Number of Boomerangs

Given n points in the plane that are all pairwise distinct, a "boomerang" is a tuple of points (i, j, k) such that the distance between iand j equals the distance between i and k (the order of the tuple matters).

Find the number of boomerangs. You may assume that n will be at most 500 and coordinates of points are all in the range [-10000, 10000](inclusive).

Example:

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

Output:
2

Explanation:
The two boomerangs are [[1,0],[0,0],[2,0]] and [[1,0],[2,0],[0,0]]

Analyse:

1. 首先想找两个点所连直线的垂直平分线，再判断其他的点是否在该垂直平分线上。O(n^3), 超时。

class Solution {
public:
    int numberOfBoomerangs(vector<pair<int, int>>& points) {
        // check all pairs accordingly
        int count = 0;
        for (int i = 0; i < points.size(); i++) {
            for (int j = i + 1; j < points.size(); j++) {
                for (int k = 0; k < points.size(); k++) {
                    if (k != i && k != j) {
                        if (isBoomerang(points[i], points[j], points[k]))
                            count++;
                    }
                }
            }
        }
        return count * 2;
    }
    
    bool isBoomerang(pair<int, int> point1, pair<int, int> point2, pair<int, int> checkMe) {
        int x1 = point1.first, y1 = point1.second, x2 = point2.first, y2 = point2.second, x = checkMe.first, y = checkMe.second;
        
        if (y1 == y2) {
            return x * 2 == x1 + x2;
        } else
            return (x1 - x2) * 1.0 / (y2 - y1) * (2 * x - x1 - x2) + y1 + y2 == 2 * y;
    }
};

2. 用hash table maintain point i到point j的距离，map该距离到count，for all i，j。对于每个i，如果该hash table的value >= 2，则取改value的全排列并更新result。O(n^2)

class Solution {
public:
    int numberOfBoomerangs(vector<pair<int, int>>& points) {
        int result = 0;
        for (int i = 0; i < points.size(); i++) {
            unordered_map<int, int> um;
            for (int j = 0; j < points.size(); j++) {
                if (i != j) {
                    int x = points[i].first - points[j].first;
                    int y = points[i].second - points[j].second;
                    um[x * x + y * y]++;
                }
            }
            for (auto count : um) {
                int n = count.second;
                if (n >= 2)
                    result += n * (n - 1); // permutation
            }
        }
        return result;
    }
};