Recursive Count Write a recursive function rc_count (L:List[int], x:int) -> int that takes a list L and an integer x. The function returns the number of occurrences of x in L. For example, rc_count([1,1,2,3], 1) returns 2 and rc_count([1,1,2,3], 4) returns 0. Your Answer: 1 # Put your answer here 2

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Exercise 3 Write a function called middle that takes a list and returns a new list that contains all but the first and last elements. For example: >>> t = [1, 2, 3, 4] >>> middle(t) [2, 3]

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…