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
is the equation of diameter.
The point (R, S) lies above the line if
- A point (R, S) lies above the line formed by joining points (X, Y) and (P, Q) if
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
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 codeint 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 approachimport 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 approachdef 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# InputX = 0Y = 0R = 5P = 5Q = 0arr = [[2, 3], [5, 6], [-1, 4], [5, 5]]print(getPointsIns(X, Y, R, P, Q, arr)) |
C#
// C# program for above approachusing 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 approachfunction 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 codelet 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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