use std::collections::HashMap; struct Solution; /* * 'Line' is represented as y = m * x + c * m & c are rational numbers because of integer coordinates, * but in general floating point numbers can be insufficient to * guarantee uniqueness as they are still finite precision. * so for most real irl cases we use numerator and denominator * to represent m & c. decimal precision!!! * if m -> +- infinity then use x-int to identify line * if m == 0 then use y-int to identify line */ #[derive(Debug, Clone, Copy, PartialEq, Eq, Hash)] enum Line { Inf(i32), //x-int MC((i32, i32), (i32, i32)), //numerator, denominator tuples for m & c Zero(i32) //y-int } impl Solution { pub fn max_points(points: Vec>) -> i32 { if points.len() == 1 { return 1; } //y - y0 = (y1 - y0) / (x1 - x0) * (x - x0) //y = dy / dx x + (dx * y0 - dy * x0) / dx let mut fm: HashMap = HashMap::new(); points.iter().enumerate().flat_map(|(i, pt1)| { points.iter().skip(i + 1).map(move |pt2| { //swap points arounto make sure dy >= 0 since we are storing //fraction numerator & denominator separately. let (x0, y0, x1, y1) = if pt2[1] > pt1[1] || (pt2[1] == pt1[1] && pt2[0] > pt2[1]) { (pt1[0], pt1[1], pt2[0], pt2[1]) } else { (pt2[0], pt2[1], pt1[0], pt1[1]) }; let mut dy = y1 - y0; let mut dx = x1 - x0; let g = Self::gcd(dy.abs(), dx.abs()); if g != 0 { dy /= g; dx /= g; } if dx != 0 && dy != 0 { /* * slope is non-zero and finite so consistent line representation * can be achieved using numerator & denominator representation * because we have already preprocessed dy & dx for sign consistency * and removed gcd from dy & dx */ Line::MC((dy, dx), (dx * y0 - dy * x0, dx)) } else if dx == 0 { //x-int is enough to identify common lines Line::Inf(x0) } else { //y-int is enough to identify common lines Line::Zero(y0) } }) }).for_each(|k| { *fm.entry(k).or_insert(0) += 1; }); let n = fm.values().max().cloned().unwrap(); /* * if 'm' points lie on a line & all points have been processed pairwise * then n = m * (m - 1) / 2 pairs of points would have matched to same * line m^2 - m - 2 * n = 0 => m = (1 + sqrt(8 * n + 1)) / 2 */ (1 + (8f64 * n as f64 + 1f64).sqrt() as i32) / 2 } fn gcd(a: i32, b: i32) -> i32 { if a == 0 || b == 0 { return 0; } let (mut a, mut b) = (a, b); while b > 0 { let tmp = a % b; a = b; b = tmp; } a } } fn main() { let points1 = vec![vec![1,1],vec![2,2],vec![3,3]]; let points2 = vec![vec![1,1],vec![3,2],vec![5,3],vec![4,1],vec![2,3],vec![1,4]]; println!("{}", Solution::max_points(points1)); //expect: 3 println!("{}", Solution::max_points(points2)); //expect: 3 }