Quick Overview

This question evaluates a candidate's ability to design and analyze efficient algorithms for detecting longest consecutive incrementing sequences in unsorted integer arrays, testing understanding of appropriate data structures and time/space complexity trade-offs.

Identify Longest Consecutive Incrementing Watch-Time Sequence

Company: Netflix

Role: Data Scientist

Category: Coding & Algorithms

Difficulty: medium

Interview Round: Onsite

##### Scenario A streaming platform records daily minutes watched per user and wants to identify engagement streaks. ##### Question Given an unsorted integer array representing daily watch-time deltas, return the length of the longest sequence of consecutive, incrementing integers (e.g., 1,2,3…). Explain the algorithm and analyze complexity. ##### Hints Think hash-set to achieve O(n) time and avoid re-scanning sequences.

Overview: This question evaluates a candidate's ability to design and analyze efficient algorithms for detecting longest consecutive incrementing sequences in unsorted integer arrays, testing understanding of appropriate data structures and time/space complexity trade-offs.

Given an unsorted integer array deltas representing daily watch-time changes, return the length of the longest sequence of values that are consecutive and strictly increment by 1 (k, k+1, k+2, ...). The sequence is formed from values only; indices and original order do not matter. Duplicates do not extend a sequence. If the array is empty, return 0.

Constraints

  • 0 <= len(deltas) <= 200000
  • -10^9 <= deltas[i] <= 10^9
  • Duplicates may appear and should be treated as a single value for sequence building
  • Expected average time: O(n) using hashing
  • Auxiliary space: O(n)

Hints

  1. Insert all numbers into a hash set for O(1) average lookups.
  2. Only start counting from numbers x where x-1 is not in the set (start of a run).
  3. From each start, incrementally check x+1, x+2, ... to count the run length.
  4. Duplicates are naturally handled by the set and do not extend runs.

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Approach

Use a hash set of all values for O(1) average membership checks. For each number x, only attempt to grow a sequence if x-1 is not present; this ensures each sequence is counted exactly once. From such a start, repeatedly check x+1, x+2, ... to find the sequence length. Duplicates are ignored by the set. The maximum length across all starts is the answer.

Time complexity:
O(n) average-case
Space complexity:
O(n)