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.
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 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.
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