Given a non-negative integer N, the task is to find count of non-negative integers

less than or equal to N whose bitwise OR and sum with N are equal.

Examples:

Input : N = 3 Output : 1 0 is the only number in [0, 3] that satisfies given property. (0 + 3) = (0 | 3) Input : 10 Output : 4 (0 + 10) = (0 | 10) (Both are 10) (1 + 10) = (1 | 10) (Both are 11) (4 + 10) = (4 | 10) (Both are 14) (5 + 10) = (5 | 10) (Both are 15)

A **simple solution** is to traverse all numbers from 0 to N and do bitwise OR and SUM with N, if both are equal increment counter.

Time complexity = O(N).

An **efficient solution **is to follow following steps.

1. Find count of zero bit in N.

2. Return pow(2,count).

The idea is based on the fact that bitwise OR and sum of a number x with N are equal, if and only if

bitwise AND of x with N will is 0

Let, N=10 =1010_{2}. Bitwise AND of a number with N will be 0, if number contains zero bit with all respective set bit(s) of N and either zero bit or set bit with all respective zero bit(s) of N (because, 0&0=1&0=0).e.g.bit : 1 0 1 0 position: 4 3 2 1 Bitwise AND of any number with N will be 0, if the number has following bit pattern 1^{st}position can be either 0 or 1 (2 ways) 2_{nd}position can be 1 (1 way) 3_{rd}position can be either 0 or 1 (2 ways) 4^{th}position can be 1 (1 way) Total count = 2*1*2*1 = 2^{2}= 4.

## C++

// C++ program to count numbers whose bitwise // OR and sum with N are equal #include <bits/stdc++.h> using namespace std; // Function to find total 0 bit in a number unsigned int CountZeroBit(int n) { unsigned int count = 0; while(n) { if (!(n & 1)) count++; n >>= 1; } return count; } // Function to find Count of non-negative numbers // less than or equal to N, whose bitwise OR and // SUM with N are equal. int CountORandSumEqual(int N ) { // count number of zero bit in N int count = CountZeroBit(N); // power of 2 to count return (1 << count); } // Driver code int main() { int N = 10; cout << CountORandSumEqual(N); return 0; }

## Java

// Java program to count numbers whose bitwise // OR and sum with N are equal class GFG { // Function to find total 0 bit in a number static int CountZeroBit(int n) { int count = 0; while(n > 0) { if ((n & 1) != 0) count++; n >>= 1; } return count; } // Function to find Count of non-negative // numbers less than or equal to N, whose // bitwise OR and SUM with N are equal. static int CountORandSumEqual(int N ) { // count number of zero bit in N int count = CountZeroBit(N); // power of 2 to count return (1 << count); } //Driver code public static void main (String[] args) { int N = 10; System.out.print(CountORandSumEqual(N)); } } // This code is contributed by Anant Agarwal.

Output:

4

**Total time complexity of above solution will be O(log _{2}(N)).**

This article is contributed by **Viranjan Kumar Akela**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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