# Find an N-length permutation that contains subarrays with sum less than Bitwise XOR

Given a positive integer **N**, the task is to find a permutation of length** N **having Bitwise OR of any of its subarray greater than or equal to the length of the subarray.

**Examples:**

Input:N = 5Output:1 3 5 2 4Explanation:

Consider the subarray {1, 3, 5} from the permutation {1, 3, 5, 2, $}.

Length of the subarray = 3

Bitwise OR of the subarray = 1 | 3 | 5 = 7 ( which is greater than 3)

Similarly, every subarray of any length smaller than equal to N will satisfy the condition.

Input:4Output:4 3 1 2

**Approach: **Actually any permutation of length** N** satisfies the required conditions based on the following facts:

- Bitwise OR of any set of numbers is greater than or equal to the maximum number present in that set.
- Since any subarray will contain at least one element greater than or equal to its length, any permutation satisfies the given condition.

Below is the implementation of the above approach:

## C++

`// C++ implementation of` `// the above approach` `#include <iostream>` `using` `namespace` `std;` `// Function to print the` `// required permutation` `void` `findPermutation(` `int` `N)` `{` ` ` `for` `(` `int` `i = 1; i <= N; i++)` ` ` `cout << i << ` `" "` `;` ` ` ` ` `cout << endl;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `N = 5;` ` ` ` ` `findPermutation(N);` ` ` ` ` `return` `0;` `}` |

## Java

`// Java implementation of` `// the above approach` `import` `java.util.*;` `class` `GFG{` ` ` `// Function to print the` `// required permutation` `static` `void` `findPermutation(` `int` `N)` `{` ` ` `for` `(` `int` `i = ` `1` `; i <= N; i++)` ` ` `System.out.print(i + ` `" "` `);` `}` ` ` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `N = ` `5` `;` ` ` ` ` `findPermutation(N);` `}` `}` `// This code is contributed by susmitakundugoaldanga` |

## Python3

`# Python3 implementation of` `# the above approach` `# Function to print the` `# required permutation` `def` `findPermutation(N):` ` ` `for` `i ` `in` `range` `(` `1` `, N ` `+` `1` `, ` `1` `):` ` ` `print` `(i, end` `=` `" "` `) ` ` ` `print` `(` `"\n"` `, end ` `=` `"")` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `N ` `=` `5` ` ` `findPermutation(N)` ` ` ` ` `# This code is contributed by ipg2016107.` |

## C#

`// C# implementation of` `// the above approach` `using` `System;` `class` `GFG{` ` ` `// Function to print the` `// required permutation` `static` `void` `findPermutation(` `int` `N)` `{` ` ` `for` `(` `int` `i = 1; i <= N; i++)` ` ` `Console.Write(i + ` `" "` `);` `}` `// Driver code` `public` `static` `void` `Main()` `{` ` ` `int` `N = 5;` ` ` ` ` `findPermutation(N);` `}` `}` `// This code is contributed by SURENDRA_GANGWAR` |

## Javascript

`<script>` `// javascript implementation of` `// the above approach` `// Function to print the` `// required permutation` `function` `findPermutation(N)` `{` ` ` `for` `(` `var` `i = 1; i <= N; i++)` ` ` `document.write( i + ` `" "` `);` ` ` ` ` `document.write(` `"<br>"` `);` `}` `var` `N = 5;` `findPermutation(N);` `//This code in contributed by SoumikMondal` `</script>` |

**Output:**

1 2 3 4 5

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