README_EN.md

comments	difficulty	edit_url	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/0400-0499/0447.Number%20of%20Boomerangs/README_EN.md	Array Hash Table Math

447. Number of Boomerangs

中文文档

Description

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.length
• 1 <= n <= 500
• points[i].length == 2
• -104 <= xi, yi <= 104
• All the points are unique.

Solutions

Solution 1: Enumeration + Counting

We can enumerate each point in points as the boomerang's point $i$, and then use a hash table $cnt$ to record the number of times the distance from other points to $i$ appears.

If there are $x$ points with equal distance to $i$, then we can arbitrarily select two of them as the boomerang's $j$ and $k$. The number of schemes is $A_x^2 = x \times (x - 1)$. Therefore, for each value $x$ in the hash table, we calculate and accumulate $A_x^2$, which gives us the total number of boomerangs that meet the problem's requirements.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$, where $n$ is the length of the array points.

Python3

class Solution:
    def numberOfBoomerangs(self, points: List[List[int]]) -> int:
        ans = 0
        for p1 in points:
            cnt = Counter()
            for p2 in points:
                d = dist(p1, p2)
                ans += cnt[d]
                cnt[d] += 1
        return ans << 1

Java

class Solution {
    public int numberOfBoomerangs(int[][] points) {
        int ans = 0;
        for (int[] p1 : points) {
            Map<Integer, Integer> cnt = new HashMap<>();
            for (int[] p2 : points) {
                int d = (p1[0] - p2[0]) * (p1[0] - p2[0]) + (p1[1] - p2[1]) * (p1[1] - p2[1]);
                ans += cnt.getOrDefault(d, 0);
                cnt.merge(d, 1, Integer::sum);
            }
        }
        return ans << 1;
    }
}

C++

class Solution {
public:
    int numberOfBoomerangs(vector<vector<int>>& points) {
        int ans = 0;
        for (auto& p1 : points) {
            unordered_map<int, int> cnt;
            for (auto& p2 : points) {
                int d = (p1[0] - p2[0]) * (p1[0] - p2[0]) + (p1[1] - p2[1]) * (p1[1] - p2[1]);
                ans += cnt[d];
                cnt[d]++;
            }
        }
        return ans << 1;
    }
};

Go

func numberOfBoomerangs(points [][]int) (ans int) {
	for _, p1 := range points {
		cnt := map[int]int{}
		for _, p2 := range points {
			d := (p1[0]-p2[0])*(p1[0]-p2[0]) + (p1[1]-p2[1])*(p1[1]-p2[1])
			ans += cnt[d]
			cnt[d]++
		}
	}
	ans <<= 1
	return
}

TypeScript

function numberOfBoomerangs(points: number[][]): number {
    let ans = 0;
    for (const [x1, y1] of points) {
        const cnt: Map<number, number> = new Map();
        for (const [x2, y2] of points) {
            const d = (x1 - x2) ** 2 + (y1 - y2) ** 2;
            ans += cnt.get(d) || 0;
            cnt.set(d, (cnt.get(d) || 0) + 1);
        }
    }
    return ans << 1;
}

Solution 2

Python3

class Solution:
    def numberOfBoomerangs(self, points: List[List[int]]) -> int:
        ans = 0
        for p1 in points:
            cnt = Counter()
            for p2 in points:
                d = dist(p1, p2)
                cnt[d] += 1
            ans += sum(x * (x - 1) for x in cnt.values())
        return ans

Java

class Solution {
    public int numberOfBoomerangs(int[][] points) {
        int ans = 0;
        for (int[] p1 : points) {
            Map<Integer, Integer> cnt = new HashMap<>();
            for (int[] p2 : points) {
                int d = (p1[0] - p2[0]) * (p1[0] - p2[0]) + (p1[1] - p2[1]) * (p1[1] - p2[1]);
                cnt.merge(d, 1, Integer::sum);
            }
            for (int x : cnt.values()) {
                ans += x * (x - 1);
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int numberOfBoomerangs(vector<vector<int>>& points) {
        int ans = 0;
        for (auto& p1 : points) {
            unordered_map<int, int> cnt;
            for (auto& p2 : points) {
                int d = (p1[0] - p2[0]) * (p1[0] - p2[0]) + (p1[1] - p2[1]) * (p1[1] - p2[1]);
                cnt[d]++;
            }
            for (auto& [_, x] : cnt) {
                ans += x * (x - 1);
            }
        }
        return ans;
    }
};

Go

func numberOfBoomerangs(points [][]int) (ans int) {
	for _, p1 := range points {
		cnt := map[int]int{}
		for _, p2 := range points {
			d := (p1[0]-p2[0])*(p1[0]-p2[0]) + (p1[1]-p2[1])*(p1[1]-p2[1])
			cnt[d]++
		}
		for _, x := range cnt {
			ans += x * (x - 1)
		}
	}
	return
}

TypeScript

function numberOfBoomerangs(points: number[][]): number {
    let ans = 0;
    for (const [x1, y1] of points) {
        const cnt: Map<number, number> = new Map();
        for (const [x2, y2] of points) {
            const d = (x1 - x2) ** 2 + (y1 - y2) ** 2;
            cnt.set(d, (cnt.get(d) || 0) + 1);
        }
        for (const [_, x] of cnt) {
            ans += x * (x - 1);
        }
    }
    return ans;
}