Given a positive integer n, count numbers x such that 0 < x <n and x^n > n where ^ is bitwise XOR operation.

Examples:

Input : n = 12 Output : 3 Numbers are 1, 2 and 3 1^12 > 12, 2^12 > 12 and 3^12 > 12 Input : n = 11 Output : 4 Numbers are 4, 5, 6 and 7

A number may x produce a greater XOR value if x has a set bit at a position where n has a 0 bit. So we traverse bits of n, and one by one consider all 0 bits. For every set bit at position k (Considering k = 0 for rightmost bit, k = 1 for second rightmost bit, ..), we add 2 2^{k} to result. For a bit at k-th position, there are 2^{k} numbers with set bit 1.

Below is the implementation of the above idea.

## C++

`// C++ program to count numbers whose XOR with n ` `// produces a value more than n. ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `int` `countNumbers(` `int` `n) ` `{ ` ` ` `/* If there is a number like m = 11110000, ` ` ` `then it is bigger then 1110xxxx. x can either ` ` ` `0 or 1. So we have pow(2, k) greater numbers ` ` ` `where k is position of rightmost 1 in m. Now ` ` ` `by using XOR bit at each position can be changed. ` ` ` `To change bit at any position, it needs to XOR ` ` ` `it with 1. */` ` ` `int` `k = 0; ` `// Position of current bit in n ` ` ` ` ` `/* Traverse bits from LSB (least significant bit) ` ` ` `to MSB */` ` ` `int` `count = 0; ` `// Initialize result ` ` ` `while` `(n > 0) ` ` ` `{ ` ` ` `// If current bit is 0, then there are ` ` ` `// 2^k numbers with current bit 1 and ` ` ` `// whose XOR with n produces greater value ` ` ` `if` `((n&1) == 0) ` ` ` `count += ` `pow` `(2, k); ` ` ` ` ` `// Increase position for next bit ` ` ` `k += 1; ` ` ` ` ` `// Reduce n to find next bit ` ` ` `n >>= 1; ` ` ` `} ` ` ` ` ` `return` `count; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `n = 11; ` ` ` `cout << countNumbers(n) << endl; ` ` ` `return` `0; ` `} ` |

## Java

`// Java program to count numbers ` `// whose XOR with n produces a ` `// value more than n. ` `import` `java.lang.*; ` `class` `GFG { ` ` ` ` ` `static` `int` `countNumbers(` `int` `n) ` ` ` `{ ` ` ` ` ` `// If there is a number like ` ` ` `// m = 11110000, then it is ` ` ` `// bigger than 1110xxxx. x ` ` ` `// can either be 0 or 1. So ` ` ` `// where k is the position of ` ` ` `// rightmost 1 in m. Now by ` ` ` `// using the XOR bit at each ` ` ` `// position can be changed. ` ` ` `// To change the bit at any ` ` ` `// position, it needs to ` ` ` `// XOR it with 1. ` ` ` `int` `k = ` `0` `; ` ` ` `// Position of current bit in n ` ` ` ` ` `// Traverse bits from LSB (least ` ` ` `// significant bit) to MSB ` ` ` ` ` `int` `count = ` `0` `; ` ` ` `// Initialize result ` ` ` `while` `(n > ` `0` `) { ` ` ` `// If the current bit is 0, then ` ` ` `// there are 2^k numbers with ` ` ` `// current bit 1 and whose XOR ` ` ` `// with n produces greater value ` ` ` `if` `((n & ` `1` `) == ` `0` `) ` ` ` `count += (` `int` `)(Math.pow(` `2` `, k)); ` ` ` ` ` `// Increase position for next bit ` ` ` `k += ` `1` `; ` ` ` ` ` `// Reduce n to find next bit ` ` ` `n >>= ` `1` `; ` ` ` `} ` ` ` ` ` `return` `count; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `int` `n = ` `11` `; ` ` ` `System.out.println(countNumbers(n)); ` ` ` `} ` `} ` ` ` `// This code is contributed by Smitha. ` |

