Given a positive integer **N**, the task is to print all the distinct even and odd values of prefix Bitwise XORs of first **N** natural numbers.

**Examples:**

Input:N = 6Output:

Even: 0 4

Odd: 1 3 7Explanation:

The prefix Bitwise XOR of the first 6 natural number si {1, 3, 0, 4, 1, 7}.

Even prefix Bitwise XORs are 0, 4.

Odd prefix Bitwise XORs are 1, 3, 7.

Input:N = 9Output:

Even: 0 4 8

Odd: 1 3 7

**Approach:** The given problem can be solved based on the below observations:

- If the value of
**N modulo 4**is**0**, then the value of Bitwise XOR of the first**N**natural numbers is**N**. - If the value of
**N modulo 4**is**1**, then the value of Bitwise XOR of the first**N**natural numbers is**1**. - If the value of
**N modulo 4**is**2**, then the value of Bitwise XOR of the first**N**natural numbers is**(N + 1)**. - If the value of
**N modulo 4**is**3**, then the value of Bitwise XOR of the first**N**natural numbers is**0**.

Therefore, from the above principle, it can be said that Bitwise XOR as even numbers will always come as **0** or multiples of **4**, and Bitwise XOR as odd numbers will always come as **1** or **1** less than multiples of **4**.

Below is the implementation of the above approach.

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Print all distinct even & odd` `// prefix Bitwise XORs from 1 to N` `void` `evenOddBitwiseXOR(` `int` `N)` `{` ` ` `cout << ` `"Even: "` `<< 0 << ` `" "` `;` ` ` `// Print the even number` ` ` `for` `(` `int` `i = 4; i <= N; i = i + 4) {` ` ` `cout << i << ` `" "` `;` ` ` `}` ` ` `cout << ` `"\n"` `;` ` ` `cout << ` `"Odd: "` `<< 1 << ` `" "` `;` ` ` `// Print the odd number` ` ` `for` `(` `int` `i = 4; i <= N; i = i + 4) {` ` ` `cout << i - 1 << ` `" "` `;` ` ` `}` ` ` `if` `(N % 4 == 2)` ` ` `cout << N + 1;` ` ` `else` `if` `(N % 4 == 3)` ` ` `cout << N;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 6;` ` ` `evenOddBitwiseXOR(N);` ` ` `return` `0;` `}` |

## Java

`// Java approach for the above approach` `class` `GFG{` `// Print all distinct even & odd` `// prefix Bitwise XORs from 1 to N` `static` `void` `evenOddBitwiseXOR(` `int` `N)` `{` ` ` `System.out.print(` `"Even: "` `+ ` `0` `+ ` `" "` `);` ` ` `// Print the even number` ` ` `for` `(` `int` `i = ` `4` `; i <= N; i = i + ` `4` `) ` ` ` `{` ` ` `System.out.print(i + ` `" "` `);` ` ` `}` ` ` `System.out.print(` `"\n"` `);` ` ` `System.out.print(` `"Odd: "` `+ ` `1` `+ ` `" "` `);` ` ` `// Print the odd number` ` ` `for` `(` `int` `i = ` `4` `; i <= N; i = i + ` `4` `)` ` ` `{` ` ` `System.out.print(i - ` `1` `+ ` `" "` `);` ` ` `}` ` ` `if` `(N % ` `4` `== ` `2` `)` ` ` `System.out.print(N + ` `1` `);` ` ` `else` `if` `(N % ` `4` `== ` `3` `)` ` ` `System.out.print(N);` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `N = ` `6` `;` ` ` `evenOddBitwiseXOR(N);` `}` `}` `// This code is contributed by abhinavjain194` |

## Python3

`# Python3 program for the above approach` `# Print all distinct even & odd` `# prefix Bitwise XORs from 1 to N` `def` `evenOddBitwiseXOR(N):` ` ` ` ` `print` `(` `"Even: "` `, ` `0` `, end ` `=` `" "` `)` ` ` `# Print the even number` ` ` `for` `i ` `in` `range` `(` `4` `, N ` `+` `1` `, ` `4` `):` ` ` `print` `(i, end ` `=` `" "` `)` ` ` ` ` `print` `()` ` ` `print` `(` `"Odd: "` `, ` `1` `, end ` `=` `" "` `)` ` ` `# Print the odd number` ` ` `for` `i ` `in` `range` `(` `4` `, N ` `+` `1` `, ` `4` `):` ` ` `print` `(i ` `-` `1` `, end ` `=` `" "` `)` ` ` ` ` `if` `(N ` `%` `4` `=` `=` `2` `):` ` ` `print` `(N ` `+` `1` `)` ` ` `elif` `(N ` `%` `4` `=` `=` `3` `):` ` ` `print` `(N)` `# Driver Code` `N ` `=` `6` `evenOddBitwiseXOR(N)` `# This code is contributed by sanjoy_62` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG{` ` ` `// Print all distinct even & odd` `// prefix Bitwise XORs from 1 to N` `static` `void` `evenOddBitwiseXOR(` `int` `N)` `{` ` ` `Console.Write(` `"Even: "` `+ 0 + ` `" "` `);` ` ` ` ` `// Print the even number` ` ` `for` `(` `int` `i = 4; i <= N; i = i + 4)` ` ` `{` ` ` `Console.Write(i + ` `" "` `);` ` ` `}` ` ` ` ` `Console.Write(` `"\n"` `);` ` ` ` ` `Console.Write(` `"Odd: "` `+ 1 + ` `" "` `);` ` ` ` ` `// Print the odd number` ` ` `for` `(` `int` `i = 4; i <= N; i = i + 4)` ` ` `{` ` ` `Console.Write(i - 1 + ` `" "` `);` ` ` `}` ` ` ` ` `if` `(N % 4 == 2)` ` ` `Console.Write(N + 1);` ` ` `else` `if` `(N % 4 == 3)` ` ` `Console.Write(N);` `}` `// Driver code` `public` `static` `void` `Main()` `{` ` ` `int` `N = 6;` ` ` `evenOddBitwiseXOR(N);` `}` `}` `// This code is contributed by splevel62` |

## Javascript

`<script>` `// Javascript implementation of the above approach` `// Print all distinct even & odd` `// prefix Bitwise XORs from 1 to N` `function` `evenOddBitwiseXOR(N)` `{` ` ` `document.write(` `"Even: "` `+ 0 + ` `" "` `);` ` ` ` ` `// Print the even number` ` ` `for` `(let i = 4; i <= N; i = i + 4)` ` ` `{` ` ` `document.write(i + ` `" "` `);` ` ` `}` ` ` ` ` `document.write(` `"<br/>"` `);` ` ` ` ` `document.write(` `"Odd: "` `+ 1 + ` `" "` `);` ` ` ` ` `// Print the odd number` ` ` `for` `(let i = 4; i <= N; i = i + 4)` ` ` `{` ` ` `document.write(i - 1 + ` `" "` `);` ` ` `}` ` ` ` ` `if` `(N % 4 == 2)` ` ` `document.write(N + 1);` ` ` `else` `if` `(N % 4 == 3)` ` ` `document.write(N);` `}` ` ` ` ` `// Driver Code` ` ` ` ` `let N = 6;` ` ` `evenOddBitwiseXOR(N);` ` ` `</script>` |

**Output:**

Even: 0 4 Odd: 1 3 7

**Time Complexity:** O(N)**Auxiliary Space:** O(1)

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