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DoorDash·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
May 2026

Summary

DoorDash coding screen, one problem, pretty standard binary search setup but they wanted the full proof of correctness which I wasn't expecting at all.

Questions Asked (1)

Q1

You have an array where each element represents a pile size, and an integer H representing total hours available. Find the minimum processing rate R (items per hour) such that you can finish all piles within H hours, processing one pile at a time and rounding up fractional hours. Walk through your algorithm, prove it's correct, and give time and space complexity.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

I got the binary search angle pretty quickly.

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AI HintsAI Generated

Suggested Approach

Recognize this as a binary search on the answer problem: the minimum rate R lies between 1 and max(piles), and the feasibility of a rate can be checked in O(n) time. Present the algorithm, prove monotonicity for correctness, and analyze complexity.

Pro tip: Mention that the upper bound can be max(piles) because any rate higher than the largest pile doesn't reduce total hours further, and use integer arithmetic to avoid floating-point precision issues.

1. Clarify and Define

Restate the problem: given piles and H, find minimal integer R such that sum(ceil(pile/R)) <= H. Confirm that R is an integer and that each pile is processed separately.

2. Identify Feasibility Function

Define a function canFinish(R) that computes total hours = sum(ceil(pile/R)) and returns true if total <= H. This function is monotonic: if R works, any larger R also works.

3. Apply Binary Search

Binary search R in the range [1, max(piles)]. For each mid, check canFinish(mid). If true, search left to find smaller R; else search right.

4. Prove Correctness

Argue that the feasibility function is monotonic, so binary search finds the minimal R. Also show that the answer is within the search range: R=1 might be too slow, but R=max(piles) always finishes in n hours (<= H if H>=n).

5. Analyze Complexity

Time: O(n log(max(piles))) because each feasibility check is O(n) and binary search takes O(log(max(piles))) iterations. Space: O(1) extra space.

Key Points to Mention

  • Monotonicity of the feasibility function: if a rate works, any higher rate also works.
  • Binary search bounds: low=1, high=max(piles) (or sum(piles) but max is tighter).
  • Feasibility check: sum of ceil(pile/R) computed efficiently using integer arithmetic: (pile + R - 1) // R.
  • Edge cases: H < number of piles (impossible, but problem likely guarantees H >= n), and piles with zero size (ignore or handle).
  • Time complexity: O(n log M) where M = max(piles); space O(1).
  • Alternative approaches: linear search (too slow) or priority queue (not optimal).

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.