Amazon·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jul 2026

Summary

Amazon SWE coding round, one meaty algorithmic problem that took up basically the whole session. The question was deceptively clean on the surface but the complexity and edge case requirements made it a real grind.

Questions Asked (1)

Given an integer array called priorities, compute delay[i] for each index i, where delay[i] is the distance to the rightmost index j (j > i) such that priorities[j] is strictly less than priorities[i]. If no such j exists, delay[i] is 0. Implement this in better than O(n^2) time, explain your complexity, and handle edge cases like duplicates, fully increasing arrays, and fully decreasing arrays.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The brute force came to me in about 30 seconds and I almost just said it out loud before catching myself.

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AI HintsAI Generated

Suggested Approach

Use a monotonic stack to efficiently find the rightmost smaller element for each index. Traverse the array from right to left, maintaining a stack of indices with increasing priorities, and for each element, pop elements that are greater than or equal to the current priority to find the rightmost smaller element. Then compute the delay as the distance between indices, or 0 if no such element exists.

Pro tip: Explicitly discuss how duplicates are handled: since we pop elements that are greater than or equal to the current priority, we ensure that we find the rightmost strictly smaller element, not just any smaller element. Also, walk through edge cases like fully increasing and decreasing arrays to demonstrate thoroughness.

1. Understand the problem and edge cases

Clarify that delay[i] is the distance to the rightmost j > i with priorities[j] < priorities[i], and 0 if none. Identify edge cases: duplicates, fully increasing, fully decreasing, single element, empty array.

2. Design an efficient algorithm

Choose a monotonic stack approach to achieve O(n) time. Explain that a stack can maintain candidates for the next smaller element to the right, and by processing from right to left, we can find the rightmost smaller element.

3. Detail the stack algorithm

Initialize an empty stack and a delay array of zeros. Iterate i from n-1 down to 0: while stack is not empty and priorities[stack.top()] >= priorities[i], pop. If stack is not empty, delay[i] = stack.top() - i. Push i onto the stack.

4. Analyze complexity and correctness

Explain that each index is pushed and popped at most once, so time is O(n) and space is O(n). Argue correctness: the stack maintains indices with strictly increasing priorities from top to bottom? Actually, after popping, the top is the nearest index to the right with priority less than current, which is also the rightmost because we process right to left.

5. Test with examples and edge cases

Walk through examples: [5,4,3,2,1] (fully decreasing) gives delays [1,1,1,1,0]; [1,2,3,4,5] (fully increasing) gives all 0; duplicates like [3,3,3] give all 0; mixed arrays demonstrate correctness.

Key Points to Mention

Monotonic stack technique for next smaller element problems
Time complexity O(n) and space complexity O(n)
Handling duplicates by using >= in the while condition to ensure strictly smaller
Edge cases: fully increasing array (all delays 0), fully decreasing array (delays 1 except last), single element, empty array
Correctness argument: why the stack top after popping is the rightmost smaller element
Comparison with brute force O(n^2) approach to highlight efficiency

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