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

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

- a XOR x > x
- 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<bits/stdc++.h>` `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

`<?php` `// PHP program to find count of values` `// whose XOR with x is greater than x` `// and values are smaller than x` `function` `countValues(` `$x` `)` `{` ` ` `$count` `= 0;` ` ` `for` `(` `$i` `= 1; ` `$i` `< ` `$x` `; ` `$i` `++)` ` ` `if` `((` `$i` `^ ` `$x` `) > ` `$x` `)` ` ` `$count` `++;` ` ` `return` `$count` `;` `}` ` ` `// Driver code` ` ` `$x` `= 10;` ` ` `echo` `countValues(` `$x` `);` `// This code is contributed by anuj_67.` `?>` |

## Javascript

`<script>` `// Javascript program to find count of values` `// whose XOR with x is greater than x` `// and values are smaller than x` `function` `countValues(x)` `{` ` ` `let count = 0;` ` ` `for` `(let i=1; i < x; i++)` ` ` `if` `((i ^ x) > x)` ` ` `count++;` ` ` `return` `count;` `}` `// Driver code` ` ` `let x = 10;` ` ` `document.write(countValues(x));` `</script>` |

**Output :**

5

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

**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 2^{i} 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<bits/stdc++.h>` `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

`<?php` `// PHP program to find count` `// of values whose XOR with` `// x is greater than x and` `// values are smaller than x` `function` `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` `$x` `= 10;` `echo` `countValues(` `$x` `);` `// This code is contributed` `// by Smitha` `?>` |

## Javascript

`<script>` `// Javascript program to find count of` `// values whose XOR with x is greater` `// than x and values are smaller than x ` `function` `countValues(x)` `{` ` ` ` ` `// Initialize result` ` ` `var` `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 = parseInt(x / 2);` ` ` `}` ` ` `return` `count;` `}` `// Driver code` `var` `x = 10;` `document.write(countValues(x));` `// This code is contributed by Princi Singh` `</script>` |

**Output :**

5

Time complexity of this solution is O(Log x)

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