Table of Contents
Number of Boomerangs – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
You are given n points in the plane that are all distinct, where points[i] = [xi, yi]. A boomerang is a tuple of points (i, j, k) such that the distance between i and j equals the distance between i and k (the order of the tuple matters).
Return the number of boomerangs.
Example 1:
Input: points = [[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]].
Example 2:
Input: points = [[1,1],[2,2],[3,3]] Output: 2
Example 3:
Input: points = [[1,1]] Output: 0
Constraints:
n == points.length1 <= n <= 500points[i].length == 2-104 <= xi, yi <= 104- All the points are unique.
C++ Number of Boomerangs LeetCode Solution
int numberOfBoomerangs(vector<pair<int, int>>& points) {
int res = 0;
// iterate over all the points
for (int i = 0; i < points.size(); ++i) {
unordered_map<long, int> group(points.size());
// iterate over all points other than points[i]
for (int j = 0; j < points.size(); ++j) {
if (j == i) continue;
int dy = points[i].second - points[j].second;
int dx = points[i].first - points[j].first;
// compute squared euclidean distance from points[i]
int key = dy * dy;
key += dx * dx;
// accumulate # of such "j"s that are "key" distance from "i"
++group[key];
}
for (auto& p : group) {
if (p.second > 1) {
/*
* for all the groups of points,
* number of ways to select 2 from n =
* nP2 = n!/(n - 2)! = n * (n - 1)
*/
res += p.second * (p.second - 1);
}
}
}
return res;
}
Java Number of Boomerangs LeetCode Solution
public int numberOfBoomerangs(int[][] points) {
int res = 0;
Map<Integer, Integer> map = new HashMap<>();
for(int i=0; i<points.length; i++) {
for(int j=0; j<points.length; j++) {
if(i == j)
continue;
int d = getDistance(points[i], points[j]);
map.put(d, map.getOrDefault(d, 0) + 1);
}
for(int val : map.values()) {
res += val * (val-1);
}
map.clear();
}
return res;
}
private int getDistance(int[] a, int[] b) {
int dx = a[0] - b[0];
int dy = a[1] - b[1];
return dx*dx + dy*dy;
}
Python 3 Number of Boomerangs LeetCode Solution
res = 0
for p in points:
cmap = {}
for q in points:
f = p[0]-q[0]
s = p[1]-q[1]
cmap[f*f + s*s] = 1 + cmap.get(f*f + s*s, 0)
for k in cmap:
res += cmap[k] * (cmap[k] -1)
return res
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