# Check if it is possible to move from (0, 0) to (X, Y) in exactly K steps

Given a point (X, Y) in a 2-D plane and an integer K, the task is to check whether it is possible to move from (0, 0) to the given point (X, Y) in exactly K moves. In a single move, the positions that are reachable from (X, Y) are (X, Y + 1), (X, Y – 1), (X + 1, Y) and (X – 1, Y).
Examples:

Input: X = 0, Y = 0, K = 2
Output: Yes
Move 1: (0, 0) -> (0, 1)
Move 2: (0, 1) -> (0, 0)
Input: X = 5, Y = 8, K = 20
Output: No

Approach: It is clear that the shortest path to reach (X, Y) from (0, 0) will be minMoves = (|X| + |Y|). So, if K < minMoves then it is impossible to reach (X, Y) but if K ? minMoves then after reaching (X, Y) in minMoves number of moves the remaining (K – minMoves) number of moves have to be even in order to remain at that point for the rest of the moves.
So it is possible to reach (X, Y) from (0, 0) only if K ? (|X| + |Y|) and (K – (|X| + |Y|)) % 2 = 0.
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 (0, 0) to``// (x, y) in exactly k moves``bool` `isPossible(``int` `x, ``int` `y, ``int` `k)``{``    ``// Minimum moves required``    ``int` `minMoves = ``abs``(x) + ``abs``(y);` `    ``// If possible``    ``if` `(k >= minMoves && (k - minMoves) % 2 == 0)``        ``return` `true``;` `    ``return` `false``;``}` `// Driver code``int` `main()``{``    ``int` `x = 5, y = 8, k = 20;` `    ``if` `(isPossible(x, y, k))``        ``cout << ``"Yes"``;``    ``else``        ``cout << ``"No"``;` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach ``class` `GFG``{``    ` `    ``// Function that returns true if it is ``    ``// possible to move from (0, 0) to ``    ``// (x, y) in exactly k moves ``    ``static` `boolean` `isPossible(``int` `x, ``int` `y, ``int` `k) ``    ``{ ``        ``// Minimum moves required ``        ``int` `minMoves = Math.abs(x) + Math.abs(y); ``    ` `        ``// If possible ``        ``if` `(k >= minMoves && (k - minMoves) % ``2` `== ``0``) ``            ``return` `true``; ``    ` `        ``return` `false``; ``    ``} ``    ` `    ``// Driver code ``    ``public` `static` `void` `main (String[] args) ``    ``{ ``        ``int` `x = ``5``, y = ``8``, k = ``20``; ``    ` `        ``if` `(isPossible(x, y, k)) ``            ``System.out.println(``"Yes"``); ``        ``else``            ``System.out.println(``"No"``); ``    ``} ``}` `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 implementation of the approach` `# Function that returns true if it is``# possible to move from (0, 0) to``# (x, y) in exactly k moves``def` `isPossible(x, y, k):``    ` `    ``# Minimum moves required``    ``minMoves ``=` `abs``(x) ``+` `abs``(y)` `    ``# If possible``    ``if` `(k >``=` `minMoves ``and` `(k ``-` `minMoves) ``%` `2` `=``=` `0``):``        ``return` `True` `    ``return` `False` `# Driver code``x ``=` `5``y ``=` `8``k ``=` `20` `if` `(isPossible(x, y, k)):``    ``print``(``"Yes"``)``else``:``    ``print``(``"No"``)` `# This code is contributed by Mohit Kumar`

## C#

 `// C# implementation of the approach ``using` `System;``class` `GFG``{``    ` `    ``// Function that returns true if it is ``    ``// possible to move from (0, 0) to ``    ``// (x, y) in exactly k moves ``    ``static` `bool` `isPossible(``int` `x, ``int` `y, ``int` `k) ``    ``{ ``        ``// Minimum moves required ``        ``int` `minMoves = Math.Abs(x) + Math.Abs(y); ``    ` `        ``// If possible ``        ``if` `(k >= minMoves && (k - minMoves) % 2 == 0) ``            ``return` `true``; ``    ` `        ``return` `false``; ``    ``} ``    ` `    ``// Driver code ``    ``public` `static` `void` `Main () ``    ``{ ``        ``int` `x = 5, y = 8, k = 20; ``    ` `        ``if` `(isPossible(x, y, k)) ``            ``Console.Write(``"Yes"``); ``        ``else``            ``Console.Write(``"No"``); ``    ``} ``}` `// This code is contributed by Nidhi`

## Javascript

 ``

Output:
`No`

Time Complexity: O(1)

Auxiliary Space: O(1)

Previous
Next
Share your thoughts in the comments
Similar Reads