Given the source point and the destination point of the path and two integers x and y. The task is to check whether it is possible to move from source to the destination with the below moves, If current position is (a, b) then the valid moves are:
- (a + x, b + y)
- (a – x, b + y)
- (a + x, b – y)
- (a – x, b – y)
Examples:
Input: Sx = 0, Sy = 0, Dx = 0, Dy = 6, x = 2, y = 3
Output: Yes
(0, 0) -> (2, 3) -> (0, 6)
Input: Sx = 1, Sy = 1, Dx = 3, Dy = 6, x = 1, y = 5
Output: No
Approach: Let’s approach this problem as if the steps were (a, b) -> (a + x, 0) or (a, b) -> (a – x, 0) or (a, b) -> (0, b + y) or (a, b) -> (0, b – y). Then the answer is Yes if |Sx – Dx| mod x = 0 and |Sy – Dy| mod y = 0.
It’s easy to see that if the answer to this problem is NO then the answer to the original problem is also NO.
Let’s return to the original problem and take a look at some sequence of steps. It ends in some point (xe, ye). Define cntx as |xe – Sx| / x and cnty as |ye – Sy| / y . The parity of cntx is the same as the parity of cnty because every type of move changes the parity of both cntx and cnty.
So the answer is Yes if |Sx – Dx| mod x = 0, |Sy – Dy| mod y = 0 and |Sx – Dx| / x mod 2 = |Sy – Dy| / y mod 2.
Below is the implementation of the above approach.
C++
#include <bits/stdc++.h>
using namespace std;
bool isPossible( int Sx, int Sy, int Dx, int Dy, int x, int y)
{
if ( abs (Sx - Dx) % x == 0
and abs (Sy - Dy) % y == 0
and ( abs (Sx - Dx) / x) % 2 == ( abs (Sy - Dy) / y) % 2)
return true ;
return false ;
}
int main()
{
int Sx = 0, Sy = 0, Dx = 0, Dy = 0;
int x = 3, y = 4;
if (isPossible(Sx, Sy, Dx, Dy, x, y))
cout << "Yes" ;
else
cout << "No" ;
return 0;
}
|
Java
import java .io.*;
class GFG
{
static boolean isPossible( int Sx, int Sy, int Dx,
int Dy, int x, int y)
{
if (Math.abs(Sx - Dx) % x == 0 &&
Math.abs(Sy - Dy) % y == 0 &&
(Math.abs(Sx - Dx) / x) % 2 ==
(Math.abs(Sy - Dy) / y) % 2 )
return true ;
return false ;
}
public static void main(String[] args)
{
int Sx = 0 , Sy = 0 , Dx = 0 , Dy = 0 ;
int x = 3 , y = 4 ;
if (isPossible(Sx, Sy, Dx, Dy, x, y))
System.out.println( "Yes" );
else
System.out.println( "No" );
}
}
|
Python3
def isPossible(Sx, Sy, Dx, Dy, x, y):
if ( abs (Sx - Dx) % x = = 0 and
abs (Sy - Dy) % y = = 0 and
( abs (Sx - Dx) / x) % 2 = =
( abs (Sy - Dy) / y) % 2 ):
return True ;
return False ;
Sx = 0 ;
Sy = 0 ;
Dx = 0 ;
Dy = 0 ;
x = 3 ;
y = 4 ;
if (isPossible(Sx, Sy, Dx, Dy, x, y)):
print ( "Yes" );
else :
print ( "No" );
|
C#
using System;
class GFG
{
static bool isPossible( int Sx, int Sy, int Dx,
int Dy, int x, int y)
{
if (Math.Abs(Sx - Dx) % x == 0 &&
Math.Abs(Sy - Dy) % y == 0 &&
(Math.Abs(Sx - Dx) / x) % 2 ==
(Math.Abs(Sy - Dy) / y) % 2)
return true ;
return false ;
}
static void Main()
{
int Sx = 0, Sy = 0, Dx = 0, Dy = 0;
int x = 3, y = 4;
if (isPossible(Sx, Sy, Dx, Dy, x, y))
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
}
|
PHP
<?php
function isPossible( $Sx , $Sy , $Dx ,
$Dy , $x , $y )
{
if ( abs ( $Sx - $Dx ) % $x == 0 &&
abs ( $Sy - $Dy ) % $y == 0 &&
( abs ( $Sx - $Dx ) / $x ) % 2 ==
( abs ( $Sy - $Dy ) / $y ) % 2)
return true;
return false;
}
$Sx = 0; $Sy = 0; $Dx = 0;
$Dy = 0; $x = 3; $y = 4;
if (isPossible( $Sx , $Sy , $Dx ,
$Dy , $x , $y ))
echo ( "Yes" );
else
echo ( "No" );
?>
|
Javascript
<script>
function isPossible(Sx, Sy, Dx, Dy, x, y)
{
if (Math.abs(Sx - Dx) % x == 0 &&
Math.abs(Sy - Dy) % y == 0 &&
(Math.abs(Sx - Dx) / x) % 2 ==
(Math.abs(Sy - Dy) / y) % 2)
return true ;
return false ;
}
let Sx = 0, Sy = 0, Dx = 0, Dy = 0;
let x = 3, y = 4;
if (isPossible(Sx, Sy, Dx, Dy, x, y))
document.write( "Yes" );
else
document.write( "No" );
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
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