Queries on count of points lie inside a circle

3.2

Given n coordinate (x, y) of points on 2D plane and Q queries. Each query contains an integer r, the task is to count the number of points lying inside or on the circumference of the circle having radius r and centered at the origin.

Examples:

Input : n = 5
Coordinates: 
1 1
2 2
3 3
-1 -1
4 4

Query 1: 3
Query 2: 32

Output :
3
5
For first query radius = 3, number of points lie
inside or on the circumference are (1, 1), (-1, -1),
(2, 2). There are only 3 points lie inside or on 
the circumference of the circle.
For second query radius = 32, all five points are
inside the circle. 

The equation for the circle centered at origin (0, 0) with radius r, x2 + y2 = r2. And condition for a point at (x1, y1) to lie inside or on the circumference, x12 + y12 <= r2.

A Naive approach can be for each query, traverse through all points and check the condition. This take O(n*Q) time complexity.

An Efficient approach is to precompute x2 + y2 for each point coordinate and store them in an array p[]. Now, sort the array p[]. Then apply binary search on the array to find last index with condition p[i] <= r2 for each query.

Below is the implementation of this approach:

C++

// C++ program to find number of points lie inside or
// on the cirumference of circle for Q queries.
#include<bits/stdc++.h>
using namespace std;

// Computing the x^2 + y^2 for each given points
// and sorting them.
void preprocess(int p[], int x[], int y[], int n)
{
    for (int i = 0; i < n; i++)
        p[i] = x[i]*x[i] + y[i]*y[i];

    sort(p, p + n);
}

// Return count of points lie inside or on circumference
// of circle using binary search on p[0..n-1]
int query(int p[], int n, int rad)
{
    int start = 0, end = n-1;
    while ((end - start)>1)
    {
        int mid = (start+end)/2;
        double tp = sqrt(p[mid]);

        if (tp > (rad*1.0))
            end = mid - 1;
        else
            start = mid;
    }

    double tp1 = sqrt(p[start]), tp2 = sqrt(p[end]);

    if (tp1 > (rad*1.0))
        return 0;
    else if (tp2 <= (rad*1.0))
        return end+1;
    else
        return start+1;
}

// Driven Program
int main()
{
    int x[] = { 1, 2, 3, -1, 4 };
    int y[] = { 1, 2, 3, -1, 4 };
    int n = sizeof(x)/sizeof(x[0]);

    // Compute distances of all points and keep
    // the distances sorted so that query can
    // work in O(logn) using Binary Search.
    int p[n];
    preprocess(p, x, y, n);

    // Print number of points in a circle of radius 3.
    cout << query(p, n, 3) << endl;

    // Print number of points in a circle of radius 32.
    cout << query(p, n, 32) << endl;
    return 0;
}

Java

// JAVA Code for Queries on count of
// points lie inside a circle
import java.util.*;

class GFG {
     
    // Computing the x^2 + y^2 for each given points
    // and sorting them.
    public static void preprocess(int p[], int x[], 
                                   int y[], int n)
    {
        for (int i = 0; i < n; i ++)
            p[i] = x[i] * x[i] + y[i] * y[i];
     
        Arrays.sort(p);
    }
     
    // Return count of points lie inside or on 
    // circumference of circle using binary  
    // search on p[0..n-1]
    public static int query(int p[], int n, int rad)
    {
        int start = 0, end = n-1;
        while ((end - start) > 1)
        {
            int mid = (start + end) / 2;
            double tp = Math.sqrt(p[mid]);
     
            if (tp > (rad * 1.0))
                end = mid - 1;
            else
                start = mid;
        }
     
        double tp1 = Math.sqrt(p[start]);       
       double tp2 = Math.sqrt(p[end]);
     
        if (tp1 > (rad * 1.0))
            return 0;
        else if (tp2 <= (rad * 1.0))
            return end + 1;
        else
            return start + 1;
    }
    
    /* Driver program to test above function */
    public static void main(String[] args) 
    {
         int x[] = { 1, 2, 3, -1, 4 };
        int y[] = { 1, 2, 3, -1, 4 };
         int n = x.length;
         
         // Compute distances of all points and keep
         // the distances sorted so that query can
         // work in O(logn) using Binary Search.
         int p[]=new int[n];
         preprocess(p, x, y, n);
         
         // Print number of points in a circle of
         // radius 3.
         System.out.println(query(p, n, 3));
         
         // Print number of points in a circle of 
         // radius 32.
         System.out.println(query(p, n, 32));
    }
  }
// This code is contributed by Arnav Kr. Mandal.


Output:

3
5

Time Complexity: O(n log n) for preprocessing and O(Q Log n) for Q queries.

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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