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 * 2N) 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 possibleint 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 codeint 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 codepublic 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 codepublic 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 tableint dp[5000][(1 << 10) + 5];// Function to obtain the mask for any integerint 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 waysint 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 uniqueint largestSubset(int a[], int n){ // Initializing dp memset(dp, -1, sizeof(dp)); return calculate(0, 0, a, n);}// Driver codeint 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 codepublic 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 codeA = [ 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 codepublic 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 tablevar dp = Array.from(Array(5000), ()=>Array((1 << 10) + 5));// Function to obtain the mask for any integerfunction 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 waysfunction 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 uniquefunction largestSubset(a, n){ // Initializing dp dp = Array.from(Array(5000), ()=>Array((1 << 10) + 5).fill(-1)); return calculate(0, 0, a, n);}// Driver codevar 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)
Auxiliary Space: O(n * 2^n)
Efficient approach : Using DP Tabulation method ( Iterative approach )
The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memorization(top-down) because memorization method needs extra stack space of recursion calls.
Steps to solve this problem :
- Initialize a DP table “dp” with size (n+1) x (2^n) and set all its values to 0.
- Fill the DP table in a bottom-up manner:
a. For each position “pos” in the array “a” (from right to left), and for each mask value “mask” (from 0 to 2^n – 1), do the following:
b. If the current element “a[pos]” has no common factor with the numbers represented by the “mask” value, then compute the new mask “new_mask” by ORing the “mask” value with the “getmask(a[pos])” value.
c. Update the size of the subset for this position and mask by taking the maximum of:- the subset size for the next position “pos+1” and the current mask value “mask”, and
- the subset size for the next position “pos+1” and the new mask value “new_mask” (if it is valid).
- Return the subset size stored in the DP table for position 0 and mask 0.
Implementation :
C++
#include <bits/stdc++.h>using namespace std;// Function to obtain the mask for any integerint 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;}int largestSubset(int a[], int n) { int dp[n+1][(1<<10)+5]; // Initialize base cases for (int i = 0; i <= n; i++) { for (int j = 0; j <= (1<<n)-1; j++) { dp[i][j] = 0; } } // Fill the DP table in a bottom-up manner for (int pos = n-1; pos >= 0; pos--) { for (int mask = 0; mask < (1<<n)-1; mask++) { int size = dp[pos+1][mask]; if ((getmask(a[pos]) & mask) == 0) { int new_mask = (mask | (getmask(a[pos]))); size = max(size, 1 + dp[pos+1][new_mask]); } dp[pos][mask] = size; } } // Return the answer return dp[0][0];}// Driver codeint 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
import java.io.*;import java.lang.*;import java.util.*;class Main { // 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; } static int largestSubset(int[] a, int n) { int[][] dp = new int[n + 1][(1 << 10) + 5]; // Initialize base cases for (int i = 0; i <= n; i++) { for (int j = 0; j <= (1 << n) - 1; j++) { dp[i][j] = 0; } } // Fill the DP table in a bottom-up manner for (int pos = n - 1; pos >= 0; pos--) { for (int mask = 0; mask < (1 << n) - 1; mask++) { int size = dp[pos + 1][mask]; if ((getmask(a[pos]) & mask) == 0) { int new_mask = (mask | (getmask(a[pos]))); size = Math.max( size, 1 + dp[pos + 1][new_mask]); } dp[pos][mask] = size; } } // Return the answer return dp[0][0]; } // 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)); }} |
Python3
# Function to obtain the mask for any integerdef 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(15): if val % prime[i] == 0: # Set this prime's bit ON in the mask mask = mask | (1 << i) # Return the mask value return maskdef largestSubset(a): n = len(a) dp = [[0 for j in range((1 << 10)+5)] for i in range(n+1)] # Fill the DP table in a bottom-up manner for pos in range(n-1, -1, -1): for mask in range((1 << n)-1): size = dp[pos+1][mask] if (getmask(a[pos]) & mask) == 0: new_mask = (mask | (getmask(a[pos]))) size = max(size, 1 + dp[pos+1][new_mask]) dp[pos][mask] = size # Return the answer return dp[0][0]# Driver codeif __name__ == '__main__': A = [2, 3, 13, 5, 14, 6, 7, 11] print(largestSubset(A)) |
C#
using System;public class GFG{ // 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; } static int LargestSubset(int[] a, int n) { int[,] dp = new int[n+1, (1<<10)+5]; // Initialize base cases for (int i = 0; i <= n; i++) { for (int j = 0; j <= (1<<n)-1; j++) { dp[i, j] = 0; } } // Fill the DP table in a bottom-up manner for (int pos = n-1; pos >= 0; pos--) { for (int mask = 0; mask < (1<<n)-1; mask++) { int size = dp[pos+1, mask]; if ((GetMask(a[pos]) & mask) == 0) { int new_mask = (mask | (GetMask(a[pos]))); size = Math.Max(size, 1 + dp[pos+1, new_mask]); } dp[pos, mask] = size; } } // Return the answer return dp[0, 0]; } // 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)); }} |
Javascript
// Function to obtain the mask for any integerfunction getmask(val) { let mask = 0; // List of prime numbers till 50 const 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 (let 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 largestSubset(a) { const n = a.length; const dp = Array.from({ length: n + 1 }, () => new Array((1 << 10) + 5).fill(0)); // Fill the DP table in a bottom-up manner for (let pos = n - 1; pos >= 0; pos--) { for (let mask = (1 << n) - 2; mask >= 0; mask--) { let size = dp[pos + 1][mask]; if ((getmask(a[pos]) & mask) === 0) { const new_mask = mask | getmask(a[pos]); size = Math.max(size, 1 + dp[pos + 1][new_mask]); } dp[pos][mask] = size; } } // Return the answer return dp[0][0];}// Driver codeconst A = [2, 3, 13, 5, 14, 6, 7, 11];console.log(largestSubset(A)); |
Output
6
Time Complexity: O(n * 2^n)
Auxiliary Space: O(n * 2^n)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
