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# Source to destination in 2-D path with fixed sized jumps

• Difficulty Level : Medium
• Last Updated : 15 Jun, 2021

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:

1. (a + x, b + y)
2. (a – x, b + y)
3. (a + x, b – y)
4. (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 ``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

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## Javascript

 ``
Output:
`Yes` My Personal Notes arrow_drop_up