3886 Count Number Of Trapezoids I

Count Number of Trapezoids I

You are given a 2D integer array points, where points[i] = [xi, yi] represents the coordinates of the ith point on the Cartesian plane.

A horizontal trapezoid is a convex quadrilateral with at least one pair of horizontal sides (i.e. parallel to the x-axis). Two lines are parallel if and only if they have the same slope.

Return the number of unique horizontal trapezoids that can be formed by choosing any four distinct points from points.

Since the answer may be very large, return it modulo 109 + 7.

 

Example 1:

Input: points = [[1,0],[2,0],[3,0],[2,2],[3,2]]

Output: 3

Explanation:

There are three distinct ways to pick four points that form a horizontal trapezoid:

Using points [1,0], [2,0], [3,2], and [2,2].
Using points [2,0], [3,0], [3,2], and [2,2].
Using points [1,0], [3,0], [3,2], and [2,2].

Example 2:

Input: points = [[0,0],[1,0],[0,1],[2,1]]

Output: 1

Explanation:

There is only one horizontal trapezoid that can be formed.

 

Constraints:

4 <= points.length <= 105
–108 <= xi, yi <= 108
All points are pairwise distinct.

use std::collections::HashMap;

impl Solution {
    pub fn count_trapezoids(points: Vec<Vec<i32>>) -> i32 {
        // helper for mC2
        fn fact(a: i64) -> i64 {
            (a * (a - 1)) / 2
        }

        let mut hs: HashMap<i32, Vec<(i32, i32)>> = HashMap::new();

        // group points by y
        for k in points {
            hs.entry(k[1]).or_insert(Vec::new()).push((k[0], k[1]));
        }

        let modulo: i64 = 1_000_000_007;

        // compute fact(m) for each group
        let mut f: Vec<i64> = hs.values().map(|v| fact(v.len() as i64)).collect();

        // linear-time sum over all pairs
        let mut total: i64 = 0;
        let mut sum_f: i64 = f.iter().sum();
        for fi in f {
            sum_f -= fi;
            total = (total + fi * sum_f) % modulo;
        }

        total as i32
    }
}