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Write a function to count the number of nodes in a singly linked list. Discuss the approach
and complexity of your solution.

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This is to be done in ML programming language. Please use the simplest code possible, nothing too advanced. Define a function rem_duplicate: ''a list -> ''a list whichtakes in input a list and removes all the duplicates. Test your code with sampleinput and report result. Examples: rem_duplicate [] = [] rem_duplicate [1,2,1] = [1,2] rem_duplicate ["a","a","a"] = ["a"] rem_duplicate [[1],[1,2],[1,2,3],[1,2],[4,5]] = [[1],[1,2],[1,2,3],[4,5]]
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Using the function provided, write a function that decides whether two lists are "identical". That is, if cycling by any number N makes the lists identical, the function should return True, otherwise, it should return False. for example identical([1,2,3,4,5], [5,4,1,2,3,]) should give True and identical([1,2,3,4,5], [1,2,3,5,4]) should give False.
Using the function provided, write a function that decides whether two lists are "identical". That is, if cycling by any number N makes the lists identical, the function should return True, otherwise, it should return False. for example identical([1,2,3,4,5], [5,4,1,2,3,]) should give True and identical([1,2,3,4,5,6], [1,2,3,5,4]) should give False.
Define a recursive function named trim; it takes as its arguments a linked list (the head node of it) and a non-negative integer representing the index of an element in the list. The function mutates the linked list such that all elements that occur after the index is removed. Additionally, the function returns a linked list of all the elements that were moved. If the index value exceeds the length of the linked list, it should raise an IndexError. For example, if we defined a list as l1 = LN(1, LN(2, LN 3))) , calling the function would return the linked list LN(2, LN(3)), and the mutated l1 would become LN(1). You must use recursion. def trim(l: LN, index: int) -> LN:
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