Given two arrays X[] and Y[], each of length 4, where (X[0], Y[0]) and (X[1], Y[1]) represents the bottom left and top right corners of one rectangle and (X[2], Y[2]) and (X[3], Y[3]) represents the bottom left and top right corners of the other rectangle, the task is to find the perimeter of the outer boundaries of the union of the two rectangles as shown below.
Examples:
Input: X[] = {-1, 2, 0, 4}, Y[] = {2, 5, -3, 3}
Output: 26
Explanation: Required Perimeter = 2 * ( (4 – (-1)) + (5 – (-3)) ) = 2*(8 + 5) = 26.Input: X[] = {-3, 1, 1, 4}, Y[] = {-2, 3, 1, 5}
Output: 26
Explanation: Required Perimeter = 2 * ( (4 – (-3)) + (5 – (-2)) ) = 2*(7 + 7) = 28.
Approach: Follow the steps below to solve the problem:
- Check if the rectangles formed by the given points intersect or not.
- If found to be intersecting, then the perimeter can be calculated by the formula 2*((X[1] – X[0]) + (X[3] – X[2]) + (Y[1] – Y[0]) + (Y[3] – Y[2])).
- Otherwise, print twice the sum of maximum differences between X and Y coordinates respectively, i.e. 2 * (max(X[]) – min(X[]) + max(Y[]) – min(Y[])).
Below is the implementation of the above approach:
C++
// C++ program for the above approachÂ
#include <bits/stdc++.h>using namespace std;Â
// Function to check if two// rectangles are intersecting or notbool 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;Â
    // If one rectangle is on the    // top of other's top edge    if (Y[0] > Y[3] || Y[2] > Y[1])        return false;Â
    return true;}Â
// Function to return the perimeter of// the Union of Two Rectanglesint getUnionPerimeter(vector<int> X,                      vector<int> Y){    // Stores the resultant perimeter    int perimeter = 0;Â
    // If rectangles do not interesect    if (!doIntersect(X, Y)) {Â
        // Perimeter of Rectangle 1        perimeter            += 2 * (abs(X[1] - X[0])                    + abs(Y[1] - Y[0]));Â
        // Perimeter of Rectangle 2        perimeter            += 2 * (abs(X[3] - X[2])                    + abs(Y[3] - Y[2]));    }Â
    // If the rectangles intersect    else {Â
        // Get width of combined figure        int w = *max_element(X.begin(),                             X.end())                - *min_element(X.begin(),                               X.end());Â
        // Get length of combined figure        int l = *max_element(Y.begin(),                             Y.end())                - *min_element(Y.begin(),                               Y.end());Â
        perimeter = 2 * (l + w);    }Â
    // Return the perimeter    return perimeter;}Â
// Driver Codeint main(){Â Â Â Â vector<int> X{ -1, 2, 4, 6 };Â Â Â Â vector<int> Y{ 2, 5, 3, 7 };Â
    cout << getUnionPerimeter(X, Y);} |
Java
// Java program for the above approachimport java.util.*;class GFG{Â
// Function to check if two// rectangles are intersecting or notstatic 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;Â
    // If one rectangle is on the    // top of other's top edge    if (Y[0] > Y[3] || Y[2] > Y[1])        return false;Â
    return true;}Â
// Function to return the perimeter of// the Union of Two Rectanglesstatic int getUnionPerimeter(int []X,                      int []Y){    // Stores the resultant perimeter    int perimeter = 0;Â
    // If rectangles do not interesect    if (!doIntersect(X, Y)) {Â
        // Perimeter of Rectangle 1        perimeter            += 2 * (Math.abs(X[1] - X[0])                    + Math.abs(Y[1] - Y[0]));Â
        // Perimeter of Rectangle 2        perimeter            += 2 * (Math.abs(X[3] - X[2])                    + Math.abs(Y[3] - Y[2]));    }Â
    // If the rectangles intersect    else {Â
