Find subsequences with maximum Bitwise AND and Bitwise OR
Given an array of n elements. The task is to print the maximum sum by selecting two subsequences of the array (not necessarily different) such that the sum of bitwise AND of all elements of the first subsequence and bitwise OR of all the elements of the second subsequence is maximum.
Examples:
Input: arr[] = {3, 5, 6, 1}
Output: 13
We get maximum AND value by choosing 6 only and maximum OR value by choosing all (3 | 5 | 6 | 1) = 7. So the result is 6 + 7 = 13.
Input: arr[] = {3, 3}
Output: 6
Approach: The maximum OR would be the or of all the numbers and the maximum AND would be the maximum element in the array. This is so because if (x | y) >= x, y and (x & y) <=x, y.
C++
#include<bits/stdc++.h>
using namespace std;
void maxSum( int a[], int n)
{
int maxAnd=0;
for ( int i=0;i<n;i++)
maxAnd=max(maxAnd,a[i]);
int maxOR=0;
for ( int i=0;i<n;i++)
{
maxOR=maxOR|a[i];
}
cout<<maxAnd+maxOR;
}
int main()
{
int a[]={3,5,6,1};
int n= sizeof (a)/ sizeof (a[0]);
maxSum(a,n);
}
|
Java
import java.util.Arrays;
class GFG {
static void maxSum( int [] a, int n) {
int maxAnd = Arrays.stream(a).max().getAsInt();
int maxOR = 0 ;
for ( int i = 0 ; i < n; i++) {
maxOR |= a[i];
}
System.out.println((maxAnd + maxOR));
}
public static void main(String[] args) {
int n = 4 ;
int [] a = { 3 , 5 , 6 , 1 };
maxSum(a, n);
}
}
|
Python3
def maxSum(a, n):
maxAnd = max (a)
maxOR = 0
for i in range (n):
maxOR| = a[i]
print (maxAnd + maxOR)
n = 4
a = [ 3 , 5 , 6 , 1 ]
maxSum(a, n)
|
C#
using System;
using System.Linq;
public class GFG {
static void maxSum( int []a, int n) {
int maxAnd = a.Max();
int maxOR = 0;
for ( int i = 0; i < n; i++) {
maxOR |= a[i];
}
Console.Write((maxAnd + maxOR));
}
public static void Main() {
int n = 4;
int [] a = {3, 5, 6, 1};
maxSum(a, n);
}
}
|
PHP
<?php
function maxSum( $a , $n )
{
$maxAnd = max( $a );
$maxOR = 0;
for ( $i = 0; $i < $n ; $i ++)
$maxOR |= $a [ $i ];
print ( $maxAnd + $maxOR );
}
$n = 4;
$a = array (3, 5, 6, 1);
maxSum( $a , $n );
?>
|
Javascript
<script>
function maxSum(a, n)
{
var maxAnd = Math.max(...a);
var maxOR = 0;
for ( var i = 0; i < n; i++)
{
maxOR |= a[i];
}
document.write((maxAnd + maxOR));
}
var n = 4;
var a = [ 3, 5, 6, 1];
maxSum(a, n);
</script>
|
Time Complexity: O(n)
Auxiliary Space: O(1)
Last Updated :
19 Jul, 2022
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