Bitwise AND of the sum of prime numbers and the sum of composite numbers in an array

Given an array of positive numbers, the task is to find the bitwise AND of the sum of non-prime numbers and the sum of prime numbers. Note that 1 is neither prime nor composite.

Examples:

Input: arr[] = {1, 3, 5, 10, 15, 7}
Output: 9
Sum of non-primes = 10 + 15 = 25
Sum of primes = 3 + 5 + 7 = 15
25 & 15 = 9



Input: arr[] = {3, 4, 6, 7}
Output: 10

Naive approach: A simple solution is to traverse the array and keep checking for every element if it is prime or not. If the number is prime, then add it to S1 which stores the sum of prime numbers from the array else add it to S2 which stores the sum of non-prime numbers. Finally, print S1 & S2.
Time complexity: O(N * sqrt(N))

Efficient approach: Generate all the primes up to the maximum element of the array using the Sieve of Eratosthenes and store them in a hash. Now, traverse the array and check if the number is prime or not. In the end, calculate and print the bitwise AND of the sum of prime numbers and the sum of composite numbers.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the bitwise AND of the
// sum of primes and the sum of non-primes
int calculateAND(int arr[], int n)
{
    // Find maximum value in the array
    int max_val = *max_element(arr, arr + n);
  
    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
    // THAN OR EQUAL TO max_val
    // Create a boolean array "prime[0..n]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    vector<bool> prime(max_val + 1, true);
  
    // Remaining part of SIEVE
    prime[0] = false;
    prime[1] = false;
    for (int p = 2; p * p <= max_val; p++) {
  
        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p] == true) {
  
            // Update all multiples of p
            for (int i = p * 2; i <= max_val; i += p)
                prime[i] = false;
        }
    }
  
    // Store the sum of primes in S1 and
    // the sum of non-primes in S2
    int S1 = 0, S2 = 0;
    for (int i = 0; i < n; i++) {
  
        if (prime[arr[i]]) {
  
            // The number is prime
            S1 += arr[i];
        }
        else if (arr[i] != 1) {
  
            // The number is not prime
            S2 += arr[i];
        }
    }
  
    // Return the bitwise AND of the sums
    return (S1 & S2);
}
  
// Driver code
int main()
{
    int arr[] = { 3, 4, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << calculateAND(arr, n);
  
    return 0;
}

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Java

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// Java implementation of the approach 
import java.util.Arrays; 
import java.util.Collections; 
import java.util.List;
  
class GFG
{
    static int getMax(int[] A)
    {
        int max = Integer.MIN_VALUE;
        for (int i: A) 
        {
            max = Math.max(max, i);
        }
        return max;
    }
  
    // Function to return the bitwise AND of the 
    // sum of primes and the sum of non-primes 
    static int calculateAND(int arr[], int n) 
    
        // using Collections.max() to find 
        // maximum element using only 1 line. 
        // Find maximum value in the array 
        int max_val = getMax(arr); 
  
        // USE SIEVE TO FIND ALL PRIME NUMBERS LESS 
        // THAN OR EQUAL TO max_val 
        // Create a boolean array "prime[0..n]". A 
        // value in prime[i] will finally be false 
        // if i is Not a prime, else true. 
        boolean prime[] = new boolean [max_val + 1];
        int i;
          
        for (i = 0; i < max_val + 1; i++)
            prime[i] = true;
              
        // Remaining part of SIEVE 
        prime[0] = false
        prime[1] = false
        for (int p = 2; p * p <= max_val; p++)
        
      
            // If prime[p] is not changed, 
            // then it is a prime 
            if (prime[p] == true
            
      
                // Update all multiples of p 
                for ( i = p * 2; i <= max_val; i += p) 
                    prime[i] = false
            
        
      
        // Store the sum of primes in S1 and 
        // the sum of non-primes in S2 
        int S1 = 0, S2 = 0
        for (i = 0; i < n; i++)
        
            if (prime[arr[i]]) 
            
      
                // The number is prime 
                S1 += arr[i]; 
            
            else if (arr[i] != 1)
            
      
                // The number is not prime 
                S2 += arr[i]; 
            
