Question: Please help with this OCaml code asap!! Note that function calls have the highest precedence in OCaml, so the last expression is the same as

Please help with this OCaml code asap!!

Note that function calls have the highest precedence in OCaml, so the last expression is the same as (f00f(n/2))+(f00f(n/2)). If we assume that the function f can compute its result in constant time and constant space, what are the (worst-case) time and space complexities of the function foo? Justify your answer. (It doesn't have to be a full mathematical proof, but it should be a convincing argument.) Assume that the integer input n is always non-negative, and assume the usual applicative-order evaluation rule

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