Given an array of N numbers where a subarray is sorted in descending order and rest of the numbers in the array are in ascending order. The task is to sort an array where a subarray of a sorted array is in reversed order.
Input: 2 5 65 55 50 70 90
Output: 2 5 50 55 65 70 90
The subarray from 2nd index to 4th index is in reverse order.
So the subarray is reversed, and the sorted array is printed.
Input: 1 7 6 5 4 3 2 8
Output: 1 2 3 4 5 6 7 8
A naive approach will be to sort the array and print the array. Time Complexity of this approach will be O(N log n).
An efficient approach will be to find and store the starting index and ending index of the reversed subarray. Since the subarray is in descending order and the rest of the elements are in ascending order, only reversing the subarray will sort the complete array. Reverse the subarray using two pointer approach.
Below is the implementation of the above approach:
1 2 3 4 5 6 7 8
Time Complexity: O(n)
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