Question

1828. Queries on Number of Points Inside a Circle

中文文档

Description

You are given an array points where points[i] = [xi, yi] is the coordinates of the ith point on a 2D plane. Multiple points can have the same coordinates.

You are also given an array queries where queries[j] = [xj, yj, rj] describes a circle centered at (xj, yj) with a radius of rj.

For each query queries[j], compute the number of points inside the jth circle. Points on the border of the circle are considered inside.

Return _an array _answer_, where _answer[j]_ is the answer to the _jth_ query_.

 

Example 1:

Input: points = [[1,3],[3,3],[5,3],[2,2]], queries = [[2,3,1],[4,3,1],[1,1,2]]
Output: [3,2,2]
Explanation: The points and circles are shown above.
queries[0] is the green circle, queries[1] is the red circle, and queries[2] is the blue circle.

Example 2:

Input: points = [[1,1],[2,2],[3,3],[4,4],[5,5]], queries = [[1,2,2],[2,2,2],[4,3,2],[4,3,3]]
Output: [2,3,2,4]
Explanation: The points and circles are shown above.
queries[0] is green, queries[1] is red, queries[2] is blue, and queries[3] is purple.

 

Constraints:

• 1 <= points.length <= 500
• points[i].length == 2
• 0 <= x​​​​​​i, y​​​​​​i <= 500
• 1 <= queries.length <= 500
• queries[j].length == 3
• 0 <= xj, yj <= 500
• 1 <= rj <= 500
• All coordinates are integers.

 

Follow up: Could you find the answer for each query in better complexity than O(n)?

Solutions

Solution 1: Enumeration

Enumerate all the circles $(x, y, r)$. For each circle, calculate the number of points within the circle to get the answer.

The time complexity is $O(m \times n)$, where $m$ and $n$ are the lengths of the arrays queries and points respectively. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

class Solution:
  def countPoints(
    self, points: List[List[int]], queries: List[List[int]]
  ) -> List[int]:
    ans = []
    for x, y, r in queries:
      cnt = 0
      for i, j in points:
        dx, dy = i - x, j - y
        cnt += dx * dx + dy * dy <= r * r
      ans.append(cnt)
    return ans

class Solution {
  public int[] countPoints(int[][] points, int[][] queries) {
    int m = queries.length;
    int[] ans = new int[m];
    for (int k = 0; k < m; ++k) {
      int x = queries[k][0], y = queries[k][1], r = queries[k][2];
      for (var p : points) {
        int i = p[0], j = p[1];
        int dx = i - x, dy = j - y;
        if (dx * dx + dy * dy <= r * r) {
          ++ans[k];
        }
      }
    }
    return ans;
  }
}

class Solution {
public:
  vector<int> countPoints(vector<vector<int>>& points, vector<vector<int>>& queries) {
    vector<int> ans;
    for (auto& q : queries) {
      int x = q[0], y = q[1], r = q[2];
      int cnt = 0;
      for (auto& p : points) {
        int i = p[0], j = p[1];
        int dx = i - x, dy = j - y;
        cnt += dx * dx + dy * dy <= r * r;
      }
      ans.emplace_back(cnt);
    }
    return ans;
  }
};

func countPoints(points [][]int, queries [][]int) (ans []int) {
  for _, q := range queries {
    x, y, r := q[0], q[1], q[2]
    cnt := 0
    for _, p := range points {
      i, j := p[0], p[1]
      dx, dy := i-x, j-y
      if dx*dx+dy*dy <= r*r {
        cnt++
      }
    }
    ans = append(ans, cnt)
  }
  return
}

function countPoints(points: number[][], queries: number[][]): number[] {
  return queries.map(([cx, cy, r]) => {
    let res = 0;
    for (const [px, py] of points) {
      if (Math.sqrt((cx - px) ** 2 + (cy - py) ** 2) <= r) {
        res++;
      }
    }
    return res;
  });
}

impl Solution {
  pub fn count_points(points: Vec<Vec<i32>>, queries: Vec<Vec<i32>>) -> Vec<i32> {
    queries
      .iter()
      .map(|v| {
        let cx = v[0];
        let cy = v[1];
        let r = v[2].pow(2);
        let mut count = 0;
        for p in points.iter() {
          if (p[0] - cx).pow(2) + (p[1] - cy).pow(2) <= r {
            count += 1;
          }
        }
        count
      })
      .collect()
  }
}

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* countPoints(int** points, int pointsSize, int* pointsColSize, int** queries, int queriesSize, int* queriesColSize,
  int* returnSize) {
  int* ans = malloc(sizeof(int) * queriesSize);
  for (int i = 0; i < queriesSize; i++) {
    int cx = queries[i][0];
    int cy = queries[i][1];
    int r = queries[i][2];
    int count = 0;
    for (int j = 0; j < pointsSize; j++) {
      if (sqrt(pow(points[j][0] - cx, 2) + pow(points[j][1] - cy, 2)) <= r) {
        count++;
      }
    }
    ans[i] = count;
  }
  *returnSize = queriesSize;
  return ans;
}