Given two integer **X** and **Y**, the task is to find the minimum number of steps to convert integer X to Y using any of the operation in each step:

- Divide the number by any natural number
- Multiply the number with any natural number

**Examples:**

Input:X = 8, Y = 12Output:2Explanation:

First divide 8 by 2: 8/2 = 4

Then multiply by 3: 4*3 = 12

Input:X = 4, Y = 8Output:1Explanation:

To convert 4 to 8 multiply 4 by 2: 4 * 2 = 8

**Approach:** To solve the problem mentioned above:

- Make sure X contains the smaller value among X and Y. Now,
**if X is greater than Y**then we know that it is always easier to change a smaller number to a larger number. Hence, we just swap the values of X and Y and then follow the steps mentioned below. **If both the integers are same**then the answer will be zero as no conversion takes place.**For example,**

If X = 4, Y = 4 Here 4 = 4 Therefore, answer = 0 (as they both are already same)

- However,
**if X is less than Y**then:- we have to check that Y % X gives 0 or not.
- If yes, then Y can be represented as
**X * (Y / X)**and we get the desired output in single-step.**For example,**

If X = 4, Y = 12 Here 12 % 4 = 0 Therefore, answer = 1 (4 * 3 = 12)

- Otherwise, the answer will be 2 as it takes two steps, one for division (X = X/X) and other for multiplication (X = X * Y).
**For example,**

If X = 8, Y = 13 Here 13 % 8 != 0 Therefore, 1. X = X/X = 8/8 = 1 2. X = X*Y = 1*13 = 13 Hence, answer = 2

Below is the implementation of the above approach:

## C++

`// C++ implementation to find minimum ` `// steps to convert X to Y by repeated ` `// division and multiplication ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `int` `solve(` `int` `X, ` `int` `Y) ` `{ ` ` ` `// Check if X is greater than Y ` ` ` `// then swap the elements ` ` ` `if` `(X > Y) { ` ` ` `int` `temp = X; ` ` ` `X = Y; ` ` ` `Y = temp; ` ` ` `} ` ` ` ` ` `// Check if X equals Y ` ` ` `if` `(X == Y) ` ` ` `cout << 0 << endl; ` ` ` ` ` `else` `if` `(Y % X == 0) ` ` ` `cout << 1 << endl; ` ` ` `else` ` ` `cout << 2 << endl; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `X = 8, Y = 13; ` ` ` `solve(X, Y); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Java

`// Java implementation to find minimum ` `// steps to convert X to Y by repeated ` `// division and multiplication ` `class` `GFG{ ` ` ` `static` `int` `solve(` `int` `X, ` `int` `Y) ` `{ ` ` ` `// Check if X is greater than Y ` ` ` `// then swap the elements ` ` ` `if` `(X > Y) ` ` ` `{ ` ` ` `int` `temp = X; ` ` ` `X = Y; ` ` ` `Y = temp; ` ` ` `} ` ` ` ` ` `// Check if X equals Y ` ` ` `if` `(X == Y) ` ` ` `System.out.println(` `0` `); ` ` ` ` ` `else` `if` `(Y % X == ` `0` `) ` ` ` `System.out.println( ` `1` `); ` ` ` `else` ` ` `System.out.println(` `2` `); ` ` ` `return` `0` `; ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` ` ` `int` `X = ` `8` `, Y = ` `13` `; ` ` ` `solve(X, Y); ` `} ` `} ` ` ` `// This code is contributed by shivanisinghss2110` |

*chevron_right*

*filter_none*

## C#

`// C# implementation to find minimum ` `// steps to convert X to Y by repeated ` `// division and multiplication ` `using` `System; ` ` ` `class` `GFG{ ` ` ` `static` `int` `solve(` `int` `X, ` `int` `Y) ` `{ ` ` ` ` ` `// Check if X is greater than Y ` ` ` `// then swap the elements ` ` ` `if` `(X > Y) ` ` ` `{ ` ` ` `int` `temp = X; ` ` ` `X = Y; ` ` ` `Y = temp; ` ` ` `} ` ` ` ` ` `// Check if X equals Y ` ` ` `if` `(X == Y) ` ` ` `Console.WriteLine(0); ` ` ` ` ` `else` `if` `(Y % X == 0) ` ` ` `Console.WriteLine(1); ` ` ` `else` ` ` `Console.WriteLine(2); ` ` ` `return` `0; ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` ` ` `int` `X = 8, Y = 13; ` ` ` `solve(X, Y); ` `} ` `} ` ` ` `// This code is contributed by amal kumar choubey ` |

*chevron_right*

*filter_none*

**Output:**

2

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.

## Recommended Posts:

- Multiply two integers without using multiplication, division and bitwise operators, and no loops
- Write you own Power without using multiplication(*) and division(/) operators
- Divide two integers without using multiplication, division and mod operator | Set2
- Minimum repeated addition of even divisors of N required to convert N to M
- Minimum cost to convert M to N by repeated addition of its even divisors
- Number of steps to convert to prime factors
- Min steps to convert N-digit prime number into another by replacing a digit in each step
- Convert 1 into X in min steps by multiplying with 2 or 3 or by adding 1
- Find minimum number K such that sum of array after multiplication by K exceed S
- Minimum steps to make sum and the product of all elements of array non-zero
- Minimum steps for increasing and decreasing Array to reach either 0 or N
- Find Quotient and Remainder of two integer without using division operators
- Find two numbers with difference and division both same as N
- Count minimum steps to get the given desired array
- Minimum steps to come back to starting point in a circular tour
- Minimum steps needed to cover a sequence of points on an infinite grid
- Minimum steps to make the product of the array equal to 1
- Generate array with minimum sum which can be deleted in P steps
- Minimum steps to color the tree with given colors
- Find the minimum number of steps to reach M from N

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.