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This is a Medium difficulty Coding & Algorithms question, commonly asked during Onsite rounds at Amazon.

What role is this question designed for?

This question is commonly asked for Data Scientist candidates at Amazon during technical interviews.

Implement Reservoir Sampling; Analyze Time and Space Complexity

Quick Overview

This question evaluates understanding of streaming randomized algorithms and probabilistic reasoning—specifically reservoir sampling—along with analysis of time and space complexity and use of hash-based structures for duplicate detection; it is in the Coding & Algorithms domain and tests both conceptual probability invariants and practical implementation skills. Such problems are commonly asked to gauge a candidate's ability to design one-pass, linear-time solutions for unbounded data streams and to reason about algorithmic correctness and resource bounds in technical interviews.

Implement Reservoir Sampling; Analyze Time and Space Complexity

Company: Amazon

Role: Data Scientist

Category: Coding & Algorithms

Difficulty: Medium

Interview Round: Onsite

##### Scenario Process an unbounded stream of user IDs and maintain a uniform random sample of k users in memory. ##### Question Implement reservoir sampling for an incoming stream; analyze its time and space complexity. Explain why the algorithm guarantees each element is selected with probability k⁄n. Given an array of integers, return indices of duplicate values using only set() and enumerate(). Provide O(n) code. ##### Hints Focus on one-pass algorithms and probability invariants.

Quick Answer: This question evaluates understanding of streaming randomized algorithms and probabilistic reasoning—specifically reservoir sampling—along with analysis of time and space complexity and use of hash-based structures for duplicate detection; it is in the Coding & Algorithms domain and tests both conceptual probability invariants and practical implementation skills. Such problems are commonly asked to gauge a candidate's ability to design one-pass, linear-time solutions for unbounded data streams and to reason about algorithmic correctness and resource bounds in technical interviews.

Given a list of integers nums, return a list of indices i such that nums[i] is a value that occurs more than once in nums. Return indices in increasing order. Implement an O(n) solution using only set() and enumerate() (no dictionaries, counters, or sorting).

Constraints

0 <= len(nums) <= 200000
nums[i] fits in 32-bit signed integer
Time complexity must be O(n)
Use only set() and enumerate(); do not use dict, Counter, or sorting
Return indices in increasing order

Solution

from typing import List

def duplicate_indices(nums: list[int]) -> list[int]:
    seen = set()
    dup_vals = set()
    for x in nums:
        if x in seen:
            dup_vals.add(x)
        else:
            seen.add(x)
    return [i for i, x in enumerate(nums) if x in dup_vals]

Explanation

Traverse nums once to identify which values occur more than once using two sets: 'seen' for values observed and 'dup_vals' for values seen again. Then, traverse nums with enumerate and collect indices whose values are in 'dup_vals'. This preserves increasing index order without sorting and uses only sets and enumerate.

Time complexity: O(n). Space complexity: O(u) additional space for sets, where u is the number of unique values; plus O(k) for the output indices where k is the number of indices returned.

Hints

First pass: use a set to track seen values and collect which values are duplicates.
Second pass: use enumerate to collect indices whose values are in the duplicates set.
No need to sort; enumerating maintains increasing index order.

Last updated: Mar 29, 2026

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Quick Overview

Implement Reservoir Sampling; Analyze Time and Space Complexity

Company: Amazon

Role: Data Scientist

Category: Coding & Algorithms

Difficulty: Medium

Interview Round: Onsite

Constraints

0 <= len(nums) <= 200000
nums[i] fits in 32-bit signed integer
Time complexity must be O(n)
Use only set() and enumerate(); do not use dict, Counter, or sorting
Return indices in increasing order

Solution

from typing import List

def duplicate_indices(nums: list[int]) -> list[int]:
    seen = set()
    dup_vals = set()
    for x in nums:
        if x in seen:
            dup_vals.add(x)
        else:
            seen.add(x)
    return [i for i, x in enumerate(nums) if x in dup_vals]

Explanation

Time complexity: O(n). Space complexity: O(u) additional space for sets, where u is the number of unique values; plus O(k) for the output indices where k is the number of indices returned.

Hints

First pass: use a set to track seen values and collect which values are duplicates.
Second pass: use enumerate to collect indices whose values are in the duplicates set.
No need to sort; enumerating maintains increasing index order.

Last updated: Mar 29, 2026

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