# 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++

 `//C++ implementation of above approach` `#include`   `using` `namespace` `std;`   `//function to find the maximum sum` `void` `maxSum(``int` `a[],``int` `n)` `{` `    ``int` `maxAnd=0;` `    `  `    ``//Maximum And is maximum element` `    ``for``(``int` `i=0;i

## Java

 `import` `java.util.Arrays;`   `// Java implementation of the above approach` `// Function to find the maximum sum` `class` `GFG {`   `    ``static` `void` `maxSum(``int``[] a, ``int` `n) {`   `        ``// Maximum AND is maximum element` `        ``int` `maxAnd = Arrays.stream(a).max().getAsInt();`   `        ``// Maximum OR is bitwise OR of all.` `        ``int` `maxOR = ``0``;` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``maxOR |= a[i];` `        ``}` `        ``System.out.println((maxAnd + maxOR));`   `// Driver code` `    ``}`   `    ``public` `static` `void` `main(String[] args) {`   `        ``int` `n = ``4``;` `        ``int``[] a = {``3``, ``5``, ``6``, ``1``};` `        ``maxSum(a, n);` `    ``}` `}`   `//This code contributed by 29AjayKumar`

## Python3

 `# Python implementation of the above approach`   `# Function to find the maximum sum` `def` `maxSum(a, n):`   `    ``# Maximum AND is maximum element` `    ``maxAnd ``=` `max``(a)`   `    ``# Maximum OR is bitwise OR of all.` `    ``maxOR ``=` `0` `    ``for` `i ``in` `range``(n):` `        ``maxOR|``=` `a[i]` `        `  `    ``print``(maxAnd ``+` `maxOR)`   `# Driver code` `n ``=` `4` `a ``=` `[``3``, ``5``, ``6``, ``1``]` `maxSum(a, n)`

## C#

 `// C# implementation of the above approach`   `using` `System;` `using` `System.Linq;` `public` `class` `GFG {` `     `  `    ``// Function to find the maximum sum` `    ``static` `void` `maxSum(``int` `[]a, ``int` `n) {` ` `  `        ``// Maximum AND is maximum element` `        ``int` `maxAnd = a.Max();` `        ``// Maximum OR is bitwise OR of all.` `        ``int` `maxOR = 0;` `        ``for` `(``int` `i = 0; i < n; i++) {` `            ``maxOR |= a[i];` `        ``}` `        ``Console.Write((maxAnd + maxOR));` ` `  `// Driver code` `    ``}` ` `  `    ``public` `static` `void` `Main() {` ` `  `        ``int` `n = 4;` `        ``int``[] a = {3, 5, 6, 1};` `        ``maxSum(a, n);` `    ``}` `}` ` `  `//This code contributed by 29AjayKumar`

## PHP

 ``

## Javascript

 ``

Output:

`13`

Time Complexity: O(n)

Auxiliary Space: O(1)

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