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Instacart·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
May 2026

Summary

Instacart software engineering interview with a simulation-style coding problem that starts simple and then throws a multi-worker follow-up at you. The problem itself was interesting but the complexity analysis at the end is where things get real.

Questions Asked (2)

Q1

You're building a last-mile delivery simulator with a single shopper. Given n orders each with an arrival time and a service time, the shopper processes one order at a time and is idle until the next one arrives. Compute the average waiting time across all customers, where waiting time is finish time minus arrival time. Return a float rounded to 5 decimal places.

Algorithms & Data Structures
Author's notes

The single-shopper version is pretty mechanical once you track when the shopper becomes free.

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AI HintsAI Generated

Suggested Approach

Sort the orders by arrival time, then simulate the shopper's timeline by tracking the current finish time. For each order, the start time is the maximum of the current finish time and the order's arrival time; add the service time to get the finish time, and accumulate the waiting time (finish time minus arrival time). Finally, divide the total waiting time by n and round to 5 decimal places.

Pro tip: Clarify that waiting time includes both queueing and service time, and mention that sorting is necessary because orders may not be given in arrival order. Also, handle edge cases like n=0 by returning 0.0.

1. Understand the problem and define waiting time

Confirm that waiting time is finish time minus arrival time, which includes service time. Note that the shopper is idle if no orders are waiting.

2. Sort orders by arrival time

Sort the list of orders based on their arrival times to process them in the correct chronological order.

3. Simulate the shopper's timeline

Initialize current_finish_time to 0. For each order, set start_time = max(current_finish_time, arrival_time), finish_time = start_time + service_time, and update current_finish_time = finish_time.

4. Accumulate waiting times

For each order, compute waiting_time = finish_time - arrival_time and add it to a running total.

5. Compute and round the average

Divide the total waiting time by the number of orders (n) to get the average. Round the result to 5 decimal places and return it.

Key Points to Mention

  • Sorting orders by arrival time is essential for correct simulation.
  • The shopper's start time for an order is the maximum of the previous finish time and the order's arrival time.
  • Waiting time includes both queueing time and service time.
  • Time complexity is O(n log n) due to sorting, and space complexity is O(1) if sorting in place or O(n) otherwise.
  • Edge cases: n=0 (return 0.0), simultaneous arrivals, and large times requiring 64-bit integers.
  • Rounding to 5 decimal places using appropriate rounding functions (e.g., round() in Python or std::round in C++).

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

Q2

Follow-up: extend the simulator to k shoppers (k >= 1). Orders must be processed strictly in arrival order and each new order goes to whichever shopper is free earliest, breaking ties by smallest shopper ID. Compute the average waiting time and analyze the time and space complexity of your solution.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

This is where I got a bit turned around.

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AI HintsAI Generated

Suggested Approach

Model the problem as a min-heap of shoppers keyed by (next_available_time, shopper_id). Process orders in arrival order, pop the earliest-available shopper, assign the order, update their next_available_time, and push them back. Accumulate waiting times and divide by the number of orders to get the average.

Pro tip: Mention that using a heap with a tuple key (time, id) elegantly handles tie-breaking by smallest shopper ID without extra logic. Also note that if orders arrive in a stream, you can process them online with O(k) space, but if all orders are given upfront, you can still process them in order with the same heap.

1. Clarify assumptions and define variables

Confirm that orders are given as a list of arrival times (and possibly service durations) and that k shoppers are initially free at time 0. Define waiting time as the difference between when an order starts being processed and its arrival time.

2. Design the data structure

Use a min-heap to store shoppers, where each element is a tuple (next_available_time, shopper_id). This ensures that when multiple shoppers are free at the same time, the one with the smallest ID is chosen.

3. Process orders in arrival order

Iterate through the orders in the given order. For each order, pop the shopper with the smallest next_available_time (and smallest ID if tied). The order's start time is max(order_arrival_time, shopper_next_available_time). Update the shopper's next_available_time to start_time + service_duration and push them back into the heap.

4. Compute average waiting time

Accumulate the waiting time (start_time - order_arrival_time) for each order. After processing all orders, divide the total waiting time by the number of orders to get the average.

5. Analyze complexity

Time complexity: O(n log k) where n is the number of orders and k is the number of shoppers, because each order involves one heap pop and one heap push. Space complexity: O(k) for the heap, plus O(n) if storing all orders, but if processing in a stream, only O(k) extra space is needed.

Key Points to Mention

  • Use a min-heap keyed by (next_available_time, shopper_id) to efficiently select the earliest available shopper and break ties by smallest ID.
  • Process orders strictly in arrival order; do not reorder them.
  • Waiting time for an order is max(0, start_time - arrival_time), where start_time is when a shopper begins processing it.
  • Time complexity is O(n log k) due to heap operations for each of n orders.
  • Space complexity is O(k) for the heap; if orders are streamed, no additional storage for orders is needed.
  • Edge cases: k=1 (single shopper, orders queue up), multiple orders arriving at the same time, and shoppers with different service durations.

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