        
      
        // Return the bitwise AND of the sums 
        return (S1 & S2); 
    
      
    // Driver code 
    public static void main (String[] args)
    
        int arr[] = { 3, 4, 6, 7 }; 
        int n = arr.length; 
      
        System.out.println(calculateAND(arr, n)); 
    
}
  
// This code is contributed by AnkitRai01

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Python3

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# Python3 implementation of the approach
  
# Function to return the bitwise AND of the
# sum of primes and the sum of non-primes
def calculateAND(arr, n):
      
    # Find maximum value in the array
    max_val = max(arr)
   
    # USE SIEVE TO FIND ALL PRIME NUMBERS 
    # LESS THAN OR EQUAL TO max_val
    # Create a boolean array "prime[0..n]". 
    # A value in prime[i] will finally be false
    # if i is Not a prime, else true.
    prime = [True for i in range(max_val + 1)]
  
    # Remaining part of SIEVE
    prime[0] = False
    prime[1] = False
    for p in range(2, max_val + 1):
  
        if p * p >= max_val:
            break
  
        # If prime[p] is not changed, 
        # then it is a prime
        if (prime[p]):
  
            # Update all multiples of p
            for i in range(2 * p, max_val + 1, p):
                prime[i] = False
  
    # Store the sum of primes in S1 and
    # the sum of non-primes in S2
    S1 = 0
    S2 = 0
    for i in range(n):
  
        if (prime[arr[i]]):
  
            # The number is prime
            S1 += arr[i]
        elif (arr[i] != 1):
  
            # The number is not prime
            S2 += arr[i]
  
    # Return the bitwise AND of the sums
    return (S1 & S2)
  
# Driver code
arr = [3, 4, 6, 7]
n = len(arr)
  
print(calculateAND(arr, n))
  
# This code is contributed by Mohit Kumar

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C#

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// C# implementation of the approach
using System;
using System.Collections.Generic; 
      
class GFG
{
    static int getMax(int[] A)
    {
        int max = int.MinValue;
        foreach (int i in A) 
        {
            max = Math.Max(max, i);
        }
        return max;
    }
  
    // Function to return the bitwise AND of the 
    // sum of primes and the sum of non-primes 
    static int calculateAND(int []arr, int n) 
    
        // using Collections.max() to find 
        // maximum element using only 1 line. 
        // Find maximum value in the array 
        int max_val = getMax(arr); 
  
        // USE SIEVE TO FIND ALL PRIME NUMBERS LESS 
        // THAN OR EQUAL TO max_val 
        // Create a boolean array "prime[0..n]". A 
        // value in prime[i] will finally be false 
        // if i is Not a prime, else true. 
        bool []prime = new bool [max_val + 1];
        int i;
          
        for (i = 0; i < max_val + 1; i++)
            prime[i] = true;
              
        // Remaining part of SIEVE 
        prime[0] = false
        prime[1] = false
        for (int p = 2; p * p <= max_val; p++)
        
      
            // If prime[p] is not changed, 
            // then it is a prime 
            if (prime[p] == true
            
      
                // Update all multiples of p 
                for (i = p * 2; i <= max_val; i += p) 
                    prime[i] = false
            
        
      
        // Store the sum of primes in S1 and 
        // the sum of non-primes in S2 
        int S1 = 0, S2 = 0; 
        for (i = 0; i < n; i++)
        
            if (prime[arr[i]]) 
            
      
                // The number is prime 
                S1 += arr[i]; 
            
            else if (arr[i] != 1)
            
      
                // The number is not prime 
                S2 += arr[i]; 
            
        
      
        // Return the bitwise AND of the sums 
        return (S1 & S2); 
    
      
    // Driver code 
    public static void Main (String[] args)
    
        int []arr = { 3, 4, 6, 7 }; 
        int n = arr.Length; 
      
        Console.WriteLine(calculateAND(arr, n)); 
    
}
      
// This code is contributed by Rajput-Ji

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Output:

10

Time Complexity: O(N * log(log(N))
Space Complexity: O(max_val) where max_val is the maximum value of an element in the given array.



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