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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program for the above approach
#include <bits/stdc++.h>
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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


 
 

Output: 

No






 

 

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

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Recommended Posts:


Code and Let Code

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.