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Count points from an array that lies inside a semi-circle

  • Last Updated : 28 May, 2021

Given two pairs (X, Y), (P, Q) and R the coordinate of the center of semi-circle, coordinate of the intersection of semicircle and diameter of the semicircle and, the radius of the semicircle, and an array arr[] of dimension N*2 consisting of the coordinates of few points, the task is to find the number of points from the array that lies inside or on the 
semicircle
Note: The semicircle above the diameter is considered.

Examples:

Input: X = 0, Y = 0, R = 5, P = 5, Q = 0, arr[][] = { {2, 3}, {5, 6}, {-1, 4}, {5, 5} }
Output: 2
Explanation: The points {2, 3} and {-1, 4} are inside the semi-circle.

Input: X = 2, Y = 3, R = 10, P = 12, Q = 3, arr[][] = { {-7, -5}, {0, 6}, {11, 4} }
Output: 2

Approach: The given problem can be solved based on the following observations: 

  • The points that lies on or inside the semicircle must be above or on the diameter of semicircle and the distance between center and that point should be ≤ R.
  • Suppose a\times x + b\times y + c           is the equation of diameter. 
    The point (R, S) lies above the line if 
    a\times R + b\times S + C>=0
     
  • A point (R, S) lies above the line formed by joining points (X, Y) and (P, Q) if(S - Q)\times(X-P) - (R-P)\times (Y-Q) >= 0

Follow the steps below to solve the problem:

  • Find the equation of line the diameter of the semi-circle from the points (X, Y) and (P, Q).
  • Initialize a variable, say ans, to store the count of required points.
  • Traverse the array arr[] and perform the following operations:
    • Calculate the distance between the points (X, Y) and (P, Q) and store it in a variable, say d.
    • Put arr[i][0] and arr[i][1] in the place of R and S respectively, in the formula 
      (S - Q)\times(X-P) - (R-P)\times (Y-Q)
      and store the result in a variable, say f.
    • Increment the count of ans by 1 if R ≤ d and f ≥ 0.
  • After completing the above steps, print the value stored in ans.

Below is the implementation of the above approach:

C++




// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
 
int getPointsIns(int x1, int y1, int radius, int x2,
                 int y2, vector<pair<int, int>> points)
{
    int ans = 0;
     
    // Traverse the array
    for(int i = 0; i < points.size(); i++)
    {
         
        // Stores if a point lies
        // above the diameter or not
        bool condOne = false, condTwo = false;
        if ((points[i].second - y2) *
              (x2 - x1) - (y2 - y1) *
             (points[i].first - x2) >= 0)
        {
            condOne = true;
        }
 
        // Stores if the R is less than or
        // equal to the distance between
        // center and point
        if (radius >= (int)sqrt(pow((y1 - points[i].second), 2) +
                                 pow(x1 - points[i].first, 2)))
        {
            condTwo = true;
        }
        if (condOne && condTwo)
        {
            ans += 1;
        }
    }
    return ans;
}
 
// Driver code
int main()
{
    int X = 0;
    int Y = 0;
    int R = 5;
    int P = 5;
    int Q = 0;
 
    vector<pair<int, int>> arr = { make_pair(2, 3),
                                   make_pair(5, 6),
                                   make_pair(-1, 4),
                                   make_pair(5, 5) };
 
    cout << getPointsIns(X, Y, R, P, Q, arr);
    return 0;
}
 
// This code is contributed by nirajgusain5

Java




// Java program for above approach
import java.io.*;
 
class Gfg {
  public static int getPointsIns(int x1, int y1,int radius,
                                 int x2,int y2, pair points[])
  {
    int ans = 0;
    // Traverse the array
    for (int i = 0; i < points.length; i++)
    {
       
      // Stores if a point lies
      // above the diameter or not
      boolean condOne = false, condTwo = false;
      if ((points[i].b - y2) *
          (x2 - x1)- (y2 - y1) *
          (points[i].a - x2)>= 0)
      {
        condOne = true;
      }
 
      // Stores if the R is less than or
      // equal to the distance between
      // center and point
      if (radius >= (int)Math.sqrt(Math.pow((y1 - points[i].b), 2)+
                                   Math.pow(x1 - points[i].a, 2)))
      {
        condTwo = true;
      }
      if (condOne && condTwo)
      {
        ans += 1;
      }
    }
    return ans;
  }
   
