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 a 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. 

Efficient Approach:

1. Create a structure Point to store the coordinates of a point.
2. Create a function findmin which takes an array of Point p and integer n as arguments.
   Declare and initialize integer variables a, b, c, and d to 0.
3. Iterate from i=0 to i=n-1:
                  a. If the x-coordinate of p[i] is less than or equal to 0, increment a by 1.
                  b. Else if the x-coordinate of p[i] is greater than or equal to 0, increment b by 1.
                  c. If the y-coordinate of p[i] is greater than or equal to 0, increment c by 1.
                  d. Else if the y-coordinate of p[i] is less than or equal to 0, increment d by 1.
4. Return the minimum value of a, b, c, and d.
5. In the main function:
                  a. Declare an array of Point p and initialize it with coordinates of some points.
                  b. Calculate the size of the array.
                  c. Call the findmin function with the array and size as arguments and print the returned value.

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;
}

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

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

C#

// C# program 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

Javascript

<script>

// JavaScript program to find minimum points to be moved
// so that all points are on same side.

    // Function to find the minimum number of points
    function findmin(p,n)
    {
        let a = 0, b = 0, c = 0, d = 0;
    for (let i = 0; i < n; i++)
    {
        // Number of points on the left of Y-axis.
        if (p[i][0] <= 0)
            a++;

        // Number of points on the right of Y-axis.
        else if (p[i][0] >= 0)
            b++;

        // Number of points above X-axis.
        if (p[i][1] >= 0)
            c++;

        // Number of points below X-axis.
        else if (p[i][1] <= 0)
            d++;
    }
    return Math.min(Math.min(a, b),
                    Math.min(c, d));
    }

    // Driver Code
    let p = [ [1, 1], [2, 2], [-1, -1], [-2, 2] ]
    let n = p.length;
    document.write(findmin(p, n));

// This code is contributed by unknown2108

</script>

Output:

1

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

Please suggest if someone has a better solution which is more efficient in terms of space and time.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above