The point 0.3 lies on the y axis true or false

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Given coordinates of bottom-left and top-right corners of a rectangle. Check if a point (x, y) lies inside this rectangle or not.

Examples: 

Input: bottom-left: (0, 0), top-right: (10, 8), point: (1, 5) 
Output: Yes

Input: bottom-left: (-1, 4), top-right:(2, 3), point:(0, 4) 
Output: No

This problem is already discussed in a previous post. In this post, we have discussed a new approach.

Approach: If we observe carefully, It will be clear that for a point to be lie inside the rectangle, it should be lie inside the x-coordinate (x1, x2) of the rectangle as well as it should also lie inside the y-coordinate (y1, y2) of the rectangle.

Algorithm:

• Check if the x-coordinate of the given point (x, y) is lie inside the x-coordinate (x1, x2) of the rectangle, i.e, x > x1 and x < x2
• Check if the y-coordinate of the given point (x, y) is lie inside the y-coordinate (y1, y2) of the rectangle, i.e, y > y1 and y < y2
• If both the above condition satisfy then, 
  • The given point completely lie inside the rectangle.
• Otherwise, not.

Below is the implementation of the above approach:  

C++

#include <bits/stdc++.h>

using namespace std;

bool FindPoint(int x1, int y1, int x2,

               int y2, int x, int y)

{

    if (x > x1 and x < x2 and y > y1 and y < y2)

        return true;

    return false;

}

int main()

{

    int x1 = 0, y1 = 0, x2 = 10, y2 = 8;

    int x = 1, y = 5;

    if (FindPoint(x1, y1, x2, y2, x, y))

        cout << "Yes";

    else

        cout << "No";

    return 0;

}

C

#include <stdio.h>

#include <stdbool.h>

bool FindPoint(int x1, int y1, int x2, int y2, int x, int y)

{

    if (x > x1 && x < x2 && y > y1 && y < y2)

        return true;

    return false;

}

int main()

{

    int x1 = 0, y1 = 0, x2 = 10, y2 = 8;

    int x = 1, y = 5;

    if (FindPoint(x1, y1, x2, y2, x, y))

        printf("Yes");

    else

        printf("No");

    return 0;

}

Java

class GFG

{

static boolean FindPoint(int x1, int y1, int x2,

                         int y2, int x, int y)

{

if (x > x1 && x < x2 &&

    y > y1 && y < y2)

    return true;

return false;

}

public static void main(String[] args)

{

    int x1 = 0, y1 = 0,

        x2 = 10, y2 = 8;

    int x = 1, y = 5;

    if (FindPoint(x1, y1, x2, y2, x, y))

        System.out.println("Yes");

    else

        System.out.println("No");

}

}

Python3

def FindPoint(x1, y1, x2,

              y2, x, y) :

    if (x > x1 and x < x2 and

        y > y1 and y < y2) :

        return True

    else :

        return False

if __name__ == "__main__" :

    x1 , y1 , x2 , y2 = 0, 0, 10, 8

    x, y = 1, 5

    if FindPoint(x1, y1, x2,

                 y2, x, y) :

        print("Yes")

    else :

        print("No")

C#

using System;

class GFG

{

static bool FindPoint(int x1, int y1, int x2,

                      int y2, int x, int y)

{

if (x > x1 && x < x2 &&

    y > y1 && y < y2)

    return true;

return false;

}

public static void Main()

{

    int x1 = 0, y1 = 0,

        x2 = 10, y2 = 8;

    int x = 1, y = 5;

    if (FindPoint(x1, y1, x2, y2, x, y))

        Console.Write("Yes");

    else

        Console.Write("No");

}

}

PHP

<?php

function FindPoint($x1, $y1, $x2,

                   $y2, $x, $y)

{

    if ($x > $x1 and $x < $x2 and

        $y > $y1 and $y < $y2)

        return true;

    return false;

}

$x1 = 0; $y1 = 0;

$x2 = 10; $y2 = 8;

$x = 1; $y = 5;

if (FindPoint($x1, $y1, $x2,

              $y2, $x, $y))

    echo "Yes";

else

    echo "No";

?>

Javascript

<script>

function FindPoint(x1, y1, x2,

            y2, x, y)

{

    if (x > x1 && x < x2 && y > y1 && y < y2)

        return true;

    return false;

}

    let x1 = 0, y1 = 0, x2 = 10, y2 = 8;

    let x = 1, y = 5;

    if (FindPoint(x1, y1, x2, y2, x, y))

        document.write("Yes");

    else

        document.write("No");

</script>

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