Minimum number of points to be removed to get remaining points on one side of axis
We are given n points in a Cartesian plane. Our task is to find the minimum number of points that should be removed in order to get the remaining points on one side of any axis.
Examples :
Input : 4
1 1
2 2
-1 -1
-2 2
Output : 1
Explanation :
If we remove (-1, -1) then all the remaining
points are above x-axis. Thus the answer is 1.
Input : 3
1 10
2 3
4 11
Output : 0
Explanation :
All points are already above X-axis. Hence the
answer is 0.
Approach :
This problem is a simple example of constructive brute force algorithm on Geometry. The solution can be approached simply by finding the number of points on all sides of the X-axis and Y-axis. The minimum of this will be the answer.
C++
// CPP program to find minimum points to be moved // so that all points are on same side. #include <bits/stdc++.h> using namespace std; typedef long long ll; // Structure to store the coordinates of a point. struct Point { int x, y; }; // Function to find the minimum number of points int findmin(Point p[], int n) { int a = 0, b = 0, c = 0, d = 0; for (int i = 0; i < n; i++) { // Number of points on the left of Y-axis. if (p[i].x <= 0) a++; // Number of points on the right of Y-axis. else if (p[i].x >= 0) b++; // Number of points above X-axis. if (p[i].y >= 0) c++; // Number of points below X-axis. else if (p[i].y <= 0) d++; } return min({a, b, c, d}); } // Driver Function int main() { Point p[] = { {1, 1}, {2, 2}, {-1, -1}, {-2, 2} }; int n = sizeof(p)/sizeof(p[0]); cout << findmin(p, n); return 0; } |
chevron_right
filter_none
Java
// Java program to find minimum points to be moved // so that all points are on same side. import java.util.*; class GFG { // Structure to store the coordinates of a point. static class Point { int x, y; public Point(int x, int y) { this.x = x; this.y = y; } }; // Function to find the minimum number of points static int findmin(Point p[], int n) { int a = 0, b = 0, c = 0, d = 0; for (int i = 0; i < n; i++) { // Number of points on the left of Y-axis. if (p[i].x <= 0) a++; // Number of points on the right of Y-axis. else if (p[i].x >= 0) b++; // Number of points above X-axis. if (p[i].y >= 0) c++; // Number of points below X-axis. else if (p[i].y <= 0) d++; } return Math.min(Math.min(a, b), Math.min(c, d)); } // Driver Code public static void main(String[] args) { Point p[] = {new Point(1, 1), new Point(2, 2), new Point(-1, -1), new Point(-2, 2)}; int n = p.length; System.out.println(findmin(p, n)); } } // This code is contributed by PrinciRaj1992 |
chevron_right
filter_none
Python3
# Python3 program to find minimum points to be # moved so that all points are on same side. # Function to find the minimum number # of points def findmin(p, n): a, b, c, d = 0, 0, 0, 0 for i in range(n): # Number of points on the left # of Y-axis. if (p[i][0] <= 0): a += 1 # Number of points on the right # of Y-axis. elif (p[i][0] >= 0): b += 1 # Number of points above X-axis. if (p[i][1] >= 0): c += 1 # Number of points below X-axis. elif (p[i][1] <= 0): d += 1 return min([a, b, c, d]) # Driver Code p = [ [1, 1], [2, 2], [-1, -1], [-2, 2] ] n = len(p) print(findmin(p, n)) # This code is contributed by Mohit Kumar |
chevron_right
filter_none
C#
// C# rogram to find minimum points to be moved // so that all points are on same side. using System; class GFG { // Structure to store the coordinates of a point. public class Point { public int x, y; public Point(int x, int y) { this.x = x; this.y = y; } }; // Function to find the minimum number of points static int findmin(Point []p, int n) { int a = 0, b = 0, c = 0, d = 0; for (int i = 0; i < n; i++) { // Number of points on the left of Y-axis. if (p[i].x <= 0) a++; // Number of points on the right of Y-axis. else if (p[i].x >= 0) b++; // Number of points above X-axis. if (p[i].y >= 0) c++; // Number of points below X-axis. else if (p[i].y <= 0) d++; } return Math.Min(Math.Min(a, b), Math.Min(c, d)); } // Driver Code public static void Main(String[] args) { Point []p = {new Point(1, 1), new Point(2, 2), new Point(-1, -1), new Point(-2, 2)}; int n = p.Length; Console.WriteLine(findmin(p, n)); } } // This code is contributed by Princi Singh |
chevron_right
filter_none
Output :
1
GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details
Recommended Posts:
- Program to check if the points are parallel to X axis or Y axis
- Number of lines from given N points not parallel to X or Y axis
- Count of Squares that are parallel to the coordinate axis from the given set of N points
- Number of Integral Points between Two Points
- Check whether two points (x1, y1) and (x2, y2) lie on same side of a given line or not
- Count of obtuse angles in a circle with 'k' equidistant points between 2 given points
- Ways to choose three points with distance between the most distant points <= L
- Minimum lines to cover all points
- Minimum area of a Polygon with three points given
- Find integral points with minimum distance from given set of integers using BFS
- Number of quadrilaterals possible from the given points
- Number of triangles that can be formed with given N points
- Maximum number of segments that can contain the given points
- Number of triangles in a plane if no more than two points are collinear
- Total number of triplets (A, B, C) in which the points B and C are Equidistant to A
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.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
