# Coordinates of rectangle with given points lie inside

• Difficulty Level : Medium
• Last Updated : 20 Feb, 2023

Given two arrays X[] and Y[] with n-elements, where (Xi, Yi) represent a point on coordinate system, find the smallest rectangle such that all points from given input lie inside that rectangle and sides of rectangle must be parallel to Coordinate axis. Print all four coordinates of an obtained rectangle.
Note: n >= 4 (we have at least 4 points as input).
Examples :

Input : X[] = {1, 3, 0, 4},  y[] = {2, 1, 0, 2}
Output : {0, 0}, {0, 2}, {4, 2}, {4, 0}

Input : X[] = {3, 6, 1, 9, 13, 0, 4}, Y[] = {4, 2, 6, 5, 2, 3, 1}
Output : {0, 1}, {0, 6}, {13, 6}, {13, 1}

For finding the smallest rectangle you may take two of basic approach :

1. Consider origin of coordinate plane as the smallest rectangle and then step by step keep expanding it as per the value of coordinates of points if they don’t lie inside the current rectangle. This concept requires a little of complex logic to find the exact smallest rectangle.
2. A very simple and efficient logic behind this is to find the smallest, as well as largest x & y coordinates among all given points and then the all possible four combinations of these values, result in the four points of the required rectangle as {Xmin, Ymin}, {Xmin, Ymax}, {Xmax, Ymax}, {Xmax, Ymin}.

## C++

 // Program to find smallest rectangle// to conquer all points#include using namespace std;  // function to print coordinate of smallest rectanglevoid printRect(int X[], int Y[], int n){    // find Xmax and Xmin    int Xmax = *max_element(X, X + n);    int Xmin = *min_element(X, X + n);      // find Ymax and Ymin    int Ymax = *max_element(Y, Y + n);    int Ymin = *min_element(Y, Y + n);      // print all four coordinates    cout << "{" << Xmin << ", " << Ymin << "}" << endl;    cout << "{" << Xmin << ", " << Ymax << "}" << endl;    cout << "{" << Xmax << ", " << Ymax << "}" << endl;    cout << "{" << Xmax << ", " << Ymin << "}" << endl;}  // driver programint main(){    int X[] = { 4, 3, 6, 1, -1, 12 };    int Y[] = { 4, 1, 10, 3, 7, -1 };    int n = sizeof(X) / sizeof(X[0]);          printRect(X, Y, n);          return 0;}

## Java

 // Program to find smallest rectangle// to conquer all pointsimport java.util.Arrays;import java.util.Collections;  class GFG {              // function to print coordinate of smallest rectangle    static void printRect(Integer X[], Integer Y[], int n)    {                  // find Xmax and Xmin        int Xmax = Collections.max(Arrays.asList(X));        int Xmin = Collections.min(Arrays.asList(X));              // find Ymax and Ymin        int Ymax = Collections.max(Arrays.asList(Y));        int Ymin = Collections.min(Arrays.asList(Y));              // print all four coordinates        System.out.println("{" + Xmin + ", " + Ymin + "}");        System.out.println("{" + Xmin + ", " + Ymax + "}");        System.out.println("{" + Xmax + ", " + Ymax + "}");        System.out.println("{" + Xmax + ", " + Ymin + "}");    }          //Driver code    public static void main (String[] args)    {                  Integer X[] = { 4, 3, 6, 1, -1, 12 };        Integer Y[] = { 4, 1, 10, 3, 7, -1 };        int n = X.length;                  printRect(X, Y, n);    }}  // This code is contributed by Anant Agarwal.

## Python 3

 # Program to find smallest rectangle# to conquer all points  # function to print coordinate of smallest rectangledef printRect(X, Y, n):      # find Xmax and Xmin    Xmax = max(X)    Xmin = min(X)      # find Ymax and Ymin    Ymax = max(Y)    Ymin = min(Y)      # print all four coordinates    print("{",Xmin,", ",Ymin,"}",sep="" )    print("{",Xmin,", ",Ymax,"}",sep="" )    print("{",Xmax,", ",Ymax,"}",sep="" )    print("{",Xmax,", ",Ymin,"}",sep="" )  # driver programX = [4, 3, 6, 1, -1, 12] Y =  [4, 1, 10, 3, 7, -1]n = len(X) printRect(X, Y, n)# This code is contributed by# Smitha Dinesh Semwal

## C#

 // Program to find smallest rectangle// to conquer all pointsusing System.Linq;using System;  public class GFG{          // function to print coordinate    // of smallest rectangle    static void printRect(int[] X,                         int[] Y, int n)    {                  // find Xmax and Xmin        int Xmax = X.Max();        int Xmin = X.Min();              // find Ymax and Ymin        int Ymax = Y.Max();        int Ymin = Y.Min();              // print all four coordinates        Console.WriteLine("{" + Xmin + ", "                             + Ymin + "}");                                       Console.WriteLine("{" + Xmin + ", "                             + Ymax + "}");                                       Console.WriteLine("{" + Xmax + ", "                             + Ymax + "}");                                       Console.WriteLine("{" + Xmax + ", "                             + Ymin + "}");    }          // Driver code    static public void Main ()    {        int[] X = { 4, 3, 6, 1, -1, 12 };        int[] Y = { 4, 1, 10, 3, 7, -1 };        int n = X.Length;                  printRect(X, Y, n);    }}  // This code is contributed by Ajit.



## Javascript



Output:

{-1, -1}
{-1, 10}
{12, 10}
{12, -1}

Time complexity: O(len(X)+len(Y))

space complexity: O(1)

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