Nested lists can get very complex very quickly, and trying to go "down every level" in a nested list is both time-consuming and hard to
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Nested lists can get very complex very quickly, and trying to go "down every level" in a nested list is both time-consuming and hard to keep track of. This is why when we're thinking recursively, we rarely go down more than one level in a nested list. To prepare for this idea, review from the readings the definition of sub-nested-list, and then answer the following question: How many sub-nested-lists does [[15, [105, 13]], [0, 1, 2, 3], 4, []] ) have? 0 8 O 13 0 4 0 6 0 1

The assumption we made in our previous tracing---that the recursive calls always returned the right values»»might seem odd. But as we've discussed in the readings, this assumption is valid as long as we're confident that the base case is correct. So to complete this exercise, select the correct base case implementation that should fill in the below: def numintegerswbj: Union[int, List]) —> int: 1r isinstnnce(nbj, int): e152: 1! Recursive part 5 = 0 for sublist in ahj: s += nurLintegers(su|flist) return 5 0 return [obj] 0 return 1 0 return 0 0 return None 0 return obj

Recall our recursive definition of a nested list: A nested list is one of two types of values: . A single integer . A list of other nested lists ([ Ist_1 , Ist_2 , ..., Ist_n ]). Each Ist_i is called a sub-nested-list of the outer list. According to this definition, select all of the values that are valid nested lists below. O [1, [2, 3], , [4, 5]] O [1, 2, 3] 0 10 O [1, [2, 3], 'hello'] O None

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