  // Driver code
  public static void main(String[] args)
  {
    int X = 0;
    int Y = 0;
    int R = 5;
    int P = 5;
    int Q = 0;
 
    pair arr[] = {new pair(2, 3), new pair(5, 6), new pair(-1, 4), new pair(5,5)};
 
    System.out.print(getPointsIns(X, Y, R, P, Q, arr));
  }
}
class pair
{
  int a;
  int b;
  pair(int a,int b)
  {   
    this.a = a;
    this.b = b;
  }
}

Python3




# Python implementation of above approach
def getPointsIns(x1, y1, radius, x2, y2, points):
    # Stores the count of ans
    ans = 0
 
    # Traverse the array
    for point in points:
 
        # Stores if a point lies
        # above the diameter or not
        condOne = (point[1] - y2) * (x2 - x1) \
                  - (y2 - y1) * (point[0] - x2) >= 0
 
        # Stores if the R is less than or
        # equal to the distance between
        # center and point
 
        condTwo = radius >= ((y1 - point[1]) ** 2 \
                  + (x1 - point[0]) ** 2) ** (0.5)
 
        if condOne and condTwo:
            ans += 1
 
    return ans
 
 
# Driver Code
# Input
X = 0
Y = 0
R = 5
P = 5
Q = 0
arr = [[2, 3], [5, 6], [-1, 4], [5, 5]]
 
print(getPointsIns(X, Y, R, P, Q, arr))

C#




// C# program for above approach
using System;
 
class Gfg
{
  public static int getPointsIns(int x1, int y1,
                                 int radius, int x2,
                                 int y2, pair[] points)
  {
    int ans = 0;
     
    // Traverse the array
    for (int i = 0; i < points.Length; i++) {
 
      // Stores if a point lies
      // above the diameter or not
      bool condOne = false, condTwo = false;
      if ((points[i].b - y2) * (x2 - x1)
          - (y2 - y1) * (points[i].a - x2)
          >= 0) {
        condOne = true;
      }
 
      // Stores if the R is less than or
      // equal to the distance between
      // center and point
      if (radius >= (int)Math.Sqrt(
        Math.Pow((y1 - points[i].b), 2)
        + Math.Pow(x1 - points[i].a, 2))) {
        condTwo = true;
      }
      if (condOne && condTwo) {
        ans += 1;
      }
    }
    return ans;
  }
 
  // Driver code
  public static void Main(string[] args)
  {
    int X = 0;
    int Y = 0;
    int R = 5;
    int P = 5;
    int Q = 0;
 
    pair[] arr = { new pair(2, 3), new pair(5, 6),
                  new pair(-1, 4), new pair(5, 5) };
 
    Console.Write(getPointsIns(X, Y, R, P, Q, arr));
  }
}
public class pair {
  public int a;
  public int b;
  public pair(int a, int b)
  {
    this.a = a;
    this.b = b;
  }
}
 
// This code is contributed by code_hunt.

Javascript




<script>
// Javascript program for above approach
 
function getPointsIns(x1,y1,radius,x2,y2,points)
{
    let ans = 0;
    // Traverse the array
    for (let i = 0; i < points.length; i++)
    {
        
      // Stores if a point lies
      // above the diameter or not
      let condOne = false, condTwo = false;
      if ((points[i][1] - y2) *
          (x2 - x1)- (y2 - y1) *
          (points[i][0] - x2)>= 0)
      {
        condOne = true;
      }
  
      // Stores if the R is less than or
      // equal to the distance between
      // center and point
      if (radius >= Math.sqrt(Math.pow((y1 - points[i][1]), 2)+
                                   Math.pow(x1 - points[i][0], 2)))
      {
        condTwo = true;
      }
       
      if (condOne && condTwo)
      {
        ans += 1;
      }
       
    }
    return ans;
}
 
// Driver code
let X = 0;
    let Y = 0;
    let R = 5;
    let P = 5;
    let Q = 0;
  
    let arr = [[2, 3], [5, 6], [-1, 4], [5, 5]];
  
    document.write(getPointsIns(X, Y, R, P, Q, arr));
   
 
 
 
// This code is contributed by avanitrachhadiya2155
</script>
Output: 
2

 

Time Complexity: O(N)
Auxiliary Space: O(1)


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