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Queries on count of points lie inside a circle
• Difficulty Level : Medium
• Last Updated : 14 Apr, 2021

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 ``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);` `    ``// 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.`

## Python 3

 `# Python 3 program to find number of``# points lie inside or on the cirumference``# 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``    ``elif` `(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#

 `// 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.`

## Javascript

 ``

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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