Apple Software Engineer Interview Questions

I would start with the iterative approach because it is the simplest to explain and the most space-efficient. I would keep three pointers: previous, current, and next, then walk through the list and flip each pointer one by one. The reason I choose this approach is that it gives me O(n) time and O(1) extra space. If the interviewer wanted a recursive version, I could explain that too, but I would mention it uses stack space. I would also call out edge cases like an empty list and a single-node list.

I would use the slow and fast pointer technique. The slow pointer moves one step at a time, the fast pointer moves two, and if there is a cycle they will eventually meet. The reason I choose this approach is that it avoids extra memory and still runs in O(n) time. If the interviewer asks for the cycle entry point, I would say I can reset one pointer to the head and move both one step at a time until they meet again. I would also mention edge cases like an empty list or a single node pointing to itself.

I would use two pointers, one for each array, and build the result from the smallest remaining element at each step. If I need to do it in place, I would work from the end of the destination array so I do not overwrite values I still need. The reason I choose this approach is that it is straightforward and runs in O(n + m) time. The space complexity is O(1) if I merge in place, or O(n + m) if I create a new array. I would also mention edge cases like one empty array or duplicate values.

I would first count the frequency of each character using a hash map. Then I would scan the string from left to right and return the first character whose count is one. The reason I choose this approach is that it is easy to explain and gives O(n) time. The space usage depends on the character set, but it is usually small or O(n) in the general case. I would also mention edge cases like an empty string and the case where every character repeats.

I would use two pointers, one at the start and one at the end of the string. While the pointers move inward, I compare the characters and stop as soon as I find a mismatch. If the problem says to ignore spaces, punctuation, or case, I would clarify that first and preprocess the string accordingly. The reason I choose two pointers is that it is simple, fast, and uses O(1) extra space. I would also mention edge cases like empty strings and single-character strings.