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Is it possible to reach N and M from 1 and 0 respectively as per given conditio
  • Last Updated : 30 Mar, 2021

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 <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;
}

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

Python3




# 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

Javascript




<script>
 
// Javascript program for the above approach
 
// Function that find given x and y
// is possible or not
function is_possible(x, 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
 
// Given X and Y
let x = 5, y = 2;
 
// Function Call
if (is_possible(x, y))
    document.write("Yes");
else
    document.write("No");
     
// This code is contributed by divyesh072019       
         
</script>
Output: 
No

 

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

 

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