Dadas dos arrays X[] e Y[] , cada una de longitud 4 , donde (X[0], Y[0]) y (X[1], Y[1]) representan las esquinas inferior izquierda y superior derecha de una rectángulo y (X[2], Y[2]) y (X[3], Y[3]) representan las esquinas inferior izquierda y superior derecha del otro rectángulo, la tarea es encontrar el perímetro de los límites exteriores de la unión de los dos rectángulos como se muestra a continuación.
Ejemplos:
Entrada: X[] = {-1, 2, 0, 4}, Y[] = {2, 5, -3, 3}
Salida: 26
Explicación: Perímetro requerido = 2 * ( (4 – (-1)) + (5 – (-3)) ) = 2*(8 + 5) = 26.Entrada: X[] = {-3, 1, 1, 4}, Y[] = {-2, 3, 1, 5}
Salida: 26
Explicación: Perímetro requerido = 2 * ( (4 – (-3)) + (5 – (-2)) ) = 2*(7 + 7) = 28.
Enfoque: siga los pasos a continuación para resolver el problema:
- Comprueba si los rectángulos formados por los puntos dados se cruzan o no.
- Si se encuentra que se cruzan, entonces el perímetro se puede calcular mediante la fórmula 2*((X[1] – X[0]) + (X[3] – X[2]) + (Y[1] – Y[ 0]) + (Y[3] – Y[2])) .
- De lo contrario, imprima el doble de la suma de las diferencias máximas entre las coordenadas X e Y respectivamente, es decir, 2 * (max(X[]) – min(X[]) + max(Y[]) – min(Y[])) .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if two
// rectangles are intersecting or not
bool doIntersect(vector<int> X,
vector<int> Y)
{
// If one rectangle is to the
// right of other's right edge
if (X[0] > X[3] || X[2] > X[1])
return false;
// If one rectangle is on the
// top of other's top edge
if (Y[0] > Y[3] || Y[2] > Y[1])
return false;
return true;
}
// Function to return the perimeter of
// the Union of Two Rectangles
int getUnionPerimeter(vector<int> X,
vector<int> Y)
{
// Stores the resultant perimeter
int perimeter = 0;
// If rectangles do not interesect
if (!doIntersect(X, Y)) {
// Perimeter of Rectangle 1
perimeter
+= 2 * (abs(X[1] - X[0])
+ abs(Y[1] - Y[0]));
// Perimeter of Rectangle 2
perimeter
+= 2 * (abs(X[3] - X[2])
+ abs(Y[3] - Y[2]));
}
// If the rectangles intersect
else {
// Get width of combined figure
int w = *max_element(X.begin(),
X.end())
- *min_element(X.begin(),
X.end());
// Get length of combined figure
int l = *max_element(Y.begin(),
Y.end())
- *min_element(Y.begin(),
Y.end());
perimeter = 2 * (l + w);
}
// Return the perimeter
return perimeter;
}
// Driver Code
int main()
{
vector<int> X{ -1, 2, 4, 6 };
vector<int> Y{ 2, 5, 3, 7 };
cout << getUnionPerimeter(X, Y);
}
Java
// Java program for the above approach
import java.util.*;
class GFG
{
// Function to check if two
// rectangles are intersecting or not
static boolean doIntersect(int []X,
int []Y)
{
// If one rectangle is to the
// right of other's right edge
if (X[0] > X[3] || X[2] > X[1])
return false;
// If one rectangle is on the
// top of other's top edge
if (Y[0] > Y[3] || Y[2] > Y[1])
return false;
return true;
}
// Function to return the perimeter of
// the Union of Two Rectangles
static int getUnionPerimeter(int []X,
int []Y)
{
// Stores the resultant perimeter
int perimeter = 0;
// If rectangles do not interesect
if (!doIntersect(X, Y)) {
// Perimeter of Rectangle 1
perimeter
+= 2 * (Math.abs(X[1] - X[0])
+ Math.abs(Y[1] - Y[0]));
// Perimeter of Rectangle 2
perimeter
+= 2 * (Math.abs(X[3] - X[2])
+ Math.abs(Y[3] - Y[2]));
}
// If the rectangles intersect
else {
// Get width of combined figure
int w = Arrays.stream(X).max().getAsInt()
- Arrays.stream(X).min().getAsInt();
// Get length of combined figure
int l = Arrays.stream(Y).max().getAsInt()
- Arrays.stream(Y).min().getAsInt();
perimeter = 2 * (l + w);
}
// Return the perimeter
return perimeter;
}
// Driver Code
public static void main(String[] args)
{
int []X = { -1, 2, 4, 6 };
int []Y = { 2, 5, 3, 7 };
