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; ` `} ` |

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## 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. ` |

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## 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. ` |

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## 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. ` |

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## 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 ` `?> ` |

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**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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