# Is it possible to reach N and M from 1 and 0 respectively as per given conditio

Given two integers N and M, the task is to check if it is possible to obtain these values from X = 1 and Y = 0 respectively by performing the two operations any number of times::

• Increase X and Y by 1, if and only if x>0.
• Increase Y by 2, if and only if y>0.

Examples:

Input: N = 3, M = 4
Output: Yes
Explanation:
Initially X = 1, Y = 0
Operation 1: X = 2, Y = 1
Operation 1: X = 3, Y = 2
Operation 2: X = 3, Y = 4, hence the final values are got so the answer is Yes.

Input: N = 5, M = 2
Output: No
Explanation :
Obtaining X = 5 and Y = 2 from X = 1 and Y = 0 is not possible.

Approach: The above problem can be solved using the below observations:

1. If N is less than 2 and M is not equal to zero, then getting the final values is not possible, hence the answer is No.
2. Otherwise, subtract N from M and if M ? 0 and M is divisible by 2 then the answer is Yes.
3. In all other cases, the answer is No.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function that find given x and y` `// is possible or not` `bool` `is_possible(``int` `x, ``int` `y)` `{` `    ``// Check if x is less than 2 and` `    ``// y is not equal to 0` `    ``if` `(x < 2 && y != 0)` `        ``return` `false``;`   `    ``// Perform subtraction` `    ``y = y - x + 1;`   `    ``// Check if y is divisible by 2` `    ``// and greater than equal to 0` `    ``if` `(y % 2 == 0 && y >= 0)` `        ``return` `true``;`   `    ``else` `        ``return` `false``;` `}`   `// Driver Code` `int` `main()` `{` `    ``// Given X and Y` `    ``int` `x = 5, y = 2;`   `    ``// Function Call` `    ``if` `(is_possible(x, y))` `        ``cout << ``"Yes"``;` `    ``else` `        ``cout << ``"No"``;` `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `class` `GFG{` ` `  `// Function that find given x and y` `// is possible or not` `static` `boolean` `is_possible(``int` `x, ``int` `y)` `{` `    ``// Check if x is less than 2 and` `    ``// y is not equal to 0` `    ``if` `(x < ``2` `&& y != ``0``)` `        ``return` `false``;` ` `  `    ``// Perform subtraction` `    ``y = y - x + ``1``;` ` `  `    ``// Check if y is divisible by 2` `    ``// and greater than equal to 0` `    ``if` `(y % ``2` `== ``0` `&& y >= ``0``)` `        ``return` `true``;` ` `  `    ``else` `        ``return` `false``;` `}` ` `  `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``// Given X and Y` `    ``int` `x = ``5``, y = ``2``;` ` `  `    ``// Function Call` `    ``if` `(is_possible(x, y))` `        ``System.out.println(``"Yes"``);` `    ``else` `        ``System.out.println(``"No"``);` `}` `}`   `// This code is contributed by rock_cool`

## Python

 `# Python3 program for the above approach`   `# Function that find given x and y` `# is possible or not` `def` `is_possible(x, y):` `  `  `    ``# Check if x is less than 2 and` `    ``# y is not equal to 0` `    ``if` `(x < ``2` `and` `y !``=` `0``):` `        ``return` `false`   `    ``# Perform subtraction` `    ``y ``=` `y ``-` `x ``+` `1`   `    ``# Check if y is divisible by 2` `    ``# and greater than equal to 0` `    ``if` `(y ``%` `2` `=``=` `0` `and` `y >``=` `0``):` `        ``return` `True` `    ``else``:` `        ``return` `False`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `  `  `    ``# Given X and Y` `    ``x ``=` `5` `    ``y ``=` `2`   `    ``# Function Call` `    ``if` `(is_possible(x, y)):` `        ``print``(``"Yes"``)` `    ``else``:` `        ``print``(``"No"``)`   `# This code is contributed by Mohit Kumar`

## C#

 `// C# program for the above approach` `using` `System;` `class` `GFG{` `  `  `// Function that find given x and y` `// is possible or not` `static` `bool` `is_possible(``int` `x, ``int` `y)` `{` `    ``// Check if x is less than 2 and` `    ``// y is not equal to 0` `    ``if` `(x < 2 && y != 0)` `        ``return` `false``;` `  `  `    ``// Perform subtraction` `    ``y = y - x + 1;` `  `  `    ``// Check if y is divisible by 2` `    ``// and greater than equal to 0` `    ``if` `(y % 2 == 0 && y >= 0)` `        ``return` `true``;` `  `  `    ``else` `        ``return` `false``;` `}` `  `  `// Driver Code` `public` `static` `void` `Main(``string``[] args)` `{` `    ``// Given X and Y` `    ``int` `x = 5, y = 2;` `  `  `    ``// Function Call` `    ``if` `(is_possible(x, y))` `        ``Console.Write(``"Yes"``);` `    ``else` `        ``Console.Write(``"No"``);` `}` `}`   `// This code is contributed by Ritik Bansal`

Output:

```No

```

Time Complexity: O(1)
Auxiliary Space: O(1)

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