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

0 1 3 minutes read

For a given number n (n > 1), we need to find the number of ways you can make a 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

For the 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 endpoints then the number on both side of 1 is 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<bits/stdc++.h>
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

<?php
// PHP program for finding 
// total bitonic array
 
// Function to calculate
// no. of ways
function maxWays( $n)
{
     
    // return (2^(n - 1)) -2 
    return (pow(2, $n - 1) - 2);
}
 
// Driver Code
$n = 6;
echo maxWays($n);
 
// This code is contributed by vt_m.
?>

Javascript

<script>
 
// Javascript program for finding 
// total bitonic array
 
// Function to calculate no. of ways
function maxWays( n)
{
    // return (2^(n - 1)) -2 
    return (Math.pow(2, n - 1) - 2);
}
 
// Driver Code
var n = 6;
document.write( maxWays(n));
 
</script>

Output:

Time Complexity: O(log n)
Auxiliary Space: O(1)

This article is contributed by Shivam Pradhan (anuj_charm). If you like zambiatek and would like to contribute, you can also write an article using contribute.zambiatek.com or mail your article to review-team@zambiatek.com. See your article appearing on the zambiatek 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.

Recommended

Solve DSA problems on GfG Practice.

Solve Problems

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!