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Check if X can be converted to Y by converting to 3 * (X / 2) or X – 1 in each operation

  • Last Updated : 11 Jun, 2021

Given two positive integers X and Y, the task is to check if X can be converted to Y or not by repeatedly changing the value of X to (3 * X / 2) (if X is even) or (X – 1). If it is possible to convert X into Y, then print “Yes”. Otherwise, print “No”.

Examples: 

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Input: X = 6, Y = 8
Output: Yes
Explanation:
Operation 1: Convert X(= 6) to 3*X/2 ( = 3 * (6 / 2) = 9).
Operation 2: Convert X(= 9) to (X – 1) (= 8).
Therefore, the total number of operations required is 2.

Input: X = 3, Y = 6
Output: No



Approach: The given problem can be solved based on the following observations: 

  • If the value of X is at least Y, then X can always be converted to Y using the second operation, i.e. reducing X by 1.
  • If X is even, then it can be converted to (3 * (X / 2)), which is greater than X for all even numbers greater than 0.
  • If the value of X is odd, then X can be converted to (X – 1), which can be converted into (3 * (X – 1)/2), which is greater than X.
  • Therefore, from the above observations, the conversion of X into Y is always possible for X > 3.
  • Following base cases are required to be considered:
    • X = 1: Conversion possible only if Y = 1.
    • X = 2 or X = 3: Conversion possible only if Y ≤ 3.
    • In all the other cases, conversion is not possible.

Follow the steps below to solve the problem: 

  • If the value X is greater than 4, then print “Yes”.
  • If the value X is 1 and Y is 1, then print “Yes”.
  • If the value X is 2 or 3 and Y is less than 4, then print “Yes”.
  • Otherwise, print “No” for all the other cases.

Below is the implementation of above approach: 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if X can be
// made equal to Y by converting
// X to (3*X/2) or (X - 1)
void check(int X, int Y)
{
    // Conditions for possible conversion
    if (X > 3) {
        cout << "Yes";
    }
 
    else if (X == 1 and Y == 1) {
        cout << "Yes";
    }
 
    else if (X == 2 and Y <= 3) {
        cout << "Yes";
    }
 
    else if (X == 3 and Y <= 3) {
        cout << "Yes";
    }
 
    // Otherwise, conversion
    // is not possible
    else {
        cout << "No";
    }
}
 
// Driver Code
int main()
{
    int X = 6, Y = 8;
    check(X, Y);
 
    return 0;
}

Java




// Java program for the above approach
class GFG{
     
// Function to check if X can be
// made equal to Y by converting
// X to (3*X/2) or (X - 1)
static void check(int X, int Y)
{
     
    // Conditions for possible conversion
    if (X > 3)
    {
        System.out.print("Yes");
    }
 
    else if (X == 1 && Y == 1)
    {
        System.out.print("Yes");
    }
 
    else if (X == 2 && Y <= 3)
    {
        System.out.print("Yes");
    }
 
    else if (X == 3 && Y <= 3)
    {
        System.out.print("Yes");
    }
 
    // Otherwise, conversion
    // is not possible
    else
    {
        System.out.print("No");
    }
}
 
// Driver Code
public static void main (String[] args)
{
    int X = 6, Y = 8;
     
    check(X, Y);
}
}
 
// This code is contributed by AnkThon

Python3




# Python3 program for the above approach
 
# Function to check if X can be
# made equal to Y by converting
# X to (3*X/2) or (X - 1)
def check(X, Y):
     
    # Conditions for possible conversion
    if (X > 3):
        print("Yes")
 
    elif (X == 1 and Y == 1):
        print("Yes")
 
    elif (X == 2 and Y <= 3):
        print("Yes")
 
    elif (X == 3 and Y <= 3):
        print("Yes")
 
    # Otherwise, conversion
    # is not possible
    else:
        print("No")
 
# Driver Code
if __name__ == '__main__':
     
    X = 6
    Y = 8
     
    check(X, Y)
 
# This code is contributed by ipg2016107

C#




// C# program for the above approach
using System;
 
class GFG{
 
// Function to check if X can be
// made equal to Y by converting
// X to (3*X/2) or (X - 1)
static void check(int X, int Y)
{
     
    // Conditions for possible conversion
    if (X > 3)
    {
        Console.WriteLine("Yes");
    }
 
    else if (X == 1 && Y == 1)
    {
        Console.WriteLine("Yes");
    }
 
    else if (X == 2 && Y <= 3)
    {
        Console.WriteLine("Yes");
    }
 
    else if (X == 3 && Y <= 3)
    {
        Console.WriteLine("Yes");
    }
 
    // Otherwise, conversion
    // is not possible
    else
    {
        Console.WriteLine("No");
    }
}
 
// Driver Code
public static void Main(string[] args)
{
    int X = 6, Y = 8;
     
    check(X, Y);
}
}
 
// This code is contributed by avijitmondal1998

Javascript




<script>
 
    // JavaScript program for the above approach
     
    // Function to check if X can be
    // made equal to Y by converting
    // X to (3*X/2) or (X - 1)
    function check(X, Y)
    {
 
        // Conditions for possible conversion
        if (X > 3)
        {
            document.write("Yes");
        }
 
        else if (X == 1 && Y == 1)
        {
            document.write("Yes");
        }
 
        else if (X == 2 && Y <= 3)
        {
            document.write("Yes");
        }
 
        else if (X == 3 && Y <= 3)
        {
            document.write("Yes");
        }
 
        // Otherwise, conversion
        // is not possible
        else
        {
            document.write("No");
        }
    }
     
    let X = 6, Y = 8;
      
    check(X, Y);
 
</script>
Output: 
Yes

 

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




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