Problem 68. Kaprekar Steps

Created by Cody Team

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6174 is the Kaprekar constant. All natural numbers less than 10,000 (except some with same digits) can be reduced to 6174 in the following steps:

Order the digits of the number in descending and ascending order and compute the difference. Repeat the process till you get 6174.

Example

n = 2376

Digits in descending order = 7632

Digits in ascending order = 2367

Step1:

 >> 7632 - 2367
 ans = 5265

Step 2:

 >> 6552 - 2556
 ans = 3996

Step3:

 >> 9963 - 3699
 ans = 6264

Step4

 >> 6642 - 2466
 ans = 4176

Step5

 >> 7641 - 1467
 ans = 6174

Total number of steps = 5.

Your function should return the number of Kaprekar steps for a given input. Numbers such as 2222 will end in zero. These numbers will never result in 6174. They should return Inf.

Solve

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2303 Solutions
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Last Solution submitted on Nov 14, 2019

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13 Comments

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Muthu Annamalai on 17 Apr 2012

The behavior at x=6174 is artifically set to 0, which taints the pure-recursive solution, but hey it was cool problem!

Jean-Marie Sainthillier on 3 Aug 2012

The K constant is 495 for 3 digits.
So the test with 691 is wrong.
Why tests with only one digit ?
Did I miss something ?

Grzegorz Knor on 14 Sep 2012

I don't understand the Test Suite for x = 3 and x = 1 too. what should I do in this case?

Tim on 14 Sep 2012

For x=3, the steps are 3000-0003=2997, 9972-2799=7173, etc.

Marcel on 25 Dec 2013

Problem description is confusing as there are different Kaprekar constants depending on the number of digits. [0 9 495 6174 for 1, 2,3 4 digits respectively.

Georges on 30 Sep 2014

getting this error:
Internal Server Error - Read

The server encountered an internal error or misconfiguration and was unable to complete your request.
Reference #3.c2c1ab8.1412090777.18623697
any ideas?

Thorsten Bartel on 22 May 2017

The problem should specify that any number with less than four digits should be filled up to four digits with leading zeros. (e.g. 3 -> 0003)

Jesús Zambrano on 29 Jan 2019

Very nice and interesting problem!

Florian Roscheck on 4 Feb 2019

I like the recursion aspect of this problem.

Ajay Pattassery on 12 Apr 2019

There is a small correction needed in the problem statement. Not all natural numbers, but 4 digit numbers can be reduced to Kaprekar number by the mentioned method. Similarly 3 digit numbers can be reduced to 495
https://en.wikipedia.org/wiki/D._R._Kaprekar

silin yang on 23 May 2019

How it works x = 1？？？？？

Sanzhar Askaruly on 21 Jul 2019

For those confused with test cases 2,3 and 5, like myself before, do conversion to 4-digit integer. Here is an example:
x = 1:
1000-0001 = 999
9990-0999 = 8991
9981-1899 = 8082
8820-0288 = 8532
8532-2358 = 6174
Therefore, y_correct = 5

Sanzhar Askaruly on 21 Jul 2019

love it!!!

Solution Comments

Solution 1712373

1 Comment

1 Comment

Serhii Tetora on 22 Jan 2019

How it works with x = 3, x = 691, x = 1?

Solution 1436614

2 Comments

2 Comments

Steffen Zelzer on 8 Feb 2018

This works for all but test 3 where it gives, in my opinion, the correct answer 8.

TurqoiseInt on 13 Feb 2018

But zero is not sorted in this solution. You should get same answer for input x = 691 and 6910. (And then no need for abs())

1) 9610-0169=9441
2) 9441-1449=7992
3) 9972-2799=7173
4) 7731-1377=6354
5) 6543-3456=3087
6) 8730-0378=8352
7) 8532-2358=6174

Solution 1431612

1 Comment

1 Comment

Chuck Saunders on 1 Feb 2018

This has been the lamest test so far. You need to pad the numbers with zeros to build it up to be a 4 digit number. Not explained in the rules.

Solution 1233350

1 Comment

1 Comment

Kwin on 18 Jul 2017

Not all test cases seem to be correct. For example x=3 would imply that the next value should be x=3-3=0, so y_correct=Inf and not 6

Solution 551096

1 Comment

1 Comment

Jon on 31 Dec 2014

An efficient lookup table solution

Solution 12492

1 Comment

1 Comment

Peter Wittenberg on 30 Mar 2012

Interesting - a recursive approach

Solution 9928

1 Comment

1 Comment

blaat on 10 Feb 2012

Could you stop doing this kind of thing? I think it would be a lot more fun if we could see the _actual_ best solution..

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