# Count smaller values whose XOR with x is greater than x

• Difficulty Level : Medium
• Last Updated : 12 Jun, 2022

Given a integer ‘x’, find the number of values of ‘a’ satisfying the following conditions:

1. a XOR x > x
2. 0 < a < x

Examples :

```Input : x = 10
Output : 5
Explanation: For x = 10, following 5 values
of 'a' satisfy the conditions:
1 XOR 10 = 11
4 XOR 10 = 14
5 XOR 10 = 15
6 XOR 10 = 12
7 XOR 10 = 13

Input : x = 2
Output : 1
Explanation: For x=2, we have just one value
1 XOR 2 = 3.```

Naive Approach
A Simple approach is to check for all values of ‘a’ between 0 and ‘x’ and calculate its XOR with x and check if the condition 1 satisfies.

## C++

 `// C++ program to find count of values``// whose XOR with x is greater than x``// and values are smaller than x``#include``using` `namespace` `std;` `int` `countValues(``int` `x)``{``    ``int` `count = 0;``    ``for` `(``int` `i=1; i < x; i++)``        ``if` `((i ^ x) > x)``            ``count++;``    ``return` `count;``}` `// Driver code``int` `main()``{``    ``int` `x = 10;``    ``cout << countValues(x);``    ``return` `0;``}`

## Java

 `// Java program to find count of values``// whose XOR with x is greater than x``// and values are smaller than x` `public` `class` `XOR``{``    ``static` `int` `countValues(``int` `x)``    ``{``        ``int` `count = ``0``;``        ``for` `(``int` `i=``1``; i < x; i++)``            ``if` `((i ^ x) > x)``                ``count++;``        ``return` `count;``    ``}``    ` `    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `x = ``10``;``        ``System.out.println(countValues(x));``    ``}``}` `// This code is contributed by Saket Kumar`

## Python3

 `# Python3 program to find``# count of values whose``# XOR with x is greater``# than x and values are``# smaller than x` `def` `countValues(x):` `    ``count ``=` `0``    ``for` `i ``in` `range``(``1` `,x):``        ``if` `((i ^ x) > x):``            ``count ``+``=` `1``    ``return` `count` `# Driver code``x ``=` `10``print``(countValues(x))` `# This code is contributed``# by Smitha`

## C#

 `// C# program to find count of values``// whose XOR with x is greater than x``// and values are smaller than x``using` `System;` `class` `GFG``{``    ``static` `int` `countValues(``int` `x)``    ``{``        ``int` `count = 0;``        ``for` `(``int` `i = 1; i < x; i++)``            ``if` `((i ^ x) > x)``                ``count++;``        ``return` `count;``    ``}``    ` `    ``public` `static` `void` `Main ()``    ``{``        ``int` `x = 10;``        ``Console.Write(countValues(x));``    ``}``}` `// This code is contributed by nitin mittal.`

## PHP

 ` ``\$x``)``            ``\$count``++;``    ``return` `\$count``;``}` `    ``// Driver code``    ``\$x` `= 10;``    ``echo` `countValues(``\$x``);` `// This code is contributed by anuj_67.``?>`

## Javascript

 ``

Output :

`5`

The time complexity of the above approach is O(x).

Auxiliary Space: O(1)

Efficient Approach
The efficient solution lies in the binary representation of the number. We consider all 0’s in binary representation. For every 0 at the i-th position, we can have 2i numbers smaller than or equal to x with greater XOR.

## C++

 `// C++ program to find count of values``// whose XOR with x is greater than x``// and values are smaller than x``#include``using` `namespace` `std;` `int` `countValues(``int` `x)``{``    ``// Initialize result``    ``int` `count = 0, n = 1;` `    ``// Traversing through all bits of x``    ``while` `(x != 0)``    ``{``        ``// If current last bit of x is set``        ``// then increment count by n. Here``        ``// n is a power of 2 corresponding``        ``// to position of bit``        ``if` `(x%2 == 0)``            ``count += n;` `        ``// Simultaneously calculate the 2^n``        ``n *= 2;` `        ``// Replace x with x/2;``        ``x /= 2;``    ``}` `    ``return` `count;``}` `// Driver code``int` `main()``{``    ``int` `x = 10;``    ``cout << countValues(x);``    ``return` `0;``}`

## Java

 `// Java program to find count of values``// whose XOR with x is greater than x``// and values are smaller than x` `class` `GFG``{``    ``static` `int` `countValues(``int` `x)``    ``{``        ``// Initialize result``        ``int` `count = ``0``, n = ``1``;``        ` `        ``// Traversing through all bits of x``        ``while` `(x != ``0``)``        ``{``            ``// If current last bit of x is set``            ``// then increment count by n. Here``            ``// n is a power of 2 corresponding``            ``// to position of bit``            ``if` `(x % ``2` `== ``0``)``                ``count += n;``                ` `            ``// Simultaneously calculate the 2^n``            ``n *= ``2``;``            ` `            ``// Replace x with x/2;``            ``x /= ``2``;``        ``}``        ``return` `count;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``int` `x = ``10``;``        ``System.out.println(countValues(x));``    ``}``    ` `}` `// This code is contributed by Saket Kumar`

## Python3

 `# Python3 program to find count``# of values whose XOR with``# x is greater than x and``# values are smaller than x` `def` `countValues(x):``    ` `    ``# Initialize result``    ``count ``=` `0``;``    ``n ``=` `1``;` `    ``# Traversing through``    ``# all bits of x``    ``while` `(x > ``0``):``        ` `        ``# If current last bit``        ``# of x is set then``        ``# increment count by``        ``# n. Here n is a power``        ``# of 2 corresponding``        ``# to position of bit``        ``if` `(x ``%` `2` `=``=` `0``):``            ``count ``+``=` `n;` `        ``# Simultaneously``        ``# calculate the 2^n``        ``n ``*``=` `2``;` `        ``# Replace x with x/2;``        ``x ``/``=` `2``;``        ``x ``=` `int``(x);` `    ``return` `count;` `# Driver code``x ``=` `10``;``print``(countValues(x));` `# This code is contributed``# by mits`

## C#

 `// C# program to find count of values``// whose XOR with x is greater than x``// and values are smaller than x``using` `System;` `class` `GFG``{``    ``static` `int` `countValues(``int` `x)``    ``{``        ``// Initialize result``        ``int` `count = 0, n = 1;``        ` `        ``// Traversing through all bits of x``        ``while` `(x != 0)``        ``{``            ``// If current last bit of x is set``            ``// then increment count by n. Here``            ``// n is a power of 2 corresponding``            ``// to position of bit``            ``if` `(x % 2 == 0)``                ``count += n;``                ` `            ``// Simultaneously calculate the 2^n``            ``n *= 2;``            ` `            ``// Replace x with x/2;``            ``x /= 2;``        ``}``        ``return` `count;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main ()``    ``{``        ``int` `x = 10;``        ``Console.Write(countValues(x));``    ``}``    ` `}` `// This code is contributed by nitin mittal`

## PHP

 ``

## Javascript

 ``

Output :

`5`

Time complexity: O(Log x).

Auxiliary Space: O(1)

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