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

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:

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)