# 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 =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.` |

## 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(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 write.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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