Nos dan n puntos en un plano cartesiano. Nuestra tarea es encontrar el número mínimo de puntos que deben eliminarse para obtener los puntos restantes en un lado de cualquier eje.
Ejemplos:
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.
Enfoque:
Este problema es un ejemplo simple de un algoritmo constructivo de fuerza bruta en Geometría. La solución se puede abordar simplemente encontrando el número de puntos en todos los lados del eje X y el eje Y. El mínimo de esto será la respuesta.
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>
Producción:
1
Complejidad temporal: O(n)
Espacio auxiliar: O(1)
Sugiera si alguien tiene una mejor solución que sea más eficiente en términos de espacio y tiempo.
Este artículo es una contribución de Aarti_Rathi . Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.
Publicación traducida automáticamente
Artículo escrito por Vineet Joshi y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA