# 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 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 * 2N) time and can be optimized further.

Below is the implementation of the above approach:

## C++

 `// CPP implementation to Find the length of the Largest ` `// subset such that all elements are Pairwise Coprime ` `#include ` `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); ` ` `  `    ``cout << largestSubset(A, N); ` ` `  `    ``return` `0; ` `} `

## 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 `

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:

 `// CPP implementation to Find the length of the Largest ` `// subset such that all elements are Pairwise Coprime ` `#include ` `using` `namespace` `std; ` ` `  `// Dynamic programming table ` `int` `dp[(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 = { 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); ` ` `  `    ``cout << largestSubset(A, N); ` ` `  `    ``return` `0; ` `} `

Output:

```6
```

Time Complexity: O(N * 15 * 215) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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