System.out.print(getUnionPerimeter(X, Y));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 program for the above approach # Function to check if two # rectangles are intersecting or not def doIntersect(X, Y): # If one rectangle is to the # right of other's right edge if (X[0] > X[3] or X[2] > X[1]): return False # If one rectangle is on the # top of other's top edge if (Y[0] > Y[3] or Y[2] > Y[1]): return False return True # Function to return the perimeter of # the Union of Two Rectangles def getUnionPerimeter(X, Y): # Stores the resultant perimeter perimeter = 0 # If rectangles do not interesect if (not doIntersect(X, Y)): # Perimeter of Rectangle 1 perimeter += 2 * (abs(X[1] - X[0]) + abs(Y[1] - Y[0])) # Perimeter of Rectangle 2 perimeter += 2 * (abs(X[3] - X[2]) + abs(Y[3] - Y[2])) # If the rectangles intersect else: # Get width of combined figure w = max(X) - min(X) # Get length of combined figure l = max(Y) - min(Y) perimeter = 2 * (l + w) # Return the perimeter return perimeter # Driver Code if __name__ == '__main__': X = [ -1, 2, 4, 6] Y = [ 2, 5, 3, 7 ] print (getUnionPerimeter(X, Y)) # This code is contributed by mohit kumar 29.
C#
// C# program for the above approach
using System;
using System.Linq;
public class GFG
{
// Function to check if two
// rectangles are intersecting or not
static bool doIntersect(int []X,
int []Y)
{
// If one rectangle is to the
// right of other's right edge
if (X[0] > X[3] || X[2] > X[1])
return false;
// If one rectangle is on the
// top of other's top edge
if (Y[0] > Y[3] || Y[2] > Y[1])
return false;
return true;
}
// Function to return the perimeter of
// the Union of Two Rectangles
static int getUnionPerimeter(int []X,
int []Y)
{
// Stores the resultant perimeter
int perimeter = 0;
// If rectangles do not interesect
if (!doIntersect(X, Y))
{
// Perimeter of Rectangle 1
perimeter
+= 2 * (Math.Abs(X[1] - X[0])
+ Math.Abs(Y[1] - Y[0]));
// Perimeter of Rectangle 2
perimeter
+= 2 * (Math.Abs(X[3] - X[2])
+ Math.Abs(Y[3] - Y[2]));
}
// If the rectangles intersect
else
{
// Get width of combined figure
int w = X.Max()
- X.Min();
// Get length of combined figure
int l = X.Max()
- Y.Min();
perimeter = 2 * (l + w);
}
// Return the perimeter
return perimeter;
}
// Driver Code
public static void Main(String[] args)
{
int []X = { -1, 2, 4, 6 };
int []Y = { 2, 5, 3, 7 };
Console.Write(getUnionPerimeter(X, Y));
}
}
// This code contributed by shikhasingrajput
Javascript
<script>
// Javascript program for the above approach
// Function to check if two
// rectangles are intersecting or not
function doIntersect(X,Y)
{
// If one rectangle is to the
// right of other's right edge
if (X[0] > X[3] || X[2] > X[1])
return false;
// If one rectangle is on the
// top of other's top edge
if (Y[0] > Y[3] || Y[2] > Y[1])
return false;
return true;
}
// Function to return the perimeter of
// the Union of Two Rectangles
function getUnionPerimeter(X,Y)
{
// Stores the resultant perimeter
let perimeter = 0;
// If rectangles do not interesect
if (!doIntersect(X, Y)) {
// Perimeter of Rectangle 1
perimeter
+= 2 * (Math.abs(X[1] - X[0])
+ Math.abs(Y[1] - Y[0]));
// Perimeter of Rectangle 2
perimeter
+= 2 * (Math.abs(X[3] - X[2])
+ Math.abs(Y[3] - Y[2]));
}
// If the rectangles intersect
else {
// Get width of combined figure
let w = Math.max(...X)
- Math.min(...X);
// Get length of combined figure
let l = Math.max(...Y)
- Math.min(...Y);
perimeter = 2 * (l + w);
}
// Return the perimeter
return perimeter;
}
// Driver Code
let X = [-1, 2, 4, 6 ];
let Y = [ 2, 5, 3, 7 ];
document.write(getUnionPerimeter(X, Y));
// This code is contributed by patel2127
</script>
Producción:
24
Tiempo Complejidad: O(1)
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por the_alphaEye y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA