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447. Number of Boomerangs

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.

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  }
  

• 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;
      }
  };
  

• 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
  
  

• 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
  }
  

• 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;
  }
  
  

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