Find vertex coordinates of all possible rectangles with a given vertex and dimensions

Given two integers L and B representing the length and breadth of a rectangle and a coordinate (X, Y) representing a point on the cartesian plane, the task is to find coordinates of all rectangles having a vertex as (X, Y) of the given dimensions.

Example:

Input: X=9, Y=9, L=5, B=3
Ouput:
(9, 9), (14, 9), (9, 12), (14, 12)
(4, 9), (9, 9), (4, 12), (9, 12)
(9, 6), (14, 6), (9, 9), (14, 9)
(4, 6), (9, 6), (4, 9), (9, 9)
(9, 9), (12, 9), (9, 14), (12, 14)
(6, 9), (9, 9), (6, 14), (9, 14)
(9, 4), (12, 4), (9, 9), (12, 9)
(6, 4), (9, 4), (6, 9), (9, 9)
Explanation: There are 8 possible rectangles such that one of their vertex is (9, 9) and the length and breadth is 5 and 3 respectively as mentioned above.

Input: X=2, Y=3, L=4, B=1
Ouput:
(2, 3), (6, 3), (2, 4), (6, 4)
(-2, 3), (2, 3), (-2, 4), (2, 4)
(2, 2), (6, 2), (2, 3), (6, 3)
(-2, 2), (2, 2), (-2, 3), (2, 3)
(2, 3), (3, 3), (2, 7), (3, 7)
(1, 3), (2, 3), (1, 7), (2, 7)
(2, -1), (3, -1), (2, 3), (3, 3)
(1, -1), (2, -1), (1, 3), (2, 3)

Approach: It can be observed that for a given length and breadth and a vertex (X, Y), eight rectangles are possible as shown in the images below:

If the given length and breadth of the rectangles are equal, both the horizontal and vertical rectangles will represent the same coordinates. Hence, only 4 unique squares are possible either shown in image 1 or in image 2.

Below is the implementation of the above approach:

C++

// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;

void printHorizontal(int X, int Y, int L, int B)
{
    cout << '(' << X << ", " << Y << "), ";
    cout << '(' << X + L << ", " << Y << "), ";
    cout << '(' << X << ", " << Y + B << "), ";
    cout << '(' << X + L << ", " << Y + B << ")"
         << endl;
}

void printVertical(int X, int Y, int L, int B)
{
    cout << '(' << X << ", " << Y << "), ";
    cout << '(' << X + B << ", " << Y << "), ";
    cout << '(' << X << ", " << Y + L << "), ";
    cout << '(' << X + B << ", " << Y + L << ")"
         << endl;
}

// Function to find all possible rectangles
void findAllRectangles(int L, int B, int X, int Y)
{

    // First four Rectangles
    printHorizontal(X, Y, L, B);
    printHorizontal(X - L, Y, L, B);
    printHorizontal(X, Y - B, L, B);
    printHorizontal(X - L, Y - B, L, B);

    // If length and breadth are same
    // i.e, it is a square
    if (L == B)
        return;

    // Next four Rectangles
    printVertical(X, Y, L, B);
    printVertical(X - B, Y, L, B);
    printVertical(X, Y - L, L, B);
    printVertical(X - B, Y - L, L, B);
}

// Driver Code
int main()
{
    int L = 5, B = 3;
    int X = 9, Y = 9;

    findAllRectangles(L, B, X, Y);
}

Java

// Java code for the above approach
class GFG{

static void printHorizontal(int X, int Y, int L, int B)
{
    System.out.print("("+ X+ ", " +  Y+ "), ");
    System.out.print("("+ (X + L)+ ", " +  Y+ "), ");
    System.out.print("("+ X+ ", " +  (Y + B)+ "), ");
    System.out.print("("+ (X + L)+ ", " +  (Y + B)+ ")"
         +"\n");
}

static void printVertical(int X, int Y, int L, int B)
{
    System.out.print("("+ X+ ", " +  Y+ "), ");
    System.out.print("("+ (X + B)+ ", " +  Y+ "), ");
    System.out.print("("+ X+ ", " +  (Y + L)+ "), ");
    System.out.print("("+ (X + B)+ ", " +  (Y + L)+ ")"
         +"\n");
}

// Function to find all possible rectangles
static void findAllRectangles(int L, int B, int X, int Y)
{

    // First four Rectangles
    printHorizontal(X, Y, L, B);
    printHorizontal(X - L, Y, L, B);
    printHorizontal(X, Y - B, L, B);
    printHorizontal(X - L, Y - B, L, B);

    // If length and breadth are same
    // i.e, it is a square
    if (L == B)
        return;

    // Next four Rectangles
    printVertical(X, Y, L, B);
    printVertical(X - B, Y, L, B);
    printVertical(X, Y - L, L, B);
    printVertical(X - B, Y - L, L, B);
}

