Skip to content
Related Articles

Related Articles

Improve Article

Numbers whose bitwise OR and sum with N are equal

  • Difficulty Level : Medium
  • Last Updated : 29 Apr, 2021

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++




// 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.

Python3




# 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#




// 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
// 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
?>

Javascript




<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 of above solution will be O(log2(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 write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :