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

Last Updated : 11 Sep, 2020

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 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:
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 

Output:

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) 
{ 
    // Intializing 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 

Output:

Time Complexity: O(N * 15 * 2¹⁵)

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