223. Rectangle Area

LeetCode medium original: C# #csharp #leetcode #math #medium #two-pointers
El texto de la tarea se traduce del ruso para el idioma seleccionado. El código no cambia.
LeetCode Static

given координаты двух прямоугольных прямоугольников на двумерной плоскости, return общую площадь, покрытую этими двумя прямоугольниками.

Первый прямоугольник определяется его нижним левым углом (ax1, ay1) и верхним правым углом (ax2, ay2).

Второй прямоугольник определяется его нижним левым углом (bx1, by1) и верхним правым углом (bx2, by2).

Ejemplo:

Input: ax1 = -3, ay1 = 0, ax2 = 3, ay2 = 4, bx1 = 0, by1 = -1, bx2 = 9, by2 = 2

Output: 45

C# solución

coincidente/original 
public class Solution {
    public int ComputeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
        int areaOfA = (ay2 - ay1) * (ax2 - ax1);
        int areaOfB = (by2 - by1) * (bx2 - bx1);
        int left = Math.Max(ax1, bx1);
        int right = Math.Min(ax2, bx2);
        int xOverlap = right - left;
        int top = Math.Min(ay2, by2);
        int bottom = Math.Max(ay1, by1);
        int yOverlap = top - bottom;
        int areaOfOverlap = 0;
        if (xOverlap > 0 && yOverlap > 0) {
            areaOfOverlap = xOverlap * yOverlap;
        }
        int totalArea = areaOfA + areaOfB - areaOfOverlap;
        return totalArea;
    }
}

C++ solución

borrador automático, revisar antes de enviar 
#include <bits/stdc++.h>
using namespace std;

// Auto-generated C++ draft from the C# solution. Review containers, LINQ and helper types before submit.
class Solution {
public:
    public int ComputeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
        int areaOfA = (ay2 - ay1) * (ax2 - ax1);
        int areaOfB = (by2 - by1) * (bx2 - bx1);
        int left = max(ax1, bx1);
        int right = min(ax2, bx2);
        int xOverlap = right - left;
        int top = min(ay2, by2);
        int bottom = max(ay1, by1);
        int yOverlap = top - bottom;
        int areaOfOverlap = 0;
        if (xOverlap > 0 && yOverlap > 0) {
            areaOfOverlap = xOverlap * yOverlap;
        }
        int totalArea = areaOfA + areaOfB - areaOfOverlap;
        return totalArea;
    }
}

Java solución

coincidente/original 
class Solution {
    public int computeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
        int areaOfA = (ay2 - ay1) * (ax2 - ax1);
        int areaOfB = (by2 - by1) * (bx2 - bx1);

        int left = Math.max(ax1, bx1);
        int right = Math.min(ax2, bx2);
        int xOverlap = right - left;

        int top = Math.min(ay2, by2);
        int bottom = Math.max(ay1, by1);
        int yOverlap = top - bottom;

        int areaOfOverlap = 0;
        if (xOverlap > 0 && yOverlap > 0) {
            areaOfOverlap = xOverlap * yOverlap;
        }

        int totalArea = areaOfA + areaOfB - areaOfOverlap;

        return totalArea;
    }
}

JavaScript solución

coincidente/original 
function computeArea(ax1, ay1, ax2, ay2, bx1, by1, bx2, by2) {
    const areaOfA = (ay2 - ay1) * (ax2 - ax1);
    const areaOfB = (by2 - by1) * (bx2 - bx1);

    const left = Math.max(ax1, bx1);
    const right = Math.min(ax2, bx2);
    const xOverlap = right - left;

    const top = Math.min(ay2, by2);
    const bottom = Math.max(ay1, by1);
    const yOverlap = top - bottom;

    let areaOfOverlap = 0;
    if (xOverlap > 0 && yOverlap > 0) {
        areaOfOverlap = xOverlap * yOverlap;
    }

    const totalArea = areaOfA + areaOfB - areaOfOverlap;

    return totalArea;
};

Python solución

coincidente/original 
class Solution:
    def computeArea(self,
                    ax1: int, ay1: int,
                    ax2: int, ay2: int,
                    bx1: int, by1: int,
                    bx2: int, by2: int) -> int:
        area_of_a = (ay2 - ay1) * (ax2 - ax1)
        area_of_b = (by2 - by1) * (bx2 - bx1)

        left = max(ax1, bx1)
        right = min(ax2, bx2)
        x_overlap = right - left

        top = min(ay2, by2)
        bottom = max(ay1, by1)
        y_overlap = top - bottom

        area_of_overlap = 0
        if x_overlap > 0 and y_overlap > 0:
            area_of_overlap = x_overlap * y_overlap

        total_area = area_of_a + area_of_b - area_of_overlap

        return total_area

Go solución

coincidente/original 
func computeArea(ax1 int, ay1 int, ax2 int, ay2 int, bx1 int, by1 int, bx2 int, by2 int) int {
    areaOfA := (ay2 - ay1) * (ax2 - ax1)
    areaOfB := (by2 - by1) * (bx2 - bx1)

    left := max(ax1, bx1)
    right := min(ax2, bx2)
    xOverlap := right - left

    top := min(ay2, by2)
    bottom := max(ay1, by1)
    yOverlap := top - bottom

    areaOfOverlap := 0
    if xOverlap > 0 && yOverlap > 0 {
        areaOfOverlap = xOverlap * yOverlap
    }

    totalArea := areaOfA + areaOfB - areaOfOverlap
    return totalArea
}

func max(x, y int) int {
    if x > y {
        return x
    }
    return y
}

func min(x, y int) int {
    if x < y {
        return x
    }
    return y
}

Algorithm

Вычислить площади двух прямоугольников:

Рассчитайте площади прямоугольников A и B, умножив их ширину на высоту.

Вычислить перекрытие:

find перекрытие по оси X и оси Y. Если перекрытие существует, вычислите площадь перекрытия.

Вычислить и вернуть общую площадь:

Вычтите площадь перекрытия из суммы площадей двух прямоугольников и return результат.

😎

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