Reduce given Array by replacing adjacent elements with their difference

Given an array arr[] consisting of N elements(such that N = 2^k for some k ? 0), the task is to reduce the array and find the last remaining element after merging all elements into one. The array is reduced by performing the following operation:

• Merge the adjacent elements i.e merge elements at indices 0 and 1 into one, 2 and 3 into one and so on. 
• Upon merging the newly formed element will become the absolute difference between the two elements merged.

Examples:

Input: N = 4, arr[] = [1, 2, 3, 4]
Output: 0
Explanation: First operation:
On merging 1st and 2nd elements we will have a element with value1. 
On merging 3rd and 4th elements, we will have a element with value1. 
Therefore, we are left with two elements where each of them having cost 1.
Second operation:
On merging the 1st and 2nd elements we will get a new element with value 0.
This is because both elements had the same value of 1.

Input: N = 1, arr[] = [20]
Output: 20
Explanation: We can't perform any operation because performing an operation requires at least 2 elements. Hence, 20 is cost of the last remaining element

Approach: This problem can be solved using the Divide and Conquer approach.

• Create a recursive function.
  • The base condition for recursion will be if the size of the array is 1 then the answer will be the only array element in it.
  • Return the absolute difference between the first half of the array and the second half of the array by calling the recursive function for both halves.
  • Merge both halves and get the answer.

Below is the implementation of the above approach:

0

Time Complexity: O(2^log₂N)
Auxiliary Space: O(N)