hhhhhhhhhhhhhhhh

Every "k" number. (Be aware, it is not index, by itself! At the end of line, somehow you must move it back to right place at left side - I know dumbly described, but I do not know if I can here give more than small hints. If not marked as spoiler - what you will not see, if not completed :D ...)

e.g. n7 k3 (first sample test) = you have 1,2,3,4,5,6,7. First you remove third position, what is number 3. Then again 3rd position from their, so skip number 4,5 and remove the 3rd in row what is 6. Then again third - you skip 6, move back to begin, skip 1 and remove number 2 ... and some magic and you have it algorithmized :D

the language does not really matter here

It does actually matter, because every language has its own tests. It is hard to validate and fix a problem if one doesn't know where to start looking.

Hi JohanWiltink,

the language does not really matter here, but I was doing the C# one.
In my opinion there should be at least two more testcases that explicitly check that a certain solution is not correct.

• all values are 5. A simple sum check would say this is valid and accept it as a valid solution
• lines and columns correct, but blocks of 3x3 not

Hope this is clearer now.

Best regards,
Robert

easy kata i think this one should be 5 kata

ok thanks

Your task is to divide n into some different integers

different

8 is given as 3,5 3*5=15 thats not the maximum product
Input : n = 8
Output : 18
8 = 2 + 3 + 3 and 2 * 3 * 3 = 18
is maximum possible product.
why would i use 3,5 instead of 2,3,3??

i got it now thanks

The kata is named Maximum Product, and the description neatly states you need to maximize the product of these integers. Isn't 3*5 > 2*6?

8 can be divided by 2,6 why would i consider 3,5?

a pointer which you should set as the length of your return array

what(size_t *outsz)is used for ??

you can just copy the name of this kata and searchh it in google plenty of explanation

Read the instructions again, the second argument is the steps value you should use.