Given an integer n, find the largest possible set of non negative integers with bitwise OR equal to n.
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 in 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 for 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.
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.
0 1 4 5
Time complexity: O(N)
- Numbers whose bitwise OR and sum with N are equal
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- Bitwise Operators in C/C++
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