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 = 8Output:YesExplanation: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 = 6Output: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` |

**Output:**

Yes

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

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