Given an array arr[] consisting of positive integers, the task is to check whether we can modify the array by adding any of the elements of the array such that it consists of only Primes.
Examples:
Input: arr[] = {3, 5, 7, 21}
Output: YES
Explanation:
Add the following elements, 3+5+21
So the array becomes {7, 29} which consists of only primes.Input: {2, 5, 12, 16}
Output: NO
Explanation:
There is no possible combination among elements which will make the array containing all primesInput: {3, 5, 13, 22}
Output: YES
Explanation:
Only one combination is possible,
Add all elements – 3+5+13+22 = {43} which is only prime
Pre-requisites: Bitmasking-DP
Approach:
- We can solve this problem using Dynamic Programming with Bitmasks. We can represent our DP state as a mask which is subset of elements.
- So let our dp array be DP[mask] which represents whether upto this mask (the chosen elements) the subset formed will elements are primes only.
- The max number of bits in mask will be the number of elements in the array.
- We keep on marking the elements encoded as a sequence in mask (If i-th index element is selected, then i-th bit is set in the mask) and keep checking whether the current chosen element (current sum) is prime, if it is prime and all other elements are visited, then we return true, and we get the answer.
- Else if other elements are not visited, then as the current sum is prime, we just need to search for other elements which can individually or by summing up to form some primes, so we can safely mark our current sum as 0 again.
- If the mask is full (All elements are visited) and the current sum is not prime, we return false, because there is atleast one sum that is not prime.
- Recurrence:
DP[mask] = solve(curr + arr[i], mask | 1<<i) where 0 <= i < n
Below is the implementation of the above approach.
C++
// C++ program to find the // array of primes#include <bits/stdc++.h>using namespace std;// DP array to store the// ans for max 20 elementsbool dp[1 << 20];// To check whether the // number is prime or notbool isprime(int n){ if (n == 1) return false; for (int i = 2; i * i <= n; i++) { if (n % i == 0) { return false; } } return true;}// Function to check whether the// array can be modify so that // there are only primesint solve(int arr[], int curr, int mask, int n){ // If curr is prime and all // elements are visited, // return true if (isprime(curr)) { if (mask == (1 << n) - 1) { return true; } // If all elements are not // visited, set curr=0, to // search for new prime sum curr = 0; } // If all elements are visited if (mask == (1 << n) - 1) { // If the current sum is // not prime return false if (!isprime(curr)) { return false; } } // If this state is already // calculated, return the // answer directly if (dp[mask]) return dp[mask]; // Try all state of mask for (int i = 0; i < n; i++) { // If ith index is not set if (!(mask & 1 << i)) { // Add the current element // and set ith index and recur if (solve(arr, curr + arr[i] , mask | 1 << i, n)) { // If subset can be formed // then return true return true; } } } // After every possibility of mask, // if the subset is not formed, // return false by memoizing. return dp[mask] = false;}// Driver codeint main(){ int arr[] = { 3, 6, 7, 13 }; int n = sizeof(arr) / sizeof(arr[0]); if(solve(arr, 0, 0, n)) { cout << "YES"; } else { cout << "NO"; } return 0; } |
Java
// Java program to find the array of primesimport java.util.*;class GFG{// dp array to store the// ans for max 20 elementsstatic boolean []dp = new boolean[1 << 20];// To check whether the // number is prime or notstatic boolean isprime(int n){ if (n == 1) return false; for(int i = 2; i * i <= n; i++) { if (n % i == 0) { return false; } } return true;}// Function to check whether the// array can be modify so that // there are only primesstatic boolean solve(int arr[], int curr, int mask, int n){ // If curr is prime and all // elements are visited, // return true if (isprime(curr)) { if (mask == (1 << n) - 1) { return true; } // If all elements are not // visited, set curr=0, to // search for new prime sum curr = 0; } // If all elements are visited if (mask == (1 << n) - 1) { // If the current sum is // not prime return false if (!isprime(curr)) { return false; } } // If this state is already // calculated, return the // answer directly if (dp[mask]) return dp[mask]; // Try all state of mask for(int i = 0; i < n; i++) { // If ith index is not set if ((mask & (1 << i)) == 0) { // Add the current element // and set ith index and recur if (solve(arr, curr + arr[i], mask | 1 << i, n)) { // If subset can be formed // then return true return true; } } } // After every possibility of mask, // if the subset is not formed, // return false by memoizing. return dp[mask] = false;}// Driver codepublic static void main(String[] args){ int arr[] = { 3, 6, 7, 13 }; int n = arr.length; if(solve(arr, 0, 0, n)) { System.out.print("YES"); } else { System.out.print("NO"); }}}// This code is contributed by Rohit_ranjan |