// Driver Code
public static void main(String[] args)
{
    int L = 5, B = 3;
    int X = 9, Y = 9;

    findAllRectangles(L, B, X, Y);
}
}

// This code is contributed by shikhasingrajput

Python3

# python code for the above approach
def printHorizontal(X, Y, L, B):

    print(f"({X}, {Y}), ", end="")
    print(f"({X + L}, {Y}), ", end="")
    print(f"('{X}, {Y + B}), ", end="")
    print(f"({X + L}, {Y + B})")

def printVertical(X, Y, L, B):

    print(f"({X}, {Y}), ", end="")
    print(f"({X + B}, {Y}), ", end="")
    print(f"({X}, {Y + L}), ", end="")
    print(f"({X + B}, {Y + L})")

# Function to find all possible rectangles
def findAllRectangles(L, B, X, Y):

    # First four Rectangles
    printHorizontal(X, Y, L, B)
    printHorizontal(X - L, Y, L, B)
    printHorizontal(X, Y - B, L, B)
    printHorizontal(X - L, Y - B, L, B)

    # If length and breadth are same
    # i.e, it is a square
    if (L == B):
        return

    # Next four Rectangles
    printVertical(X, Y, L, B)
    printVertical(X - B, Y, L, B)
    printVertical(X, Y - L, L, B)
    printVertical(X - B, Y - L, L, B)

# Driver Code
if __name__ == "__main__":

    L = 5
    B = 3

    X = 9
    Y = 9

    findAllRectangles(L, B, X, Y)

    # This code is contributed by rakeshsahni

C#

// C# code for the above approach
using System;

class GFG{

static void printHorizontal(int X, int Y, int L, int B)
{
    Console.Write("(" + X + ", " +  Y + "), ");
    Console.Write("(" + (X + L) + ", " +  Y + "), ");
    Console.Write("(" + X + ", " +  (Y + B) + "), ");
    Console.Write("(" + (X + L) + ", " +  (Y + B) + ")" + "\n");
}

static void printVertical(int X, int Y, int L, int B)
{
    Console.Write("(" + X + ", " +  Y + "), ");
    Console.Write("(" + (X + B) + ", " +  Y + "), ");
    Console.Write("(" + X + ", " +  (Y + L) + "), ");
    Console.Write("(" + (X + B) + ", " +  (Y + L) + ")" + "\n");
}

// Function to find all possible rectangles
static void findAllRectangles(int L, int B, int X, int Y)
{

    // First four Rectangles
    printHorizontal(X, Y, L, B);
    printHorizontal(X - L, Y, L, B);
    printHorizontal(X, Y - B, L, B);
    printHorizontal(X - L, Y - B, L, B);

    // If length and breadth are same
    // i.e, it is a square
    if (L == B)
        return;

    // Next four Rectangles
    printVertical(X, Y, L, B);
    printVertical(X - B, Y, L, B);
    printVertical(X, Y - L, L, B);
    printVertical(X - B, Y - L, L, B);
}

// Driver Code
public static void Main(String[] args)
{
    int L = 5, B = 3;
    int X = 9, Y = 9;

    findAllRectangles(L, B, X, Y);
}
}

// This code is contributed by shikhasingrajput

Javascript

<script>

// JavaScript code for the above approach

function printHorizontal(X, Y, L, B)
{
    document.write('(' + X + ", " + Y + "), ");
    document.write('(' + (X + L) + ", " + Y + "), ");
    document.write('(' + X + ", " + (Y + B) + "), ");
    document.write('(' + (X + L) + ", " +
                         (Y + B) + ")" + '<br>');
}

function printVertical(X, Y, L, B)
{
    document.write('(' + X + ", " + Y + "), ");
    document.write('(' + (X + B) + ", " + Y + "), ");
    document.write('(' + X + ", " + (Y + L) + "), ");
    document.write('(' + (X + B) + ", " +
                         (Y + L) + ")" + '<br>');
}

// Function to find all possible rectangles
function findAllRectangles(L, B, X, Y)
{

    // First four Rectangles
    printHorizontal(X, Y, L, B);
    printHorizontal(X - L, Y, L, B);
    printHorizontal(X, Y - B, L, B);
    printHorizontal(X - L, Y - B, L, B);

    // If length and breadth are same
    // i.e, it is a square
    if (L == B)
        return;

    // Next four Rectangles
    printVertical(X, Y, L, B);
    printVertical(X - B, Y, L, B);
    printVertical(X, Y - L, L, B);
    printVertical(X - B, Y - L, L, B);
}

// Driver Code
let L = 5, B = 3;
let X = 9, Y = 9;

findAllRectangles(L, B, X, Y);

// This code is contributed by Potta Lokesh

</script>

Output:

(9, 9), (14, 9), (9, 12), (14, 12)
(4, 9), (9, 9), (4, 12), (9, 12)
(9, 6), (14, 6), (9, 9), (14, 9)
(4, 6), (9, 6), (4, 9), (9, 9)
(9, 9), (12, 9), (9, 14), (12, 14)
(6, 9), (9, 9), (6, 14), (9, 14)
(9, 4), (12, 4), (9, 9), (12, 9)
(6, 4), (9, 4), (6, 9), (9, 9)

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