Amazon·Software Engineer·Technical Phone Screen·Senior

Senior

Jun 2026

Summary

Amazon Applied Scientist interview with a coding problem around pair counting. Pretty algorithmic, nothing behavioral from what I can tell. The hashmap angle was the key insight and if you miss it you're probably spinning your wheels.

Questions Asked (1)

Given an integer array and a target value, count the number of distinct unordered value pairs (a, b) where a + b equals the target and a, b come from elements at different indices. A value can pair with itself only if it appears more than once in the array.

Algorithms & Data Structures

Author's notes

The duplicate case is where people mess up.

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AI HintsAI Generated

Suggested Approach

Use a hash map to store the frequency of each element, then iterate through the unique elements and for each element x, check if target - x exists in the map. To avoid double-counting, only count each pair once by ensuring x <= target - x, and handle the case where x == target - x by checking if the frequency is at least 2.

Pro tip: Clarify whether the array can contain duplicates and whether the pairs are based on values or indices. Mention that the solution runs in O(n) time and O(n) space, which is optimal for this problem.

1. Understand the problem

Restate the problem in your own words to ensure you understand that we need distinct unordered pairs of values from different indices, and that a value can pair with itself only if it appears at least twice.

2. Choose the right data structure

Decide to use a hash map to count the frequency of each element, which allows O(1) lookups for the complement.

3. Iterate and count pairs

Iterate through the unique elements. For each element x, compute complement = target - x. If x < complement and complement exists in the map, add the product of their frequencies to the count. If x == complement and frequency[x] >= 2, add frequency[x] choose 2 to the count.

4. Handle edge cases

Consider cases like empty array, no pairs, all elements same, and target such that complement equals x. Ensure the logic correctly counts pairs without double-counting.

5. Analyze complexity

State that the time complexity is O(n) because we traverse the array once to build the frequency map and once through unique elements, and space complexity is O(n) for the map.

Key Points to Mention

Use a hash map to store frequencies for O(1) complement lookups.
Avoid double-counting by only considering x <= complement and handling x == complement separately.
For x == complement, the number of pairs is C(freq, 2) = freq * (freq - 1) / 2.
Time complexity: O(n) where n is the number of elements.
Space complexity: O(n) for the hash map.
Edge cases: empty array, no valid pairs, all elements identical, target requiring self-pairing.

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