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.
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. The 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 the 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 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 k, then this is the final answer. Else no subset exists which satisfies the condition.
Steps to solve this problem:
1. declare a vector v of integers.
2. iterate through i=0 till n:
*check if arr[i] bitwise or k is equal to k than push arr[i] in v.
3. declare a variable ans=0.
4. iterate through i=0 till size of v:
*update ans to ans bitwise or v[i].
5. check if ans is not equal to k than subset doesn’t exist.
6. iterate through i=0 till size of v:
*print v[i].
C++
// CPP Program to find the maximum subset// with bitwise OR equal to k#include <bits/stdc++.h>using namespace std;// function to find the maximum subset with// bitwise OR equal to kvoid 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 Codeint 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 kimport 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<Integer> v = new ArrayList<Integer>(); 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 kdef 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 Codek = 3arr = [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 kusing 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
<?php// PHP Program to find the // maximum subset with bitwise// OR equal to k// function to find the maximum // subset with bitwise OR equal to kfunction subsetBitwiseORk($arr, $n, $k){ $v = array(); for ($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) == $k) array_push($v, $arr[$i]); } // Store the bitwise OR // of elements in v $ans = 0; for ($i = 0; $i < count($v); $i++) $ans |= $v[$i]; // If ans is not equal to // k, subset doesn't exist if ($ans != $k) { echo ("Subset does not exist\n"); return; } for ($i = 0; $i < count($v); $i++) echo ($v[$i]." ");}// Driver Code$k = 3;$arr = array(1, 4, 2);$n = count($arr);subsetBitwiseORk($arr, $n, $k);// This code is contributed by // Manish Shaw(manishshaw1)?> |
Javascript
<script>// Javascript Program to find the maximum subset// with bitwise OR equal to k// function to find the maximum subset with// bitwise OR equal to kfunction subsetBitwiseORk(arr, n, k){ var v = []; for (var 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(arr[i]); } // Store the bitwise OR of elements in v var ans = 0; for (var i = 0; i < v.length; i++) ans |= v[i]; // If ans is not equal to k, subset doesn't exist if (ans != k) { document.write( "Subset does not exist" ); return; } for (var i = 0; i < v.length; i++) document.write( v[i] + ' ');}// Driver Codevar k = 3;var arr = [1, 4, 2];var n = arr.length;subsetBitwiseORk(arr, n, k);</script> |
Output :
1 2
Time complexity: O(N), where N is the size of the array.
Auxiliary Space : O(N)
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