How do you solve Find Books with Polarized Opinions on LeetCode?

Find Books with Polarized Opinions (LeetCode #3642) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Books need both high and low ratings to be considered polarized.

What is the time complexity of Find Books with Polarized Opinions?

The optimal solution for Find Books with Polarized Opinions runs in O(n) time and uses O(n) space. We only traverse the reading_sessions once and use a dictionary to store results, leading to linear time complexity.

What is the brute force solution for Find Books with Polarized Opinions?

The brute force approach for Find Books with Polarized Opinions is Brute Force, with O(n²) time complexity and O(1) space complexity. Check each book's ratings individually to see if it has both high and low ratings. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Find Books with Polarized Opinions?

Clarify the problem requirements. Think about edge cases, like books with no ratings. Discuss your thought process aloud.

What are common mistakes when solving Find Books with Polarized Opinions?

Not considering all ratings for a book. Overlooking the need for both high and low ratings.

#3642

Find Books with Polarized Opinions

Medium

Hash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Use a single pass to track ratings for each book, allowing us to determine polarized opinions efficiently.

⚙️

Algorithm

3 steps

1Step 1: Create a dictionary to track the highest and lowest ratings for each book.
2Step 2: Iterate through reading_sessions and update the ratings in the dictionary.
3Step 3: Filter books that have at least one rating >= 4 and one rating <= 2.

solution.py1 lines

1SELECT book_id FROM (SELECT book_id, MAX(session_rating) AS max_rating, MIN(session_rating) AS min_rating FROM reading_sessions GROUP BY book_id) AS ratings WHERE max_rating >= 4 AND min_rating <= 2;

ℹ

Complexity note: We only traverse the reading_sessions once and use a dictionary to store results, leading to linear time complexity.

1Books need both high and low ratings to be considered polarized.
2Using aggregation functions helps simplify the problem.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.