Numbers whose bitwise OR and sum with N are equal

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 =10102. 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 1st position can be either 0 or 1 (2 ways) 2nd position can be 1 (1 way) 3rd position can be either 0 or 1 (2 ways) 4th position can be 1 (1 way) Total count = 2*1*2*1 = 22 = 4.

// 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 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. 

# Python3 program to count numbers whose  # bitwise OR and sum with N are equal # Function to find total 0 bit in a number def CountZeroBit(n): count = 0 while(n): if (not(n & 1)): count += 1 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. def CountORandSumEqual(N): # count number of zero bit in N count = CountZeroBit(N) # power of 2 to count return (1 << count) # Driver code N = 10 print(CountORandSumEqual(N)) # This code is contributed by Anant Agarwal. 

// C# program to count numbers whose // bitwise OR and sum with N are equal using System; class GFG {    // Function to find total   // 0 bit in a number  static int CountZeroBit(int n)  {  int count = 0;  while(n>0)  {   if (n%2!=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()  {  int N = 10;  Console.Write(CountORandSumEqual(N));  } } // This code is contributed by Anant Agarwal. 

<?php // PHP program to count  // numbers whose bitwise // OR and sum with N are equal // Function to find total // 0 bit in a number function CountZeroBit($n) { $count = 0; while($n) { if (!($n & 1)) $count++; $n >>= 1; } return $count; } // Function to find Count of  // non-negative numbers less  // than or equal to N, whose  // bitwise OR and SUM with N  // are equal. function CountORandSumEqual($N ) { // count number of  // zero bit in N $count = CountZeroBit($N); // power of 2 to count return (1 << $count); } // Driver code $N = 10; echo CountORandSumEqual($N); // This code is contributed by Ajit ?> 

<script>  // Javascript program to count numbers whose  // bitwise OR and sum with N are equal    // Function to find total   // 0 bit in a number  function CountZeroBit(n)  {  let count = 0;  while(n>0)  {   if (n%2!=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.  function CountORandSumEqual(N)  {    // count number of zero bit in N  let count = CountZeroBit(N);  // power of 2 to count  return (1 << count);  }    let N = 10;  document.write(CountORandSumEqual(N)); </script> 

Output : 
 

 4

Total time complexity: O(log₂(N)) 

Auxiliary Space: O(1)