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++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function that returns true if // it is possible to move from source // to the destination with the given moves 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 ; } // Driver code 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
// Java implementation of the approach import java .io.*; class GFG { // Function that returns true if // it is possible to move from source // to the destination with the given moves 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 ; } // Driver code 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" ); } } // This code is contributed by inder_verma.. |
Python3
# Python3 implementation of the approach # Function that returns true if it is # possible to move from source to the # destination with the given moves 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 ; # Driver code 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" ); # This code is contributed by mits |
C#
// C# implementation of the approach using System; class GFG { // Function that returns true if // it is possible to move from source // to the destination with the given moves 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 ; } // Driver code 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" ); } } // This code is contributed by chandan_jnu |
PHP
<?php // PHP implementation of the approach // Function that returns true if // it is possible to move from source // to the destination with the given moves 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; } // Driver code $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" ); // This code is contributed // by Code_Mech ?> |
Javascript
<script> // Javascript implementation of the approach // Function that returns true if // it is possible to move from source // to the destination with the given moves 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 ; } // Driver code 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" ); // This code is contributed by mukesh07 </script> |
Yes
Time Complexity: O(1)
Auxiliary Space: O(1)
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!