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Repeated subtraction among two numbers
• Last Updated : 02 Mar, 2021

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```

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

 ``

## Javascript

 ``

Output :

`13`

Time Complexity: O(log(n))

Auxiliary Space: O(1)
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