GO Classes DPP | GATE DA | Python & DSA | Nested Mutation

Answer 1

Answer can not be specified unless deep_map is defined. AI says its a part of a course at UCB, CS61A. But function definition will decide whether it works recursively or only at the first level, even if the name suggests recursive traversal till final depth. If it is assumed to be recursive, B,C and D are correct. else only C, D are correct.

Comments

@agentsmith in this specific scenario, we don't need to see the underlying code bcoz the problem statement explicitly defines the function's contract for us.

Look at the very first sentence of the problem: "replaces every non-list element inside a nested list." The word every is the key. If the function only checked the first level, it would fail to replace every element.

Bcoz the problem guarantees this deep traversal behavior as a premise, we don't have to guess if it stops at the first level. We must evaluate the code under the rule that it successfully updates the entire nested structure.

Therefore, the deep traversal behavior is not an assumption, but a given fact of the problem, making B, C, and D the definitively correct statements.

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Right, decoding assumptions is essential.

Answer 2

The function $\texttt{deep_map}$ modifies the original nested list in place.

Initially:

s = [3, [1, [4, [1]]]]

Applying $\texttt{lambda x: x + 1}$ to every non-list element gives:

3 becomes 4
1 becomes 2
4 becomes 5
1 becomes 2

So $\texttt{s}$ becomes:

[4, [2, [5, [2]]]]

Hence option B is correct.

Since $\texttt{deep_map}$ modifies in place and returns $\texttt{None}$, option C is also correct.

The function should not create new nested lists, so the old references $\texttt{s1}$, $\texttt{s2}$, and $\texttt{s3}$ should still point to the same inner lists.

Therefore, option D is correct.

Option A is incorrect because $\texttt{result}$ is $\texttt{None}$, not the modified list.

Correct Options: B, C and D