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 <bits/stdc++.h>
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

<script>

// javascript implementation of the approach


// Function that returns true if it is
// possible to move from (0, 0) to
// (x, y) in exactly k moves
function isPossible(x , y , k)
{
    // Minimum moves required
    var minMoves = Math.abs(x) + Math.abs(y);

    // If possible
    if (k >= minMoves && (k - minMoves) % 2 == 0)
        return true;

    return false;
}

// Driver code

var x = 5, y = 8, k = 20;

if (isPossible(x, y, k))
    document.write("Yes");
else
    document.write("No");

// This code contributed by shikhasingrajput

</script>

Output:

No

Time Complexity: O(1)

Auxiliary Space: O(1)