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Check if a destination is reachable from source with two movements allowed
• Difficulty Level : Medium
• Last Updated : 04 Jun, 2018

Given coordinates of a source point (x1, y1) determine if it is possible to reach the destination point (x2, y2). From any point (x, y) there only two types of valid movements:
(x, x + y) and (x + y, y). Return a boolean true if it is possible else return false.
Note: All coordinates are positive.
Asked in: Expedia, Telstra

Examples:

```Input : (x1, y1) = (2, 10)
(x2, y2) = (26, 12)
Output : True
(2, 10)->(2, 12)->(14, 12)->(26, 12)
is a valid path.

Input : (x1, y1) = (20, 10)
(x2, y2) = (6, 12)
Output : False
No such path is possible because x1 > x2
and coordinates are positive
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The problem can be solved using simple recursion. Base case would be to check if current x or y coordinate is greater than that of destination, in which case we return false. If it is not the destination point yet we make two calls for both valid movements from that point.
If any of them yields a path we return true else return false.

## C++

 `// C++ program to check if a destination is reachable ` `// from source with two movements allowed ` `#include ` `using` `namespace` `std; ` ` `  `bool` `isReachable(``int` `sx, ``int` `sy, ``int` `dx, ``int` `dy) ` `{ ` `    ``// base case ` `    ``if` `(sx > dx || sy > dy) ` `        ``return` `false``; ` ` `  `    ``// current point is equal to destination ` `    ``if` `(sx == dx && sy == dy) ` `        ``return` `true``; ` ` `  `    ``// check for other 2 possibilities ` `    ``return` `(isReachable(sx + sy, sy, dx, dy) ||  ` `            ``isReachable(sx, sy + sx, dx, dy)); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `source_x = 2, source_y = 10; ` `    ``int` `dest_x = 26, dest_y = 12; ` `    ``if` `(isReachable(source_x, source_y, dest_x, dest_y)) ` `        ``cout << ``"True\n"``; ` `    ``else` `        ``cout << ``"False\n"``; ` `    ``return` `0; ` `} `

## Java

 `// Java program to check if a destination is  ` `// reachable from source with two movements  ` `// allowed ` ` `  `class` `GFG { ` `     `  `    ``static` `boolean` `isReachable(``int` `sx, ``int` `sy, ` `                                 ``int` `dx, ``int` `dy) ` `    ``{ ` `         `  `        ``// base case ` `        ``if` `(sx > dx || sy > dy) ` `            ``return` `false``; ` `     `  `        ``// current point is equal to destination ` `        ``if` `(sx == dx && sy == dy) ` `            ``return` `true``; ` `     `  `        ``// check for other 2 possibilities ` `        ``return` `(isReachable(sx + sy, sy, dx, dy) ||  ` `                ``isReachable(sx, sy + sx, dx, dy)); ` `    ``} ` `     `  `    ``//driver code ` `    ``public` `static` `void` `main(String arg[]) ` `    ``{ ` `        ``int` `source_x = ``2``, source_y = ``10``; ` `        ``int` `dest_x = ``26``, dest_y = ``12``; ` `        ``if` `(isReachable(source_x, source_y, dest_x, ` `                                           ``dest_y)) ` `            ``System.out.print(``"True\n"``); ` `        ``else` `            ``System.out.print(``"False\n"``); ` `    ``} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python3

 `  `  `# Python3 program to check if ` `# a destination is reachable ` `# from source with two movements allowed ` ` `  `def` `isReachable(sx, sy, dx, dy): ` ` `  `    ``# base case ` `    ``if` `(sx > dx ``or` `sy > dy): ` `        ``return` `False` ` `  `    ``# current point is equal to destination ` `    ``if` `(sx ``=``=` `dx ``and` `sy ``=``=` `dy): ` `        ``return` `True` ` `  `    ``# check for other 2 possibilities ` `    ``return` `(isReachable(sx ``+` `sy, sy, dx, dy) ``or` `            ``isReachable(sx, sy ``+` `sx, dx, dy)) ` ` `  `# Driver code ` `source_x, source_y ``=` `2``, ``10` `dest_x, dest_y ``=` `26``, ``12` `if` `(isReachable(source_x, source_y, dest_x, dest_y)): ` `    ``print``(``"True"``) ` `else``: ` `    ``print``(``"False"``) ` `     `  `# This code is contributed by Anant Agarwal. `

## C#

 `// C# program to check if a destination is  ` `// reachable from source with two movements  ` `// allowed ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``static` `bool` `isReachable(``int` `sx, ``int` `sy, ` `                             ``int` `dx, ``int` `dy) ` `    ``{ ` `         `  `        ``// base case ` `        ``if` `(sx > dx || sy > dy) ` `            ``return` `false``; ` `      `  `        ``// current point is equal to destination ` `        ``if` `(sx == dx && sy == dy) ` `            ``return` `true``; ` `      `  `        ``// check for other 2 possibilities ` `        ``return` `(isReachable(sx + sy, sy, dx, dy) ||  ` `                ``isReachable(sx, sy + sx, dx, dy)); ` `    ``} ` `     `  `    ``//driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `source_x = 2, source_y = 10; ` `        ``int` `dest_x = 26, dest_y = 12; ` `        ``if` `(isReachable(source_x, source_y, dest_x,  ` `                                           ``dest_y)) ` `            ``Console.Write(``"True\n"``); ` `        ``else` `            ``Console.Write(``"False\n"``); ` `    ``} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## PHP

 ` ``\$dx` `|| ``\$sy` `> ``\$dy``) ` `        ``return` `false; ` ` `  `    ``// current point is equal  ` `    ``// to destination ` `    ``if` `(``\$sx` `== ``\$dx` `&& ``\$sy` `== ``\$dy``) ` `        ``return` `true; ` ` `  `    ``// check for other 2 possibilities ` `    ``return` `(isReachable(``\$sx` `+ ``\$sy``, ``\$sy``, ``\$dx``, ``\$dy``) ||  ` `            ``isReachable(``\$sx``, ``\$sy` `+ ``\$sx``, ``\$dx``, ``\$dy``)); ` `} ` ` `  `    ``// Driver code ` `    ``\$source_x` `= 2; ` `    ``\$source_y` `= 10; ` `    ``\$dest_x` `= 26; ` `    ``\$dest_y` `= 12; ` `    ``if` `(isReachable(``\$source_x``, ``\$source_y``,  ` `                       ``\$dest_x``, ``\$dest_y``)) ` `        ``echo` `"True\n"``; ` `    ``else` `        ``echo` `"False\n"``; ` `         `  `// This code is contributed by Sam007 ` `?> `

Output:

```True
```

This article is contributed by Aditi Sharma. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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