## Python3

`# Python program to count numbers whose ` `# XOR with n produces a value more than n. ` ` ` `def` `countNumbers(n): ` ` ` ` ` `''' If there is a number like m = 11110000, ` ` ` `then it is bigger then 1110xxxx. x can either ` ` ` `0 or 1. So we have pow(2, k) greater numbers ` ` ` `where k is position of rightmost 1 in m. Now ` ` ` `by using XOR bit at each position can be changed. ` ` ` `To change bit at any position, it needs to XOR ` ` ` `it with 1. '''` ` ` ` ` `# Position of current bit in n ` ` ` `k ` `=` `0` ` ` ` ` `# Traverse bits from ` ` ` `# LSB to MSB ` ` ` `count ` `=` `0` `# Initialize result ` ` ` `while` `(n > ` `0` `): ` ` ` ` ` `# If current bit is 0, then there are ` ` ` `# 2^k numbers with current bit 1 and ` ` ` `# whose XOR with n produces greater value ` ` ` `if` `((n & ` `1` `) ` `=` `=` `0` `): ` ` ` `count ` `+` `=` `pow` `(` `2` `, k) ` ` ` ` ` `# Increase position for next bit ` ` ` `k ` `+` `=` `1` ` ` ` ` `# Reduce n to find next bit ` ` ` `n >>` `=` `1` ` ` ` ` `return` `count ` ` ` `# Driver code ` `n ` `=` `11` `print` `(countNumbers(n)) ` ` ` `# This code is contributed by Anant Agarwal. ` |

## C#

`// C# program to count numbers ` `// whose XOR with n produces a ` `// value more than n. ` `using` `System; ` ` ` `class` `GFG { ` ` ` ` ` `static` `int` `countNumbers(` `int` `n) ` ` ` `{ ` ` ` ` ` `// If there is a number like ` ` ` `// m = 11110000, then it is ` ` ` `// bigger than 1110xxxx. x ` ` ` `// can either be 0 or 1. So ` ` ` `// where k is the position of ` ` ` `// rightmost 1 in m. Now by ` ` ` `// using the XOR bit at each ` ` ` `// position can be changed. ` ` ` `// To change the bit at any ` ` ` `// position, it needs to ` ` ` `// XOR it with 1. ` ` ` `int` `k = 0; ` ` ` `// Position of current bit in n ` ` ` ` ` `// Traverse bits from LSB (least ` ` ` `// significant bit) to MSB ` ` ` ` ` `int` `count = 0; ` ` ` `// Initialize result ` ` ` `while` `(n > 0) { ` ` ` `// If the current bit is 0, then ` ` ` `// there are 2^k numbers with ` ` ` `// current bit 1 and whose XOR ` ` ` `// with n produces greater value ` ` ` `if` `((n & 1) == 0) ` ` ` `count += (` `int` `)(Math.Pow(2, k)); ` ` ` ` ` `// Increase position for next bit ` ` ` `k += 1; ` ` ` ` ` `// Reduce n to find next bit ` ` ` `n >>= 1; ` ` ` `} ` ` ` ` ` `return` `count; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `n = 11; ` ` ` `Console.WriteLine(countNumbers(n)); ` ` ` `} ` `} ` ` ` `// This code is contributed by Anant Agarwal. ` |

## PHP

`<?php ` `// PHP program to count ` `// numbers whose XOR with n ` `// produces a value more than n. ` ` ` `function` `countNumbers(` `$n` `) ` `{ ` ` ` ` ` `/* If there is a number ` ` ` `like m = 11110000, ` ` ` `then it is bigger ` ` ` `then 1110xxxx. x ` ` ` `can either 0 or 1. ` ` ` `So we have pow(2, k) ` ` ` `greater numbers where ` ` ` `k is position of ` ` ` `rightmost 1 in m. ` ` ` `Now by using XOR bit ` ` ` `at each position can ` ` ` `be changed. To change ` ` ` `bit at any position, ` ` ` `it needs to XOR it ` ` ` `with 1. */` ` ` ` ` `// Position of current bit in n ` ` ` `$k` `= 0; ` ` ` ` ` `/* Traverse bits from LSB ` ` ` `(least significant bit) ` ` ` `to MSB */` ` ` ` ` `// Initialize result ` ` ` `$count` `= 0; ` ` ` `while` `(` `$n` `> 0) ` ` ` `{ ` ` ` ` ` `// If current bit is 0, ` ` ` `// then there are 2^k ` ` ` `// numbers with current ` ` ` `// bit 1 and whose XOR ` ` ` `// with n produces greater ` ` ` `// value ` ` ` `if` `((` `$n` `& 1) == 0) ` ` ` `$count` `+= pow(2, ` `$k` `); ` ` ` ` ` `// Increase position ` ` ` `// for next bit ` ` ` `$k` `+= 1; ` ` ` ` ` `// Reduce n to ` ` ` `// find next bit ` ` ` `$n` `>>= 1; ` ` ` `} ` ` ` ` ` `return` `$count` `; ` `} ` ` ` ` ` `// Driver code ` ` ` `$n` `= 11; ` ` ` `echo` `countNumbers(` `$n` `); ` ` ` `// This code is contributed by anuj_67. ` `?> ` |

Output:

4

Time complexity : O(Log n)

This article is contributed by **Smarak Chopdar**. 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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