (dp[pos][mask] != -1)
        return dp[pos][mask];

    int size = 0;

    // Excluding current element in the subset
    size = Math.max(size, calculate(pos + 1,
                                    mask, a, n));

    // Check if there are no common prime factors
    // then only this element can be included
    if ((getmask(a[pos]) & mask) == 0)
    {

        // Calculate the new mask if this
        // element is included
        int new_mask = (mask | (getmask(a[pos])));

        size = Math.max(size, 1 + calculate(pos + 1,
                                            new_mask,
                                            a, n));
    }

    // Store and return the answer
    return dp[pos][mask] = size;
}

// Function to find the count of
// subarray with all digits unique
static int largestSubset(int a[], int n)
{
    for(int i = 0; i < 5000; i++)
        Arrays.fill(dp[i], -1);

    return calculate(0, 0, a, n);
}

// Driver code
public static void main(String args[])
{
    int A[] = { 2, 3, 13, 5, 14, 6, 7, 11 };

    int N = A.length;

    System.out.println(largestSubset(A, N));
}
}

// This code is contributed by Stream_Cipher

Python

# Python implementation to find the
# length of the Largest subset such
# that all elements are Pairwise Coprime

# Dynamic programming table
dp = [[-1] * ((1 << 10) + 5)] * 5000

# Function to obtain the mask for any integer
def getmask(val):

    mask = 0

    # List of prime numbers till 50
    prime = [ 2, 3, 5, 7, 11, 13, 17, 19,
              23, 29, 31, 37, 41, 43, 47 ]

    # Iterate through all prime numbers
    # to obtain the mask
    for i in range(1, 15):
        if val % prime[i] == 0:

            # Set this prime's bit ON in the mask
            mask = mask | (1 << i)

    # Return the mask value
    return mask

# Function to count the number of ways
def calculate(pos, mask, a, n):

    if ((pos == n) or (mask == (1 << n - 1))):
        return 0

    # Check if subproblem has been solved
    if dp[pos][mask] != -1:
        return dp[pos][mask]

    size = 0

    # Excluding current element in the subset
    size = max(size, calculate(pos + 1,
                               mask, a, n))

    # Check if there are no common prime factors
    # then only this element can be included
    if (getmask(a[pos]) & mask) == 0:

        # Calculate the new mask if this
        # element is included
        new_mask = (mask | (getmask(a[pos])))
        size = max(size, 1 + calculate(pos + 1,
                                       new_mask,
                                       a, n))
    # Store and return the answer
    dp[pos][mask] = size
    return dp[pos][mask]

# Function to find the count of
# subarray with all digits unique
def largestSubset(A, n):

    return calculate(0, 0, A, n);

# Driver code
A = [ 2, 3, 13, 5, 14, 6, 7, 11 ]
N = len(A)

print(largestSubset(A, N))

# This code is contributed by Stream_Cipher

C#

// C# implementation to find the length
// of the largest subset such that all
// elements are Pairwise Coprime
using System;

class GFG{

// Dynamic programming table
static int [,] dp = new int [5000, 1029];

// Function to obtain the mask for any integer
static int getmask(int val)
{
    int mask = 0;

    // List of prime numbers till 50
    int []prime = { 2, 3, 5, 7, 11, 13, 17, 19,
                    23, 29, 31, 37, 41, 43, 47 };

    // Iterate through all prime
    // numbers to obtain the mask
    for(int i = 0; i < 15; i++)
    {
        if (val % prime[i] == 0)
        {

            // Set this prime's bit ON in the mask
            mask = mask | (1 << i);
        }
    }

    // Return the mask value
    return mask;
}

// Function to count the number of ways
static int calculate(int pos, int mask,
                     int []a, int n)
{
    if (pos == n ||
       mask == (int)(1 << n - 1))
        return 0;

    // Check if subproblem has been solved
    if (dp[pos, mask] != -1)
        return dp[pos, mask];

    int size = 0;

    // Excluding current element in the subset
    size = Math.Max(size, calculate(pos + 1,
                                    mask, a, n));

    // Check if there are no common prime factors
    // then only this element can be included
    if ((getmask(a[pos]) & mask) == 0)
    {

        // Calculate the new mask if
        // this element is included
        int new_mask = (mask | (getmask(a[pos])));

        size = Math.Max(size, 1 + calculate(pos + 1,
                                            new_mask,
                                            a, n));
    }

    // Store and return the answer
    return dp[pos, mask] = size;
}

// Function to find the count of
// subarray with all digits unique
static int largestSubset(int []a, int n)
{
    for(int i = 0; i < 5000; i++)
    {
        for(int j = 0; j < 1029; j++)
            dp[i, j] = -1;
    }
    return calculate(0, 0, a, n);
}

// Driver code
public static void Main()
{
    int []A = { 2, 3, 13, 5, 14, 6, 7, 11 };

    int N = A.Length;

    Console.WriteLine(largestSubset(A, N));
}
}

// This code is contributed by Stream_Cipher

Javascript

<script>

// JavaScript implementation to
// Find the length of the Largest
// subset such that all elements
// are Pairwise Coprime

// Dynamic programming table
var dp = Array.from(Array(5000), ()=>Array((1 << 10) + 5));

// Function to obtain the mask for any integer
function getmask( val)
{
    var mask = 0;

    // List of prime numbers till 50
    var prime = [2, 3, 5, 7, 11, 13, 17, 19,
                      23, 29, 31, 37, 41, 43, 47];

    // Iterate through all prime numbers to obtain the mask
    for (var i = 0; i < 15; i++) {
        if (val % prime[i] == 0) {
            // Set this prime's bit ON in the mask
            mask = mask | (1 << i);
        }
    }

    // Return the mask value
    return mask;
}

// Function to count the number of ways
function calculate(pos, mask, a, n)
{
    if (pos == n || mask == (1 << n - 1))
        return 0;

    // Check if subproblem has been solved
    if (dp[pos][mask] != -1)
        return dp[pos][mask];

    var size = 0;

    // Excluding current element in the subset
    size = Math.max(size, calculate(pos + 1, mask, a, n));

    // Check if there are no common prime factors
    // then only this element can be included
    if ((getmask(a[pos]) & mask) == 0) {

        // Calculate the new mask if this element is included
        var new_mask = (mask | (getmask(a[pos])));

        size = Math.max(size,
        1 + calculate(pos + 1, new_mask, a, n));
    }

    // Store and return the answer
    return dp[pos][mask] = size;
}

// Function to find the count of
// subarray with all digits unique
function largestSubset(a, n)
{
    // Initializing dp
    dp = Array.from(Array(5000),
    ()=>Array((1 << 10) + 5).fill(-1));

    return calculate(0, 0, a, n);
}

// Driver code
var A = [2, 3, 13, 5, 14, 6, 7, 11 ];
var N = A.length;
document.write( largestSubset(A, N));


</script>

Output:

6

Time Complexity: O(N * 15 * 2¹⁵)