Describe a fast recursive algorithm for reversing a singly linked list.

Java language NOTE THAT WE ARE USING **LINKED LISTS** SEE PICTURES FOR ALL INFORMATION Solve method public boolean inSort(double elt) { //code goes here } please and thank you!!!!

**Cenaage Python** Question: A sequential search of a sorted list can halt when the target is less than a given element in the list. Modify the program to stop when the target becomes less than the current value being compared. In a sorted list, this would indicate that the target is not in the list and searching the remaining values is unnecessary. CODE: def sequentialSearch(target, lyst): """Returns the position of the target item if found, or -1 otherwise. The lyst is assumed to be sorted in ascending order.""" position = 0 while position < len(lyst): if target == lyst[position]: return position position += 1 return -1 def main(): """Tests with three lists.""" print(sequentialSearch(3, [0, 1, 2, 3, 4])) print(sequentialSearch(3, [0, 1, 2])) # Should stop at second position. print(sequentialSearch(3, [0, 4, 5, 6])) if __name__ == "__main__": main()

Quick Sort We choose an element from the list, called the pivot. We'll use it to divide the list into two sub-lists. We reorder all the elements around the pivot The ones with smaller value are placed before it All the elements greater than the pivot after it. After this step, the pivot is in its final position. This is the important partition step. We apply the above steps recursively to both sub-lists on the left and right of the pivot. Quick Sort (Example) Consider the following array Arr[] = {5, 9, 4, 6, 5, 3} Let's suppose we pick 5 as the pivot for simplicity We'll first put all elements less than 5 in the first position of the array: {3, 4, 5, 6, 5, 9} We'll then repeat it for the left sub-array {3,4}, taking 3 as the pivot There are no elements less than 3 We apply quicksort on the sub-array in the right of the pivot, i.e. {4} This sub-array consists of only one sorted element We continue with the right part of the original array, {6, 5, 9} until we get the final ordered...

Complete the recursive function remove_last which creates a new list identical to the input list s but with the last element in the sequence that is equal to x removed. Hint: Remember that you can use negative indexing on lists! For example 1st [-1] refers to the last element in a list 1st, lst [-2] refers to the second to last element... def remove_last(x, s): """Create a new list that is identical to s but with the last element from the list that is equal to x removed.>>> remove_last(1, []) []>>> remove_last(1, [1]) []>>> [1]>>> [2] remove_last(1, [1,1]) remove_last(1, [2,1])>>> remove_last(1, [3,1,2]) [3, 2]>>> remove_last(1, [3,1,2,1]) [3, 1, 2]>>> remove_last (5, [3, 5, 2, 5, 11]) [3, 5, 2, 11] IIIII "*** YOUR CODE HERE ***" Illustrated here is a more complete doctest that shows good testing methodology. It is a little cumbersome as documentation, but you'll want to think about it for your projects. Test every condition that might come up. Then you won't be surprised when...

Lists and tuples are two distinct data structures.