# Maximum subset with bitwise OR equal to k

Given an array of non negative integers and an integer k, find the subset of maximum length with bitwise OR equal to k.

Examples :

```Input : arr[] = [1, 4, 2]
k = 3
Output : [1, 2]
Explanation: The bitwise OR of
1 and 2 equals 3. It is not possible to obtain
a subset of length greater than 2.

Input : arr[] = [1, 2, 5]
k = 4
Output : []
No subset's bitwise OR equals 4.
```

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

Method 1(Simple):
The naive method would be to consider all the subsets. While considering a subset, compute its bitwise OR. If it equals k, compare the subset’s length with the maximum length so far and update the maximum length if required.

Method 2(Efficient):
0 OR 0 = 0
1 OR 0 = 1
1 OR 1 = 1
Hence, for all the positions in the binary representation of k with the bit equal to 0, the corresponding position in the binary representations of all the elements in the resulting subset should necessarily be 0.
On the other hand, for positions in k with the bit equal to 1, there has to be at least one element with a 1 in the corresponding position. Rest of the elements can have either 0 or 1 in that position, it does not matter.

Therefore, to obtain the resulting subset, traverse the initial array. While deciding if the element should be in the resulting subset or not, check whether there is any position in the binary representation of k which is 0 and the corresponding position in that element is 1. If there exists such a position, then ignore that element, else include it in the resulting subset.

How to determine if there exists a position in the binary representation of k which is 0 and the corresponding position in an element is 1?
Simply take bitwise OR of k and that element. If it does not equal to k, then there exists such a position and the element has to be ignored. If their bitwise OR equals to k, then include the current element in the resulting subset.

The final step is to determine if there is at least one element with a 1 in a position with 1 in the corresponding position in k.
Simply compute the bitwise OR of the resulting subset. If it equals to k, then this is the final answer. Else no subset exists which satisfies the condition.

## C++

 `// CPP Program to find the maximum subset ` `// with bitwise OR equal to k ` `#include ` `using` `namespace` `std; ` ` `  `// function to find the maximum subset with ` `// bitwise OR equal to k ` `void` `subsetBitwiseORk(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``vector<``int``> v; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// If the bitwise OR of k and element ` `        ``// is equal to k, then include that element ` `        ``// in the subset ` `        ``if` `((arr[i] | k) == k) ` `            ``v.push_back(arr[i]); ` `    ``} ` ` `  `    ``// Store the bitwise OR of elements in v ` `    ``int` `ans = 0; ` ` `  `    ``for` `(``int` `i = 0; i < v.size(); i++) ` `        ``ans |= v[i]; ` ` `  `    ``// If ans is not equal to k, subset doesn't exist ` `    ``if` `(ans != k) { ` `        ``cout << ``"Subset does not exist"` `<< endl; ` `        ``return``; ` `    ``} ` ` `  `    ``for` `(``int` `i = 0; i < v.size(); i++) ` `        ``cout << v[i] << ``' '``; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `k = 3; ` `    ``int` `arr[] = { 1, 4, 2 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` `  `    ``subsetBitwiseORk(arr, n, k); ` `    ``return` `0; ` `} `

## Java

 `// Java Program to find the maximum subset ` `// with bitwise OR equal to k ` `import` `java.util.*; ` ` `  `class` `GFG { ` ` `  `    ``// function to find the maximum subset  ` `    ``// with bitwise OR equal to k ` `    ``static` `void` `subsetBitwiseORk(``int` `arr[], ` `                              ``int` `n, ``int` `k) ` `    ``{ ` `        ``ArrayList v =  ` `                  ``new` `ArrayList(); ` `     `  `        ``for` `(``int` `i = ``0``; i < n; i++) { ` `     `  `            ``// If the bitwise OR of k and  ` `            ``// element is equal to k, then ` `            ``// include that element in the  ` `            ``// subset ` `            ``if` `((arr[i] | k) == k){ ` `                ``v.add(arr[i]); ` `            ``} ` `        ``} ` `     `  `        ``// Store the bitwise OR of elements ` `        ``// in v ` `        ``int` `ans = ``0``; ` `     `  `        ``for` `(``int` `i = ``0``; i < v.size(); i++) ` `            ``ans = ans|v.get(i); ` `     `  `        ``// If ans is not equal to k, subset ` `        ``// doesn't exist ` `        ``if` `(ans != k) { ` `            ``System.out.println(``"Subset does"` `                           ``+ ``" not exist"` `); ` `            ``return``; ` `        ``} ` `     `  `        ``for` `(``int` `i = ``0``; i < v.size(); i++) ` `            ``System.out.print(v.get(i) + ``" "` `); ` `    ``} ` `     `  `    ``// main function ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `k = ``3``; ` `        ``int` `arr[] = { ``1``, ``4``, ``2` `}; ` `        ``int` `n = arr.length; ` `     `  `        ``subsetBitwiseORk(arr, n, k); ` `         `  `    ``} ` `} ` ` `  `// This code is contributed by Arnab Kundu. `

