Máximo OR suma de subarreglos de dos arreglos diferentes

Dadas dos arrays de enteros positivos. Seleccione dos sub-arrays de igual tamaño de cada array y calcule la suma OR máxima posible de las dos sub-arrays. 

Nota: Sea f(x, l, r) la suma OR de todos los elementos en el rango [l, r] en el arreglo x. 

Ejemplos: 

Input : A[] = {1, 2, 4, 3, 2}
        B[] = {2, 3, 3, 12, 1}
Output : 22
Explanation: Here, one way to get maximum
sum is to select sub-array [l = 2, r = 4]
f(A, 2, 4) = 2|4|3 = 7
f(B, 2, 4) = 3|3|12 = 15
So, f(A, 2, 4) + f(B, 2, 4) = 7 + 15 = 22.
This sum can be achieved in many other ways.

Input : A[] = {1, 2, 2}
        B[] = {2, 1, 3}
Output : 6

Observe el funcionamiento del operador OR bit a bit. Si tomamos dos enteros X e Y, entonces (X|Y >= X). Se puede probar tomando algunos ejemplos. Vamos a derivar una fórmula usando la ecuación anterior. 

y también 
de las dos ecuaciones anteriores, 
por lo tanto, obtenemos la suma máxima cuando tomamos el OR de toda la array -> 

A continuación se muestra la implementación del enfoque anterior: 

// CPP program to find maximum OR sum
#include <bits/stdc++.h>
using namespace std;
 
// function to find maximum OR sum
void MaximumSum(int a[], int b[], int n)
{
    int sum1 = 0, sum2 = 0;
     
    // OR sum of all the elements
    // in both arrays
    for (int i = 0; i < n; i++) {
        sum1 |= a[i];
        sum2 |= b[i];
    }
    cout << sum1 + sum2 << endl;
}
 
// Driver Code
int main()
{
    int A[] = { 1, 2, 4, 3, 2 };
    int B[] = { 2, 3, 3, 12, 1 };
    int n = sizeof(A) / sizeof(A[0]);
    MaximumSum(A, B, n);
    return 0;
}

// Java program to find maximum OR sum
 
class GFG {
     
// function to find maximum OR sum
static void MaximumSum(int a[], int b[], int n)
{
    int sum1 = 0, sum2 = 0;
 
    // OR sum of all the elements
    // in both arrays
    for (int i = 0; i < n; i++) {
    sum1 |= a[i];
    sum2 |= b[i];
    }
    System.out.println(sum1 + sum2);
}
 
// Driver code
public static void main(String arg[])
{
    int A[] = {1, 2, 4, 3, 2};
    int B[] = {2, 3, 3, 12, 1};
    int n = A.length;
    MaximumSum(A, B, n);
}
}
 
// This code is contributed by Anant Agarwal.

# Python 3 program to
# find maximum OR sum
 
# function to find
# maximum OR sum
def MaximumSum(a, b, n):
 
    sum1 = 0
    sum2 = 0
     
    # OR sum of all the
    # elements in both arrays
    for i in range(0, n):
        sum1 |= a[i]
        sum2 |= b[i]
     
    print(sum1 + sum2)
 
# Driver Code
A = [ 1, 2, 4, 3, 2 ]
B = [ 2, 3, 3, 12, 1 ]
n = len(A)
 
MaximumSum(A, B, n)
 
# This code is contributed by Smitha Dinesh Semwal

// C# program to find maximum OR sum
using System;
 
class GFG {
     
    // function to find maximum OR sum
    static void MaximumSum(int []a, int []b, int n)
    {
        int sum1 = 0, sum2 = 0;
     
        // OR sum of all the elements
        // in both arrays
        for (int i = 0; i < n; i++)
        {
            sum1 |= a[i];
            sum2 |= b[i];
        }
        Console.WriteLine(sum1 + sum2);
    }
     
    // Driver code
    public static void Main()
    {
        int []A = {1, 2, 4, 3, 2};
        int []B = {2, 3, 3, 12, 1};
        int n = A.Length;
        MaximumSum(A, B, n);
    }
}
 
// This code is contributed by Vt_m.

<?php
// PHP program to find maximum OR sum
 
// function to find maximum OR sum
function MaximumSum($a, $b, $n)
{
    $sum1 = 0;
    $sum2 = 0;
     
    // OR sum of all the elements
    // in both arrays
    for ($i = 0; $i < $n; $i++)
    {
        $sum1 |= $a[$i];
        $sum2 |= $b[$i];
    }
    echo ($sum1 + $sum2)."\n";
}
 
// Driver Code
$A = array(1, 2, 4, 3, 2 );
$B = array(2, 3, 3, 12, 1 );
$n = sizeof($A) / sizeof($A[0]);
MaximumSum($A, $B, $n);
 
// This code is contributed by mits
 
?>

<script>
 
// JavaScript program to find maximum OR sum
 
// function to find maximum OR sum
function MaximumSum(a, b, n)
{
    let sum1 = 0, sum2 = 0;
   
    // OR sum of all the elements
    // in both arrays
    for (let i = 0; i < n; i++) {
    sum1 |= a[i];
    sum2 |= b[i];
    }
    document.write(sum1 + sum2);
}
 
// Driver code
         
    let A = [1, 2, 4, 3, 2];
    let B = [2, 3, 3, 12, 1];
    let n = A.length;
    MaximumSum(A, B, n);
                   
</script>

22

Complejidad temporal: O(n)
Espacio auxiliar: O(1)