Python3
# Python3 program to find the array# of primes # DP array to store the # ans for max 20 elements dp = [0] * (1 << 20)# To check whether the # number is prime or not def isprime(n): if (n == 1): return False for i in range(2, n + 1): if (n % i == 0): return False return True# Function to check whether the # array can be modify so that # there are only primes def solve(arr, curr, mask, n): # If curr is prime and all # elements are visited, # return true if (isprime(curr)): if (mask == (1 << n) - 1): return True # If all elements are not # visited, set curr=0, to # search for new prime sum curr = 0 # If all elements are visited if (mask == (1 << n) - 1): # If the current sum is # not prime return false if (isprime(curr) == False): return False # If this state is already # calculated, return the # answer directly if (dp[mask] != False): return dp[mask] # Try all state of mask for i in range(n): # If ith index is not set if ((mask & 1 << i) == False): # Add the current element # and set ith index and recur if (solve(arr, curr + arr[i], mask | 1 << i, n)): # If subset can be formed # then return true return True # After every possibility of mask, # if the subset is not formed, # return false by memoizing. return (dp[mask] == False)# Driver code arr = [ 3, 6, 7, 13 ] n = len(arr) if (solve(arr, 0, 0, n)): print("YES")else: print("NO") # This code is contributed by code_hunt |
C#
// C# program to find the array of primesusing System;class GFG{// dp array to store the// ans for max 20 elementsstatic bool []dp = new bool[1 << 20];// To check whether the // number is prime or notstatic bool isprime(int n){ if (n == 1) return false; for(int i = 2; i * i <= n; i++) { if (n % i == 0) { return false; } } return true;}// Function to check whether the// array can be modify so that // there are only primesstatic bool solve(int []arr, int curr, int mask, int n){ // If curr is prime and all // elements are visited, // return true if (isprime(curr)) { if (mask == (1 << n) - 1) { return true; } // If all elements are not // visited, set curr=0, to // search for new prime sum curr = 0; } // If all elements are visited if (mask == (1 << n) - 1) { // If the current sum is // not prime return false if (!isprime(curr)) { return false; } } // If this state is already // calculated, return the // answer directly if (dp[mask]) return dp[mask]; // Try all state of mask for(int i = 0; i < n; i++) { // If ith index is not set if ((mask & (1 << i)) == 0) { // Add the current element // and set ith index and recur if (solve(arr, curr + arr[i], mask | 1 << i, n)) { // If subset can be formed // then return true return true; } } } // After every possibility of mask, // if the subset is not formed, // return false by memoizing. return dp[mask] = false;}// Driver codepublic static void Main(String[] args){ int []arr = { 3, 6, 7, 13 }; int n = arr.Length; if(solve(arr, 0, 0, n)) { Console.Write("YES"); } else { Console.Write("NO"); }}}// This code is contributed by Rohit_ranjan |
Javascript
<script>// Javascript program to find the // array of primes // DP array to store the // ans for max 20 elements dp = Array(1 << 20).fill(0); // To check whether the // number is prime or not function isprime(n) { if (n == 1) return false; for(var i = 2; i * i <= n; i++) { if (n % i == 0) { return false; } } return true; } // Function to check whether the // array can be modify so that // there are only primes function solve(arr, curr, mask, n) { // If curr is prime and all // elements are visited, // return true if (isprime(curr)) { if (mask == (1 << n) - 1) { return true; } // If all elements are not // visited, set curr=0, to // search for new prime sum curr = 0; } // If all elements are visited if (mask == (1 << n) - 1) { // If the current sum is // not prime return false if (!isprime(curr)) { return false; } } // If this state is already // calculated, return the // answer directly if (dp[mask]) return dp[mask]; // Try all state of mask for(var i = 0; i < n; i++) { // If ith index is not set if (!(mask & 1 << i)) { // Add the current element // and set ith index and recur if (solve(arr, curr + arr[i], mask | 1 << i, n)) { // If subset can be formed // then return true return true; } } } // After every possibility of mask, // if the subset is not formed, // return false by memoizing. return dp[mask] = false; } // Driver code var arr = [ 3, 6, 7, 13 ]var n = arr.lengthif (solve(arr, 0, 0, n)) { document.write("YES"); } else{ document.write("NO"); } // This code is contributed by rutvik_56 </script> |
Output:
YES
Time Complexity: O(N2)
Auxiliary Space: O(219).
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