Kth Smallest Element in a sorted array formed by reversing subarrays from a random index

Kth Smallest Element in a sorted array formed by reversing subarrays from a random index

Given a sorted array arr[] of size N and an integer K, the task is to find K^th smallest element present in the array. The given array has been obtained by reversing subarrays {arr[0], arr[R]} and {arr[R + 1], arr[N – 1]} at some random index R. If the key is not present in the array, print -1.

Examples:

Input: arr[] = { 4, 3, 2, 1, 8, 7, 6, 5 }, K = 2
Output: 2
Explanation: Sorted form of the array arr[] is { 1, 2, 3, 4, 5, 6, 7, 8 }. Therefore, the 2nd smallest element in the array arr[] is 2.

Input: arr[] = { 10, 8, 6, 5, 2, 1, 13, 12 }, K = 3
Output: 5

Naive Approach: The simplest approach to solve the problem is to sort the given array arr[] in increasing order and print the K^th smallest element in the array.
Time Complexity: O(N*log(N))
Auxiliary Space: O(1)

Efficient Approach: The optimal idea is based on the observation that the Rth element is the smallest element because the elements in the range [1, R] are reversed. Now, if the random index is R, it means subarray [1, R] and [R + 1, N] are sorted in decreasing order. Therefore, the task reduceS to finding the value of R which can be obtained using binary search. Finally, print the K^th smallest element.

Follow the steps below to solve the problem:

• Initialize l as 1 and h as N to store the boundary elements index of the search space for the binary search.
• Loop while the value of l+1 < h
  • Store the middle element in a variable, mid as (l+h)/2.
  • If arr[l] ≥ arr[mid]. If it is true then check on the right side of mid by updating l to mid.
  • Otherwise, update r to mid.
• Now after finding R, if K ≤  R, then the answer is arr[R-K+1]. Otherwise, arr[N-(K-R)+1].

Below is the implementation of the above approach: 

Python3

def findkthElement(arr, n, K):                 l = 0     h = n-1             while l+1 < h:                             mid = (l+h)//2                     if arr[l] >= arr[mid]:             l = mid                     else:             h = mid             if arr[l] < arr[h]:         r = l     else:         r = h             if K <= r+1:         return arr[r+1-K]     else:         return arr[n-(K-(r+1))]       if __name__ == "__main__":                 arr = [10, 8, 6, 5, 2, 1, 13, 12]     n = len(arr)     K = 3                 print(findkthElement(arr, n, K) )

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

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