Number of bitonic arrays of length n and consisting of elements from 1 to n

For a given number n (n > 1), we need to find the number of ways you can make bitonic array of length n, consisting of all elements from 1 to n.
Note: [1, 2,…n] and [n, n – 1…2, 1] are not considered as bitonic array.

Examples :

Input : n = 3
Output : 2

Explanation : [1, 3, 2] & [2, 3, 1]
are only two ways of bitonic array
formation for n = 3.

Input : n = 4
Output : 6

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

For creation of a bitonic array, let’s say that we have an empty array of length n and we want to put the numbers from 1 to n in this array in bitonic form, now let’s say we want to add the number 1, we have only 2 possible ways to put the number 1, both are the end positions because if we should put 1 at any place other than end points then number on both side of 1 are greater than 1. After that we can imagine that we have an array of length n-1 and now we want to put the number 2, again for the same reasons we have two ways and so on, until we want to put the number n, we will only have 1 way instead of 2, so we have n-1 numbers that have 2 ways to put, so by multiplication rule of combinatorics the answer is 2^n-1, finally we should subtract 2 from the answer because permutations 1 2 3 4 …. n and n n-1 … 3 2 1 should not be counted.

C++

 // C++ program for finding  // total bitonic array #include using namespace std;    // Function to calculate no. of ways long long int maxWays( int n) {     // return (2^(n - 1)) -2      return (pow(2, n - 1) - 2); }    // Driver Code int main() {     int n = 6;     cout << maxWays(n);     return 0; }

Java

 // Java program for finding  // total bitonic array class GFG {            // Function to calculate no. of ways     static int maxWays( int n)     {                    // return (2 ^ (n - 1)) -2          return (int)(Math.pow(2, n - 1) - 2);     }            // Driver Code     public static void main (String[] args)     {         int n = 6;                    System.out.print(maxWays(n));     } }    // This code is contributed by Anant Agarwal.

Python3

 # python program for finding  # total bitonic array    # Function to calculate no. of ways def maxWays(n):            # return (2^(n - 1)) -2      return (pow(2, n - 1) - 2);     # Driver Code n = 6; print(maxWays(n))     # This code is contributed by Sam007

C#

 // C# program for finding  // total bitonic array using System;    class GFG {            // Function to calculate no. of ways     static int maxWays( int n)     {                    // return (2 ^ (n - 1)) -2          return (int)(Math.Pow(2, n - 1) - 2);     }            // Driver Code     public static void Main ()     {         int n = 6;                    Console.Write(maxWays(n));     } }    // This code is contributed by nitin mittal.

PHP



Output:

30

This article is contributed by Shivam Pradhan (anuj_charm). 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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

My Personal Notes arrow_drop_up

Improved By : nitin mittal, vt_m, Sam007

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.