Given an integer n, find the largest possible set of non-negative integers with bitwise OR equal to n.
Examples:
Input : n = 5 Output : arr[] = [0, 1, 4, 5] The bitwise OR of 0, 1, 4 and 5 equals 5. It is not possible to obtain a set larger than this. Input : n = 8 Output : arr[] = [0, 8]
Prerequisite: Maximum subset with bitwise OR equal to k
The difference between the above-referenced article and this post is the number of elements to be checked. In the above-referenced article, we have an array of n numbers and in this post, we have the entire set of non-negative numbers.
Traversing an array was simple with the time complexity of O(N), but traversing the boundless set of non-negative numbers is not possible. So how do we limit ourselves to a smaller set of numbers?
The answer lies in the concept used. For any number, x greater than n, the bitwise OR of x and n will never be equal to n.
Hence we only need to traverse from 0 to n to obtain our answer.
The second difference is that there will always be an answer to this question. On the other hand, there was no certainty in the existence of an answer in the above-referenced article. This is because we can always include n in the resulting set.
Algorithm:
Traverse the numbers from 0 to n, checking its bitwise OR with n. If the bitwise OR equals n, then include that number in the resulting set.
1. declare a vector v of integer elements.
2. iterate through i=0 till n:
*check if i bitwise or n is equal to n than push i in v.
3. iterate through i=0 till size of v:
*print v[i].
C++
// CPP Program to find the largest set// with bitwise OR equal to n#include <bits/stdc++.h>using namespace std;// function to find the largest set with// bitwise OR equal to nvoid setBitwiseORk(int n){ vector<int> v; for (int i = 0; i <= n; i++) { // If the bitwise OR of n and i // is equal to n, then include i // in the set if ((i | n) == n) v.push_back(i); } for (int i = 0; i < v.size(); i++) cout << v[i] << ' ';}// Driver Codeint main(){ int n = 5; setBitwiseORk(n); return 0;} |
Java
// Java Program to find the largest set// with bitwise OR equal to nimport java.util.*;class GFG { // function to find the largest set with // bitwise OR equal to n static void setBitwiseORk(int n) { Vector<Integer> v = new Vector<Integer>(); for (int i = 0; i <= n; i++) { // If the bitwise OR of n and i // is equal to n, then include i // in the set if ((i | n) == n) { v.add(i); } } for (int i = 0; i < v.size(); i++) { System.out.print(v.get(i) + " "); } } // Driver Code public static void main(String[] args) { int n = 5; setBitwiseORk(n); }}/* This code contributed by PrinciRaj1992 */ |
Python3
# Python 3 Program to find the largest # set with bitwise OR equal to n# function to find the largest set # with bitwise OR equal to ndef setBitwiseORk(n): v = [] for i in range(0, n + 1, 1): # If the bitwise OR of n and i # is equal to n, then include i # in the set if ((i | n) == n): v.append(i) for i in range(0, len(v), 1): print(v[i], end = ' ')# Driver Codeif __name__ == '__main__': n = 5 setBitwiseORk(n)# This code is contributed by # Surendra_Gangwar |
C#
// C# Program to find the largest set// with bitwise OR equal to nusing System; using System.Collections.Generic; class GFG { // function to find the largest set with // bitwise OR equal to n static void setBitwiseORk(int n) { List<int> v = new List<int>(); for (int i = 0; i <= n; i++) { // If the bitwise OR of n and i // is equal to n, then include i // in the set if ((i | n) == n) { v.Add(i); } } for (int i = 0; i < v.Count; i++) { Console.Write(v[i] + " "); } } // Driver Code public static void Main(String[] args) { int n = 5; setBitwiseORk(n); }}// This code has been contributed by 29AjayKumar |
Javascript
<script>// JavaScript Program to find the largest set// with bitwise OR equal to n// function to find the largest set with// bitwise OR equal to nfunction setBitwiseORk(n){ var v = []; for (var i = 0; i <= n; i++) { // If the bitwise OR of n and i // is equal to n, then include i // in the set if ((i | n) == n) v.push(i); } for (var i = 0; i < v.length; i++) document.write( v[i] + ' ');}// Driver Codevar n = 5;setBitwiseORk(n);</script> |
Output:
0 1 4 5
Time complexity: O(N)
Auxiliary Space:O(N)
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