        // Get width of combined figure        int w = Arrays.stream(X).max().getAsInt()                - Arrays.stream(X).min().getAsInt();Â
        // Get length of combined figure        int l = Arrays.stream(Y).max().getAsInt()                - Arrays.stream(Y).min().getAsInt();Â
        perimeter = 2 * (l + w);    }Â
    // Return the perimeter    return perimeter;}Â
// Driver Codepublic static void main(String[] args){Â Â Â Â int []X = { -1, 2, 4, 6 };Â Â Â Â int []Y = { 2, 5, 3, 7 };Â
    System.out.print(getUnionPerimeter(X, Y));}}Â
// This code is contributed by 29AjayKumar |
Python3
# Python3 program for the above approachÂ
# Function to check if two# rectangles are intersecting or notdef 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 FalseÂ
    # 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 TrueÂ
# Function to return the perimeter of# the Union of Two Rectanglesdef getUnionPerimeter(X, Y):       # Stores the resultant perimeter    perimeter = 0Â
    # If rectangles do not interesect    if (not doIntersect(X, Y)):Â
        # Perimeter of Rectangle 1        perimeter += 2 * (abs(X[1] - X[0]) + abs(Y[1] - Y[0]))Â
        # Perimeter of Rectangle 2        perimeter += 2 * (abs(X[3] - X[2]) + abs(Y[3] - Y[2]))         # If the rectangles intersect    else:Â
        # Get width of combined figure        w = max(X) - min(X)                 # Get length of combined figure        l = max(Y) - min(Y)        perimeter = 2 * (l + w)Â
    # Return the perimeter    return perimeterÂ
# Driver Codeif __name__ == '__main__':Â Â Â Â X = [ -1, 2, 4, 6]Â Â Â Â Y = [ 2, 5, 3, 7 ]Â
    print (getUnionPerimeter(X, Y))Â
# This code is contributed by mohit kumar 29. |
C#
// C# program for the above approachusing System;using System.Linq;public class GFG{Â
// Function to check if two// rectangles are intersecting or notstatic 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;Â
    // If one rectangle is on the    // top of other's top edge    if (Y[0] > Y[3] || Y[2] > Y[1])        return false;Â
    return true;}Â
// Function to return the perimeter of// the Union of Two Rectanglesstatic int getUnionPerimeter(int []X,                      int []Y){       // Stores the resultant perimeter    int perimeter = 0;Â
    // If rectangles do not interesect    if (!doIntersect(X, Y))     {Â
        // Perimeter of Rectangle 1        perimeter            += 2 * (Math.Abs(X[1] - X[0])                    + Math.Abs(Y[1] - Y[0]));Â
        // Perimeter of Rectangle 2        perimeter            += 2 * (Math.Abs(X[3] - X[2])                    + Math.Abs(Y[3] - Y[2]));    }Â
    // If the rectangles intersect    else    {Â
        // Get width of combined figure        int w = X.Max()                - X.Min();Â
        // Get length of combined figure        int l = X.Max()                - Y.Min();Â
        perimeter = 2 * (l + w);    }Â
    // Return the perimeter    return perimeter;}Â
// Driver Codepublic static void Main(String[] args){Â Â Â Â int []X = { -1, 2, 4, 6 };Â Â Â Â int []Y = { 2, 5, 3, 7 };Â
    Console.Write(getUnionPerimeter(X, Y));}}Â
// This code contributed by shikhasingrajput |
Javascript
<script>// Javascript program for the above approachÂ
// Function to check if two// rectangles are intersecting or notfunction 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;}Â
// Function to return the perimeter of// the Union of Two Rectanglesfunction 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;}Â
// Driver Codelet X = [-1, 2, 4, 6 ];let Y = [ 2, 5, 3, 7 ];document.write(getUnionPerimeter(X, Y));Â
// This code is contributed by patel2127</script> |
Output:Â
24
Â
Time Complexity: O(1)
Auxiliary Space: O(1)
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