## Python3

 `# Python3 Program to find the  ` `# maximum subset with bitwise ` `# OR equal to k ` `   `  `# function to find the maximum  ` `# subset with bitwise OR equal to k ` `def` `subsetBitwiseORk(arr, n, k) : ` `    ``v ``=` `[] ` `   `  `    ``for` `i ``in` `range``(``0``, n) :  ` `        ``# If the bitwise OR of k  ` `        ``# and element is equal to k, ` `        ``# then include that element ` `        ``# in the subset ` `        ``if` `((arr[i] | k) ``=``=` `k) : ` `            ``v.append(arr[i]) ` `   `  `    ``# Store the bitwise OR ` `    ``# of elements in v ` `    ``ans ``=` `0` `   `  `    ``for` `i ``in` `range``(``0``, ``len``(v)) : ` `        ``ans |``=` `v[i] ` `   `  `    ``# If ans is not equal to ` `    ``# k, subset doesn't exist ` `    ``if` `(ans !``=` `k) : ` `        ``print` `(``"Subset does not exist\n"``) ` `        ``return` `   `  `    ``for` `i ``in` `range``(``0``, ``len``(v)) : ` `        ``print` `(``"{} "``.``format``(v[i]), end``=``"") ` `   `  `# Driver Code ` `k ``=` `3` `arr ``=` `[``1``, ``4``, ``2``] ` `n ``=` `len``(arr) ` `   `  `subsetBitwiseORk(arr, n, k) ` `   `  `# This code is contributed by  ` `# Manish Shaw(manishshaw1) `

## C#

 `// C# Program to find the maximum subset ` `// with bitwise OR equal to k ` `using` `System; ` `using` `System.Collections; ` ` `  `class` `GFG { ` ` `  `    ``// function to find the maximum subset ` `    ``// with bitwise OR equal to k ` `    ``static` `void` `subsetBitwiseORk(``int` `[]arr,  ` `                              ``int` `n, ``int` `k) ` `    ``{ ` `        ``ArrayList v = ``new` `ArrayList(); ` `     `  `        ``for` `(``int` `i = 0; i < n; i++) { ` `     `  `            ``// If the bitwise OR of k and  ` `            ``// element is equal to k, then ` `            ``// include that element in the ` `            ``// subset ` `            ``if` `((arr[i] | k) == k){ ` `                ``v.Add(arr[i]); ` `            ``} ` `        ``} ` `     `  `        ``// Store the bitwise OR of  ` `        ``// elements in v ` `        ``int` `ans = 0; ` `     `  `        ``for` `(``int` `i = 0; i < v.Count; i++) ` `            ``ans = ans|(``int``)v[i]; ` `     `  `        ``// If ans is not equal to k, subset ` `        ``// doesn't exist ` `        ``if` `(ans != k) { ` `            ``Console.WriteLine(``"Subset does"` `                          ``+ ``" not exist"` `); ` `            ``return``; ` `        ``} ` `     `  `        ``for` `(``int` `i = 0; i < v.Count; i++) ` `            ``Console.Write((``int``)v[i] + ``" "` `); ` `    ``} ` `     `  `    ``// main function ` `    ``static` `public` `void` `Main(String []args) ` `    ``{ ` `        ``int` `k = 3; ` `        ``int` `[]arr = { 1, 4, 2 }; ` `        ``int` `n = arr.Length; ` `     `  `        ``subsetBitwiseORk(arr, n, k); ` `         `  `    ``} ` `} ` ` `  `// This code is contributed by Arnab Kundu `

## PHP

 ` `

Output :

```1 2
```

Time complexity : O(N), where N is the size of array.

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