# 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

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

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++ implementation of the approach ` `#include ` `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; ` `} `

 `// 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 `

 `# 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 `

 `// 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 `

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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