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I knew the trick was slope comparison but fumbled the edge cases for a while.
Start by clarifying edge cases (e.g., duplicate points, fewer than 3 points) and then propose an O(N^2) solution: for each point, compute slopes to all other points and find the most frequent slope. Use a hash map to count slopes, handling vertical lines and duplicates separately.
Pro tip: Mention that using a reduced fraction (dx/gcd, dy/gcd) as the key avoids floating-point precision issues and is more robust than using doubles. Also, discuss how to handle duplicate points by counting them separately and adding to the max count.
Ask about input size, duplicate points, and whether points are integers. Discuss handling of N < 3, all points collinear, and vertical lines.
Mention that brute force O(N^3) is too slow, and propose the O(N^2) slope-based approach using a hash map.
For each point i, compute slopes to all j > i, using reduced fractions (dx/gcd, dy/gcd) as keys. Track the maximum count of points with the same slope, including duplicates.
Explain how to handle vertical lines (dx=0), horizontal lines (dy=0), and duplicate points (dx=0, dy=0) by counting them separately and adding to the max.
State time complexity O(N^2) and space O(N). Walk through a small example to verify correctness.
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