Find maximum and minimum distance between magnets
Given coordinates of two pivot points (x0, y0) & (x1, y1) in coordinates plane. Along with each pivot, two different magnets are tied with the help of a string of length r1 and r2 respectively. Find the distance between both magnets when they repel each other and when they are attracting each other.
Examples :
Input : x1=0, y1=0, x2=5, y2=0, r1=2, r2=2
Output : Distance while repulsion = 9, Distance while attraction = 1
Input : x1=0, y1=0, x2=5, y2=0, r1=3, r2=3
Output : Distance while repulsion = 11, Distance while attraction = 0
As we all know about the properties of magnet that they repel each other when they are facing each other with the same pole and attract each other when they are facing each other with opposite pole. Also, the force of attraction, as well as repulsion, always work in a straight line.
We have two pivots points on coordinates, so distance between these points are D = ((x1-x2)2 +(y1-y2)2 )1/2.
Also, we can conclude that distance between magnet is maximum while repulsion and that too should be the distance between pivots + sum of the length of both strings.
In case of attraction we have two cases to take care of:
Either the minimum distance is the distance between pivots – the sum of the length of both strings
Or minimum distance should be zero in case if the sum of the length of strings is greater than the distance between pivot points.
Illustration with help of diagram:
C++
#include <bits/stdc++.h>
using namespace std;
int pivotDis( int x0, int y0, int x1, int y1)
{
return sqrt ((x1 - x0) * (x1 - x0) +
(y1 - y0) * (y1 - y0));
}
int minDis( int D, int r1, int r2)
{
return max((D - r1 - r2), 0);
}
int maxDis( int D, int r1, int r2)
{
return D + r1 + r2;
}
int main()
{
int x0 = 0, y0 = 0, x1 = 8, y1 = 0, r1 = 4, r2 = 5;
int D = pivotDis(x0, y0, x1, y1);
cout << "Distance while repulsion = " << maxDis(D, r1, r2);
cout << "\nDistance while attraction = " << minDis(D, r1, r2);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static int pivotDis( int x0, int y0,
int x1, int y1)
{
return ( int )Math.sqrt((x1 - x0) *
(x1 - x0) +
(y1 - y0) *
(y1 - y0));
}
static int minDis( int D, int r1, int r2)
{
return Math.max((D - r1 - r2), 0 );
}
static int maxDis( int D, int r1, int r2)
{
return D + r1 + r2;
}
public static void main (String[] args)
{
int x0 = 0 , y0 = 0 , x1 = 8 ,
y1 = 0 , r1 = 4 , r2 = 5 ;
int D = pivotDis(x0, y0, x1, y1);
System.out.print( "Distance while " +
"repulsion = " +
maxDis(D, r1, r2));
System.out.print( "\nDistance while " +
"attraction = " +
minDis(D, r1, r2));
}
}
|
Python3
import math
def pivotDis(x0, y0, x1, y1):
return math.sqrt((x1 - x0) * (x1 - x0)
+ (y1 - y0) * (y1 - y0))
def minDis( D, r1, r2):
return max ((D - r1 - r2), 0 )
def maxDis( D, r1, r2):
return D + r1 + r2
x0 = 0
y0 = 0
x1 = 8
y1 = 0
r1 = 4
r2 = 5
D = pivotDis(x0, y0, x1, y1)
print ( "Distance while repulsion = " ,
int (maxDis(D, r1, r2)))
print ( "Distance while attraction = " ,
minDis(D, r1, r2))
|
C#
using System;
class GFG {
static int pivotDis( int x0, int y0,
int x1, int y1)
{
return ( int )Math.Sqrt((x1 - x0) *
(x1 - x0) +
(y1 - y0) *
(y1 - y0));
}
static int minDis( int D, int r1, int r2)
{
return Math.Max((D - r1 - r2), 0);
}
static int maxDis( int D, int r1, int r2)
{
return D + r1 + r2;
}
public static void Main ()
{
int x0 = 0, y0 = 0, x1 = 8,
y1 = 0, r1 = 4, r2 = 5;
int D = pivotDis(x0, y0, x1, y1);
Console.WriteLine( "Distance while " +
"repulsion = " +
maxDis(D, r1, r2));
Console.WriteLine( "Distance while " +
"attraction = " +
minDis(D, r1, r2));
}
}
|
PHP
<?php
function pivotDis( $x0 , $y0 ,
$x1 , $y1 )
{
return sqrt(( $x1 - $x0 ) *
( $x1 - $x0 ) +
( $y1 - $y0 ) *
( $y1 - $y0 ));
}
function minDis( $D , $r1 , $r2 )
{
return max(( $D - $r1 - $r2 ), 0);
}
function maxDis( $D , $r1 , $r2 )
{
return $D + $r1 + $r2 ;
}
$x0 = 0; $y0 = 0;
$x1 = 8; $y1 = 0;
$r1 = 4; $r2 = 5;
$D = pivotDis( $x0 , $y0 ,
$x1 , $y1 );
echo "Distance while repulsion = "
, maxDis( $D , $r1 , $r2 );
echo "\nDistance while attraction = "
, minDis( $D , $r1 , $r2 );
?>
|
Javascript
<script>
function pivotDis(x0, y0, x1, y1)
{
return Math.sqrt((x1 - x0) * (x1 - x0) +
(y1 - y0) * (y1 - y0));
}
function minDis(D, r1, r2)
{
return Math.max((D - r1 - r2), 0);
}
function maxDis(D, r1, r2)
{
return D + r1 + r2;
}
let x0 = 0, y0 = 0, x1 = 8,
y1 = 0, r1 = 4, r2 = 5;
let D = pivotDis(x0, y0, x1, y1);
document.write( "Distance while repulsion = " +
maxDis(D, r1, r2) + "</br>" );
document.write( "Distance while attraction = " +
minDis(D, r1, r2));
</script>
|
Output:
Distance while repulsion = 17
Distance while attraction = 0
Time Complexity: O(?((x1-x0)2 + (y1-y0)2))
Auxiliary Space: O(1), As constant extra space is used.
Last Updated :
30 Dec, 2022
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