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++ 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 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 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.
# 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 Code
if __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 circle
using 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(n) it is using extra space for array p
