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Apple 2025 Fall SDE — Two Rounds Pure Coding Interview Review

I recently helped a classmate review their Apple 2025 Fall SDE interview, which consisted of two fast-paced coding rounds.
Both problems were classic patterns, but the interview heavily focused on the full chain of:

writing code → explaining logic clearly → surviving follow-ups.

Here is the full breakdown of core ideas, pitfalls, and high-frequency follow-up questions.

Round 1 — Subarray Sum Equals K

Problem

Given an array nums and an integer k, return the number of continuous subarrays whose sum equals k.

🚫 Common Pitfall

Do NOT use sliding window.
The array may contain negative numbers or zeros, which completely breaks the shrink/expand logic of sliding windows.

✅ Correct Solution: Prefix Sum + HashMap

Maintain the running prefix sum s.

For each s, check how many times s - k appeared before.

Store counts, not just existence.

Initialize {0: 1} so subarrays starting at index 0 are included.

Python Solution
from collections import defaultdict

def subarray_sum(nums, k):
cnt = defaultdict(int)
cnt[0] = 1 # prefix sum = 0 appears once

s = 0
ans = 0

for x in nums:
    s += x
    ans += cnt[s - k]   # valid subarrays ending here
    cnt[s] += 1         # record prefix sum

return ans
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Complexity

Time: O(n)

Space: O(n)

High-Frequency Follow-Up Questions

  1. Do negative numbers or zeros affect the algorithm?

No. They only break sliding window logic — prefix sum remains valid.

  1. Can we optimize to O(1) space?

Only if the problem asks “does such a subarray exist?”
Then a set of prefix sums is enough.

But counting requires storing frequencies, so O(n) is optimal.

  1. Why store prefix sum counts instead of just existence?

Because if s - k appears 3 times, it means 3 different valid starting points.
Storing only the existence would miss those.

Round 2 — Binary Tree Inorder Iterator (Lazy Traversal)

Problem

Design an iterator that performs a lazy inorder traversal, returning one node at a time via next() and checking availability with hasNext().

You cannot pre-traverse the entire tree.

Core Idea

Convert recursive inorder traversal into:

explicit stack + “push-left-path” helper function

Iterator Logic

Initialization
Push all left children from the root into the stack.

next()

Pop the top node.

If it has a right subtree, push its entire left path.

Return node value.

hasNext()
Return whether the stack is non-empty.

Python Implementation
class InorderIterator:
def init(self, root):
self.st = []
self._push_left(root)

def _push_left(self, node):
    while node:
        self.st.append(node)
        node = node.left

def hasNext(self):
    return bool(self.st)

def next(self):
    node = self.st.pop()
    if node.right:
        self._push_left(node.right)
    return node.val
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Complexity

next() amortized O(1)
(each node pushed and popped exactly once)

Space: O(h), where h is tree height.

High-Frequency Follow-Up Questions

  1. Multiple iterators reading the same tree — any issue?

If the tree is read-only: OK; each iterator keeps its own stack.

If the tree is mutable:
Explain options like:

versioning (fail-fast)

read-write locks

snapshot (but expensive)

Being able to discuss pros/cons shows strong system thinking.

  1. What if the tree degenerates into a linked list?

Then the stack stores all nodes → O(n).
This is the optimal lower bound for lazy inorder traversal.
You cannot do better.

  1. Advantages of an iterator over recursive traversal?

Recursive traversal is not lazy; must walk the entire tree.

Iterator fetches nodes on demand → suitable for large trees.

Avoids system stack limits, preventing overflow.

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