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

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to count of steps until one
// of the two numbers become 0.
#include<bits/stdc++.h>
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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to count of 
// steps until one of the 
// two numbers become 0.
  
// Returns count of steps 
// before one of the numbers 
// become 0 after repeated
// subtractions.
function countSteps($x, $y)
{
    // If y divides x, then 
    // simply return x/y.
    if ($x % $y == 0)
        return floor(((int)$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 floor(((int)$x / $y) + 
                   countSteps($y, $x % $y));
}
  
// Driver code
$x = 100;
$y = 19;
echo countSteps($x, $y);
  
// This code is contributed by aj_36
?>

chevron_right



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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up

Improved By : jit_t, Mithun Kumar



Article Tags :
Practice Tags :


Be the First to upvote.


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