Perimeter of the Union of Two Rectangles

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 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>

Output:

24

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