Amazon·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Apr 2026

Summary

Amazon SWE coding round with a greedy array partitioning problem. Pretty clean problem once you see the sorting trick, but I fumbled around for a bit before getting there.

Questions Asked (1)

Given an integer array and a value maxdiff, partition the array into the minimum number of groups such that the difference between any two elements within any group is at most maxdiff. Return the number of groups.

Algorithms & Data Structures

Author's notes

Spent probably two minutes staring at it before realizing you just sort first.

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AI HintsAI Generated

Suggested Approach

Sort the array first, then use a greedy strategy: start a new group with the smallest remaining element and include all subsequent elements within maxdiff of it. This minimizes the number of groups because each group is as large as possible.

Pro tip: After presenting the greedy solution, mention that sorting is O(n log n) and the greedy pass is O(n), and briefly explain why this is optimal (exchange argument). This shows you understand both correctness and efficiency.

1. Clarify and Restate

Restate the problem in your own words and ask clarifying questions (e.g., can groups be empty? are elements distinct? what if maxdiff is negative?).

2. Sort the Array

Sort the array in non-decreasing order. This allows grouping contiguous elements and simplifies the difference check.

3. Greedy Grouping

Iterate through the sorted array, starting a new group with the first ungrouped element, and extend it as long as the difference between the current element and the group's first element is ≤ maxdiff.

4. Count Groups

Increment the group count each time you start a new group. Return the total count after processing all elements.

5. Analyze Complexity and Prove Optimality

State time complexity O(n log n) due to sorting and O(n) for the pass, and space O(1) or O(n) depending on sort. Explain why greedy is optimal using an exchange argument.

Key Points to Mention

Sorting the array is crucial for the greedy approach to work.
Greedy strategy: each group is formed by taking the smallest available element and including all elements within maxdiff of it.
Proof of optimality: any optimal solution can be transformed into the greedy solution without increasing the number of groups (exchange argument).
Time complexity: O(n log n) for sorting, O(n) for grouping; overall O(n log n).
Space complexity: O(1) extra space if sorting in-place, otherwise O(n) for the sorted copy.
Edge cases: empty array, maxdiff negative, all elements within maxdiff, etc.

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