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Amazon·Software Engineer·Onsite - Coding / Algorithms·Intermediate

Intermediate
Apr 2026

Summary

Amazon SWE coding round with two algorithmic problems back to back. Both were medium difficulty but had some tricky constraints that made them more interesting than they looked on the surface.

Questions Asked (2)

Q1

Given a string of ASCII characters, return a new string where characters are sorted by decreasing frequency. If two characters have the same frequency, break ties by ascending character code. Walk through your data structure choices and give time and space complexity.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

The sorting logic clicked pretty fast but I fumbled the tie-breaking part initially.

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AI HintsAI Generated

Suggested Approach

First, clarify the problem constraints (e.g., ASCII means 128 possible characters, input size). Then, propose using a frequency map (array or hash map) to count occurrences, and sort the characters based on frequency descending and character code ascending. Finally, analyze the time and space complexity, discussing trade-offs between different sorting approaches.

Pro tip: Mention that since the character set is fixed (128 ASCII), you can use a bucket sort approach to achieve O(n) time, which is optimal. This shows you consider constraints and can optimize beyond the obvious O(n log n) sorting solution.

1. Clarify requirements and constraints

Ask about input size, character set (ASCII), and whether the output should be stable. Confirm that ties are broken by ascending character code.

2. Choose data structures

Use an array of size 128 for frequency counting (since ASCII), or a hash map if the character set is larger. For sorting, consider sorting the distinct characters based on frequency and character code.

3. Outline algorithm

Count frequencies, then create a list of (char, freq) pairs, sort them with a custom comparator (freq descending, char ascending), and build the result string by repeating each character freq times.

4. Analyze complexity

Time: O(n + k log k) where k is the number of distinct characters (≤128), which is effectively O(n). Space: O(k) for the frequency map and output string. Mention that bucket sort can achieve O(n) time.

5. Discuss trade-offs and edge cases

Compare sorting vs. bucket sort. Handle empty string, single character, and all same characters. Consider if input can contain non-ASCII (then use hash map).

Key Points to Mention

  • Use an array of size 128 for O(1) frequency counting due to fixed ASCII set.
  • Sorting distinct characters takes O(k log k) where k ≤ 128, so overall O(n) time.
  • Bucket sort can achieve O(n) time by grouping characters by frequency.
  • Space complexity is O(n) for the output string, but auxiliary space is O(k) or O(1) if using fixed array.
  • Tie-breaking by ascending character code requires a custom comparator.
  • Edge cases: empty string, all characters same, and characters with same frequency.

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

Q2

Given a sorted (nondecreasing) integer array and a target sum, find all unique index pairs (i, j) where i < j and the two elements sum to the target. Do it in O(n) time and O(1) extra space using two pointers, and explain how you handle duplicate values to avoid returning the same pair twice.

Algorithms & Data Structures
Author's notes

Two pointers is second nature at this point so the basic structure came out fine.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem and constraints, then explain the two-pointer technique: initialize left at 0 and right at n-1, move pointers based on the sum compared to the target, and skip duplicates to ensure unique pairs. Walk through a small example to illustrate, and analyze time and space complexity.

Pro tip: Emphasize that the array is sorted and nondecreasing, which allows the two-pointer approach; explicitly state how you handle duplicates by skipping identical values after finding a valid pair to avoid redundant pairs.

1. Clarify and Confirm

Restate the problem to ensure understanding: sorted array, target sum, unique index pairs (i, j) with i < j, O(n) time, O(1) space. Ask if the array can contain duplicates and if the output should be pairs of indices or values.

2. Explain Two-Pointer Approach

Describe initializing two pointers at the start and end of the array. While left < right, compute the sum; if sum equals target, record the pair and move both pointers inward while skipping duplicates; if sum < target, move left forward; if sum > target, move right backward.

3. Handle Duplicates

After finding a valid pair, advance left while the next element is the same as the current left, and decrement right while the previous element is the same as the current right. This ensures each unique pair is recorded only once.

4. Walk Through Example

Choose a small array with duplicates, e.g., [1,1,2,3,4,4,5] and target 6, and trace the algorithm step by step to demonstrate correctness and duplicate handling.

5. Analyze Complexity

State that each element is visited at most once, so time complexity is O(n). Space complexity is O(1) extra space since only a few variables are used, aside from the output list.

Key Points to Mention

  • The array is sorted (nondecreasing), which is crucial for the two-pointer technique.
  • Two pointers start at opposite ends and move inward based on the sum comparison.
  • Duplicate handling: skip identical values after finding a pair to avoid duplicate pairs.
  • Time complexity is O(n) because each element is processed at most once.
  • Space complexity is O(1) extra space, excluding the output.
  • Edge cases: empty array, no valid pairs, all elements same, target smaller than smallest sum or larger than largest sum.

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