# High-Score DoorDash SWE Interview Experience (Bugfree Users): Coding + System Design Playbook

# High-Score DoorDash SWE Interview Experience (Bugfree Users)

<img src="https://hcti.io/v1/image/019e0000-bb77-7a0e-a245-011880df0c50" alt="DoorDash interview cover" style="max-width:700px; width:100%; height:auto;" />

Posted by Bugfree users — a concise, practical playbook from a high-scoring DoorDash Software Engineer interview.

This post summarizes the interview flow, highlights the most interesting problems, explains solution ideas and complexity, and shares tips for system design and behavioral rounds.

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## Interview flow (what to expect)

1. Online Assessment — usually short coding challenge(s) to filter candidates.
2. Virtual onsite — three rounds mixing coding and behavioral questions.
3. Extra technical round — sometimes an additional system-design or architecture problem.

Quick tips:
- Read the prompt carefully and clarify constraints up front (input sizes, allowed time, edge cases).
- Talk while you code: explain approach, trade-offs, and complexity.
- For system-design rounds, start with requirements, then high-level architecture, then scale/consistency concerns.

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## Highlights & problem playbook

Below are the key problems encountered and concise approaches that scored well.

### 1) Geometry / area-split: find a horizontal line that balances total cake area above vs below

Problem summary:
- Given a polygon (or a composite cake shape), find the horizontal line y = h such that the area above the line equals the area below the line (i.e., split total area in half).

Approach:
- Observe that the area above a horizontal line is a monotonic function of h. Use binary search on h.
- For a candidate h, clip the polygon to the half-plane y >= h (or y <= h) and compute the polygon area with the shoelace formula.
- Binary search until the area above is total_area / 2 within tolerance.

Implementation details & complexity:
- Polygon clipping to a horizontal line is O(n) (walk edges and compute intersections with y = h when necessary).
- Each binary-search iteration costs O(n); ~50 iterations gives good floating-point precision.
- Overall: O(n * iterations), with iterations bounded by log(precision) (practical constant ~40–60).

Edge cases / tips:
- Handle horizontal edges and vertices lying exactly on y = h carefully.
- If shapes are multiple disjoint polygons, sum clipped areas.

When to use this pattern:
- Any problem that asks for a threshold on a monotonic geometric measure — binary search + clipping/integration is a powerful pattern.

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### 2) Restaurant waitlist design (serve the first group with size ≥ table)

Problem summary:
- You maintain a waitlist of arriving groups (with group sizes). When a table becomes available (has a capacity), serve the earliest-arrived group whose size ≤ table capacity.

Design considerations & options:
- Naive: keep a FIFO queue and scan until you find a group that fits — O(n) per seat.
- Better: maintain multiple queues keyed by group size and a data structure to find the smallest eligible size quickly.

Data structures & tradeoffs:
- Sorted list of unique group sizes + queues: keep a sorted container of sizes present (e.g., TreeSet). When a table of capacity c arrives, find the largest size ≤ c in the set and take the front of that size's queue.
  - Complexity: O(log m) to find size (m = distinct sizes) and O(1) dequeue. Good when size values are small or limited.
- Balanced BST / ordered map (size -> queue): same idea but supports dynamic inserts/deletes in O(log m).
- Hash with buckets + linear scanning across possible sizes: useful if group sizes are bounded and small (e.g., 1..k), complexity O(k) per seat.

Concurrency & production notes:
- Use locking per bucket or optimistic concurrency (CAS) for scalability.
- For high throughput, partition by restaurant or by size ranges to avoid contention.
- Persistent store: store queue order in durable logs if required; use in-memory queue with periodic checkpointing.

When asked on an interview:
- Clarify arrival rates, max group size, and latency requirements.
- Sketch simple approach (buckets + queues) and then explain scaling options (sharding, caching, consistency).

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### 3) Collinearity check: detect any 3 points on one line

Problem summary:
- Given n points, determine whether any three are collinear.

Efficient approach (O(n^2)):
- For each point p_i, compute slopes to all other points p_j.
- Normalize slopes as reduced integer pairs (dy, dx) or use atan2 for floating comparison; handle vertical lines (dx = 0) specially.
- Use a hashmap from slope -> count. If any slope count >= 2 (i.e., at least two other points share the same slope relative to p_i), you have 3 collinear points.

Complexity:
- For each base point: O(n) slope computations and hashmap ops. Total O(n^2) time and O(n) extra space per iteration.
- Avoid O(n^3) brute-force.

Edge cases:
- Duplicate points: treat duplicates as increasing counts for any slope; define the required behavior first.
- Precision: it's safer to reduce dx/dy by gcd to integer pair, to avoid floating-point inaccuracies.

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### Final coding problem: matrix longest path (DFS + backtracking) and return the path

Problem summary:
- Given a matrix/grid, find the longest path under some movement constraints (e.g., moving to adjacent cells with certain rules) and return the actual path.

Common approach (DFS + backtracking):
- Use DFS from each cell, maintain a visited set (or boolean visited grid) to avoid revisiting nodes in the current path.
- Track current path (stack of coordinates) and best path found so far.
- When exploring a neighbor, mark visited, push to path, recurse; after returning, pop and unmark (backtrack).

Optimizations / variants:
- If the graph is a DAG (e.g., strictly increasing values and you only move to higher values), you can memoize the longest path starting from each cell to reduce complexity to O(n) overall.
- If revisits are allowed but with constraints, the problem might be NP-hard; expect small n or additional constraints.

Pseudo outline:
- best_path = []
- for each cell (i, j):
    dfs(i, j)

- dfs(i, j):
    mark visited[i][j] = true
    append (i,j) to current_path
    if current_path.length > best_path.length: best_path = copy(current_path)
    for each neighbor (ni, nj) allowed:
        if not visited[ni][nj] and rule_allows_move((i,j),(ni,nj)):
            dfs(ni, nj)
    pop current_path
    mark visited[i][j] = false

Complexity:
- Worst-case exponential without memoization. With memoization (in DAG-like constraints), can be polynomial.

Return value:
- Return best_path as list of coordinates in visiting order.

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## Behavioral & system design tips for DoorDash interviews

- For behavioral rounds: use STAR (Situation, Task, Action, Result). Be concise and emphasize impact and trade-offs.
- For system-design: clarify requirements (functional + non-functional), pick key flows, show a high-level diagram, discuss data models, scaling, caching, consistency, and failure modes. Quantify where possible (qps, latency, storage).
- When asked for data structures, show both a simple correct approach and how you'd optimize for scale.

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## Final notes & resources

- Practice pattern-based problems: two-pointers, sliding window, binary search on answer, DFS/backtracking, graph BFS/DFS, hash-based counting.
- System-design resources: System Design Primer (GitHub), High Scalability blog, and real-world talk videos.
- Coding resources: LeetCode, HackerRank, and Cracking the Coding Interview.

Good luck preparing — clarify constraints early, communicate trade-offs, and practice writing clean code on a whiteboard or shared editor.

#SoftwareEngineering #SystemDesign #CodingInterview