Queries on count of points lie inside a circle

22 Jun, 2022
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++
Java
Python 3
C#
Javascript
// C++ program to find number of points lie inside or// on the circumference 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 Programint 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 Code for Queries on count of// points lie inside a circleimport 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.
# Python 3 program to find number of# points lie inside or on the circumference# of circle for Q queries.import math # Computing the x^2 + y^2 for each# given points and sorting them.def preprocess(p, x, y, n):    for i in range(n):        p[i] = x[i] * x[i] + y[i] * y[i]     p.sort() # Return count of points lie inside# or on circumference of circle using# binary search on p[0..n-1]def query(p, n, rad):     start = 0    end = n - 1    while ((end - start) > 1):        mid = (start + end) // 2        tp = math.sqrt(p[mid])         if (tp > (rad * 1.0)):            end = mid - 1        else:            start = mid     tp1 = math.sqrt(p[start])    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 Codeif __name__ == "__main__":         x = [ 1, 2, 3, -1, 4 ]    y = [ 1, 2, 3, -1, 4 ]    n = len(x)     # Compute distances of all points and keep    # the distances sorted so that query can    # work in O(logn) using Binary Search.    p = [0] * n    preprocess(p, x, y, n)     # Print number of points in a    # circle of radius 3.    print(query(p, n, 3))     # Print number of points in a    # circle of radius 32.    print(query(p, n, 32)) # This code is contributed by ita_c
// C# Code for Queries on count of// points lie inside a circleusing System; 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];         Array.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()    {        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.        Console.WriteLine(query(p, n, 3));         // Print number of points in a circle of        // radius 32.        Console.WriteLine(query(p, n, 32));    }} // This code is contributed by vt_m.
<script> // Javascript Code for Queries on count of// points lie inside a circle         // Computing the x^2 + y^2 for each given points    // and sorting them.    function preprocess(p,x,y,n)    {        for (let i = 0; i < n; i++)            p[i] = x[i] * x[i] + y[i] * y[i];                 p.sort(function(a,b){return a-b;});             }         // Return count of points lie inside or on    // circumference of circle using binary    // search on p[0..n-1]    function query(p,n,rad)    {        let start = 0, end = n - 1;        while ((end - start) > 1) {            let mid = Math.floor((start + end) / 2);            let tp = Math.sqrt(p[mid]);               if (tp > (rad * 1.0))                end = mid - 1;            else                start = mid;        }           let tp1 = Math.sqrt(p[start]);        let 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 */    let x=[1, 2, 3, -1, 4 ];    let y=[1, 2, 3, -1, 4];    let n = x.length;         // Compute distances of all points and keep    // the distances sorted so that query can    // work in O(logn) using Binary Search.    let p=new Array(n);    for(let i=0;i<n;i++)    {        p[i]=0;    }    preprocess(p, x, y, n);     // Print number of points in a circle of    // radius 3.    document.write(query(p, n, 3)+"<br>");     // Print number of points in a circle of    // radius 32.    document.write(query(p, n, 32)+"<br>");         // This code is contributed by rag2127     </script>
Output: 
3
5
Time Complexity: O(n log n) for preprocessing and O(Q Log n) for Q queries.Auxiliary Space: O(1)
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