Amazon·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

May 2026

Summary

Amazon SWE interview with a greedy/simulation problem that looks deceptively approachable but has some real edge cases hiding in it. Nothing behavioral, just the one algorithmic question.

Questions Asked (1)

You have an array of positive integers representing product quantities. On day i (1-indexed), you must pick exactly i distinct products and take one unit from each. Quantities can't go below zero. What is the maximum number of days this process can continue?

Algorithms & Data Structures

Author's notes

Took me a minute to even parse what was being asked.

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AI HintsAI Generated

Suggested Approach

First, reframe the problem as finding the largest k such that we can satisfy the daily requirements for k days. Then, sort the array in descending order and check if for each day i (1-indexed), the i-th largest quantity is at least i. The maximum k is the largest index where this condition holds.

Pro tip: Always clarify that the process stops when you cannot pick i distinct products with positive quantity on day i. Mention that sorting is key, and that the greedy check works because we only need to ensure the i-th largest has enough units.

1. Understand the problem

Restate the problem: each day i, pick i distinct products and decrement each by 1. Quantities cannot go negative. Find the maximum number of days.

2. Identify the condition for k days

For k days to be possible, for every day i from 1 to k, there must be at least i products with quantity at least i. This is necessary and sufficient.

3. Sort and check

Sort the array in descending order. For each index i (1-indexed), check if the i-th element is at least i. The largest i satisfying this is the answer.

4. Return the result

The maximum number of days is the largest i such that the i-th largest quantity is at least i. If no such i, return 0.

Key Points to Mention

Sorting the array in descending order simplifies the condition check.
The condition for k days is that for all i ≤ k, the i-th largest quantity ≥ i.
This is a greedy approach that works because we only need to ensure enough products for each day.
Time complexity: O(n log n) due to sorting, space complexity: O(1) or O(n) depending on sort.
Edge cases: empty array, all quantities zero, or quantities insufficient for even one day.
The process is monotonic: if k days are possible, any smaller number is also possible.

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