Amazon·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jul 2026

Summary

Amazon SWE interview with two algorithmic problems, both on the harder side. The interval coverage one was conceptually tricky and the GCD subarray problem had a neat twist with the k-modification constraint. Came away feeling like I need to sharpen my greedy intuition.

Questions Asked (2)

You're given n closed intervals on a number line. What is the minimum number of intervals you need to remove so that, among what remains, at least one interval fully contains all the others? Describe your algorithm with time and space complexity.

Algorithms & Data Structures

Author's notes

My first instinct was to sort by left endpoint and then brute force something, which was wrong.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Sort intervals by start ascending and end descending, then find the longest chain where each interval contains the next. The minimum removals equal n minus the length of the longest such chain. Use dynamic programming or greedy with binary search to compute the longest chain efficiently.

Pro tip: Clarify that 'fully contains' means both endpoints inclusive, and handle duplicate intervals by treating them as containing each other. Also, mention that the problem reduces to finding the longest sequence of nested intervals.

1. Understand the problem

Restate the goal: remove minimum intervals so that one interval contains all remaining. This is equivalent to finding the largest subset of intervals that form a containment chain (each contains the next).

2. Sort intervals strategically

Sort by start ascending, and for equal starts, sort by end descending. This ensures that if two intervals have the same start, the longer one comes first, which is necessary for containment.

3. Find longest containment chain

Compute the longest sequence where each interval contains the next. This can be done with DP in O(n^2) time, or optimized to O(n log n) using a greedy approach with binary search on the end points.

4. Compute minimum removals

The minimum number of intervals to remove is n minus the length of the longest containment chain. Return this value.

5. Analyze complexity

State time and space complexity: O(n log n) time for sorting and O(n) space for the DP or greedy array. If using O(n^2) DP, mention that it's less optimal but simpler.

Key Points to Mention

Containment condition: interval A contains B if A.start <= B.start and A.end >= B.end.
Sorting by start ascending and end descending ensures that any containing interval appears before the contained one.
The problem reduces to finding the longest chain of nested intervals.
Dynamic programming: dp[i] = 1 + max(dp[j]) for all j < i where interval i contains interval j.
Greedy with binary search: maintain a list of end points of the longest chain, and for each interval, find the first end point that is >= current end, replace it, or append if none.
Edge cases: duplicate intervals, intervals with same start but different ends, and intervals that are disjoint.

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

Given an integer array and a value k representing the max number of elements you can change inside a chosen subarray, find the shortest contiguous subarray where you can modify at most k elements to make the entire subarray's GCD greater than 1. For example, nums = [2, 2, 4, 9, 6] with k = 1 should return 1.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The example output being 1 actually helped me sanity-check my thinking fast.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

First clarify the problem and constraints, then propose an efficient algorithm. A sliding window approach can work by maintaining a window and checking if the number of elements not divisible by any prime factor of the window's GCD is ≤ k, but since GCD changes dynamically, consider enumerating prime factors and using a two-pointer technique to find the shortest subarray where at most k elements are not divisible by that prime. Return the minimum length across all primes.

Pro tip: Mention that the answer is at most k+1 because you can always pick a subarray of length k+1 and change k elements to match one element's value, making GCD > 1. This shows insight and can help prune the search.

1. Clarify the problem and constraints

Ask about input size, value ranges, and whether k can be 0. Confirm that modifying an element means changing it to any integer, and that the subarray must be contiguous.

2. Identify key observations

Note that for GCD > 1, all elements must share a common prime factor. Thus, we can fix a prime p and find the shortest subarray where at most k elements are not divisible by p.

3. Design an algorithm

For each prime p that appears in the array, use a sliding window to find the shortest subarray with at most k elements not divisible by p. Track the minimum length overall.

4. Analyze complexity and trade-offs

Discuss time complexity: O(n * number of distinct primes) with sliding window, which is efficient for typical constraints. Mention space complexity and potential optimizations.

5. Test with examples and edge cases

Walk through the given example and consider edge cases like k=0, all elements already sharing a factor, or no solution (though k+1 bound ensures a solution).

Key Points to Mention

GCD > 1 implies a common prime factor.
Sliding window technique for each prime factor.
Time complexity: O(n * distinct primes) or O(n * log(max_value)) with prime factorization.
Space complexity: O(n) for storing prime factors or O(1) extra.
Edge cases: k=0, k >= n, array with all 1s.
The answer is at most k+1, providing an upper bound.

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