Summary of "Longest Consecutive Sequence | Google Interview Question | Brute Better Optimal"

Summary of “Longest Consecutive Sequence | Google Interview Question | Brute Better Optimal”

This video from The Stride Bus A to Z DSA course explains the Longest Consecutive Sequence problem, a common interview question. It presents three approaches to solve the problem: Brute Force, Better (Sorting-based), and Optimal (Hash Set-based). The instructor explains the problem, walks through the logic and code for each solution, and analyzes their time and space complexities.

Problem Statement

Given an array of integers, find the length of the longest consecutive elements sequence.
The sequence must consist of consecutive numbers (e.g., 1, 2, 3, 4).
The order of elements in the array can be rearranged or ignored for the purpose of finding the sequence.

Approaches

1. Brute Force Solution

Idea: For each element x in the array, check if x+1, x+2, … exist in the array by linear search.
Steps:
- Initialize longest to 1 (minimum sequence length).
- For each element in the array:
  - Set count to 1.
  - While x+1 exists in the array, increment count and x.
- Update longest if count is greater.
Code Outline: cpp longest = 1; for (int i = 0; i < n; i++) { int x = arr[i]; int count = 1; while (linear_search(arr, x+1)) { x++; count++; } longest = max(longest, count); }
Time Complexity: O(n²) due to nested loops and linear searches.
Space Complexity: O(1).

2. Better Solution (Sorting-based)

Idea: Sort the array so that consecutive elements appear next to each other, then scan through the sorted array to count consecutive sequences.
Steps:
- Sort the array.
- Initialize:
  - longest = 1
  - count_current = 0
  - last_smaller = INT_MIN (to track last element in current sequence)
- Iterate through the sorted array:
  - If current element equals last_smaller, skip (duplicate).
  - Else if current element is last_smaller + 1, increment count_current.
  - Else (non-consecutive), reset count_current to 1.
  - Update last_smaller to current element.
  - Update longest if count_current is greater.
Code Outline: cpp sort(arr.begin(), arr.end()); longest = 1; count_current = 0; last_smaller = INT_MIN; for (int i = 0; i < n; i++) { if (arr[i] == last_smaller) continue; // skip duplicates else if (arr[i] == last_smaller + 1) count_current++; else count_current = 1; last_smaller = arr[i]; longest = max(longest, count_current); }
Time Complexity: O(n log n) due to sorting.
Space Complexity: O(1) or O(n) depending on sorting implementation.
Note: The interviewer may ask to avoid modifying the input array or reduce time complexity further.

3. Optimal Solution (Hash Set-based)

Idea: Use a hash set to achieve O(n) time by:
- Inserting all elements into a set for O(1) average lookup.
- For each element, only start counting a sequence if the element is the start of a sequence (i.e., x-1 is not in the set).
- Then count how long the consecutive sequence extends by checking x+1, x+2, etc.
Steps:
- Insert all elements into an unordered set (C++) or hash set (Java).
- Initialize longest = 0.
- For each element x in the set:
  - If x-1 is not in the set (meaning x could be start of sequence):
    - Initialize count = 1.
    - While x+count is in the set, increment count.
    - Update longest if count is greater.
Code Outline: cpp unordered_set<int> s(arr.begin(), arr.end()); int longest = 0; for (int x : s) { if (s.find(x - 1) == s.end()) { // x is start of sequence int count = 1; while (s.find(x + count) != s.end()) { count++; } longest = max(longest, count); } }
Time Complexity: O(n) average case.
Space Complexity: O(n) for the set.
Note: Worst case complexity can degrade if hash collisions occur, but this is rare.

Additional Notes

The instructor emphasizes the importance of explaining time and space complexities in interviews.
The video provides code snippets in C++, Java, and Python (links in description).
The instructor encourages practicing the problem using the provided resources.
The problem link and article are shared for further study.
The video concludes with motivational remarks and social media links.

Speakers / Sources

Primary Speaker: Instructor from The Stride Bus A to Z DSA course (name not mentioned).
No other speakers or external sources featured explicitly.