# Repeated subtraction among two numbers

Given a pair of positive numbers x and y. We repeatedly subtract the smaller of the two integers from greater one until one of the integers becomes 0. The task is to count number of steps to before we stop (one of the numbers become 0).

Examples :

```Input : x = 5, y = 13
Output : 6
Explanation : There are total 6 steps before
we reach 0:
(5,13) --> (5,8) --> (5,3) --> (2,3)
--> (2,1) --> (1,1) --> (1,0).

Input : x = 3, y = 5
Output : 4
Explanation : There are 4 steps:
(5,3) --> (2,3) --> (2,1) --> (1,1) --> (1,0)

Input : x = 100, y = 19
Output : 13
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

A simple solution is to actually follow the process and count the number of steps.

A better solution is to use below steps. Let y be the smaller of two numbers
1) if y divides x then return (x/y)
2) else return ( (x/y) + solve(y, x%y) )

Illustration :
If we start with (x, y) and y divides x then the answer will be (x/y) since we can subtract y form x exactly (x/y) times.

For the other case, we take an example to see how it works: (100, 19)

We can subtract 19 from 100 exactly [100/19] = 5 times to get (19, 5).

We can subtract 5 from 19 exactly [19/5] = 3 times to get (5, 4).

We can subtract 4 from 5 exactly [5/4] = 1 times to get (4, 1).

We can subtract 1 from 4 exactly [4/1] = 4 times to get (1, 0)

hence a total of 5 + 3 + 1 + 4 = 13 steps.

Below is implementation based on above idea.

## C++

 `// C++ program to count of steps until one ` `// of the two numbers become 0. ` `#include ` `using` `namespace` `std; ` ` `  `// Returns count of steps before one ` `// of the numbers become 0 after repeated ` `// subtractions. ` `int` `countSteps(``int` `x, ``int` `y) ` `{ ` `    ``// If y divides x, then simply return ` `    ``// x/y. ` `    ``if` `(x%y == 0) ` `        ``return` `x/y; ` ` `  `    ``// Else recur. Note that this function ` `    ``// works even if x is smaller than y because ` `    ``// in that case first recursive call exchanges ` `    ``// roles of x and y. ` `    ``return` `x/y + countSteps(y, x%y); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `   ``int` `x = 100, y = 19; ` `   ``cout << countSteps(x, y); ` `   ``return` `0; ` `} `

## Java

 `// Java program to count of  ` `// steps until one of the ` `// two numbers become 0. ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` `     `  `// Returns count of steps ` `// before one of the numbers  ` `// become 0 after repeated ` `// subtractions. ` `static` `int` `countSteps(``int` `x,  ` `                      ``int` `y) ` `{ ` `    ``// If y divides x, then  ` `    ``// simply return x/y. ` `    ``if` `(x % y == ``0``) ` `        ``return` `x / y; ` ` `  `    ``// Else recur. Note that this ` `    ``// function works even if x is  ` `    ``// smaller than y because ` `    ``// in that case first recursive ` `    ``// call exchanges roles of x and y. ` `    ``return` `x / y + countSteps(y, x % y); ` `} ` ` `  `// Driver code  ` `public` `static` `void` `main (String[] args)  ` `{ ` `    ``int` `x = ``100``, y = ``19``; ` `    ``System.out.println(countSteps(x, y)); ` `     `  `} ` `} ` ` `  `// This code is contributed by aj_36 `

## Python3

 `# Python3 program to count of steps until  ` `# one of the two numbers become 0. ` `import` `math ` ` `  `# Returns count of steps before one of  ` `# the numbers become 0 after repeated ` `# subtractions. ` `def` `countSteps(x, y): ` `     `  `    ``# If y divides x, then simply ` `    ``# return x/y. ` `    ``if` `(x ``%` `y ``=``=` `0``): ` `        ``return` `math.floor(x ``/` `y); ` ` `  `    ``# Else recur. Note that this function  ` `    ``# works even if x is smaller than y  ` `    ``# because in that case first recursive  ` `    ``# call exchanges roles of x and y. ` `    ``return` `math.floor((x ``/` `y) ``+`  `           ``countSteps(y, x ``%` `y)); ` ` `  `# Driver code ` `x ``=` `100``; ` `y ``=` `19``; ` `print``(countSteps(x, y)); ` ` `  `# This code is contributed by mits `

## C#

 `// C# program to count of  ` `// steps until one of the ` `// two numbers become 0. ` `using` `System; ` ` `  `class` `GFG ` `{ ` `// Returns count of steps ` `// before one of the numbers  ` `// become 0 after repeated ` `// subtractions. ` `static` `int` `countSteps(``int` `x,  ` `                      ``int` `y) ` `{ ` `    ``// If y divides x, then  ` `    ``// simply return x/y. ` `    ``if` `(x % y == 0) ` `        ``return` `x / y; ` ` `  `    ``// Else recur. Note that this ` `    ``// function works even if x is  ` `    ``// smaller than y because ` `    ``// in that case first recursive ` `    ``// call exchanges roles of x and y. ` `    ``return` `x / y + countSteps(y, x % y); ` `} ` ` `  `// Driver Code ` `static` `public` `void` `Main () ` `{ ` `int` `x = 100, y = 19; ` `Console.WriteLine(countSteps(x, y)); ` `} ` `} ` ` `  `// This code is contributed by m_kit `

## PHP

 ` `

Output :

```13
```

This article is contributed by Shubham Agrawal. 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.

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.

My Personal Notes arrow_drop_up

Improved By : jit_t, Mithun Kumar

Article Tags :
Practice Tags :

1

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.