meetforever/GRPO_CL huggingface dataset:
user: This kata is a continuation of [Part 1](http://www.codewars.com/kata/the-fibfusc-function-part-1). The `fibfusc` function is defined recursively as follows:

    fibfusc(0) = (1, 0)
    fibfusc(1) = (0, 1)
    fibfusc(2n) = ((x + y)(x - y), y(2x + 3y)), where (x, y) = fibfusc(n)
    fibfusc(2n + 1) = (-y(2x + 3y), (x + 2y)(x + 4y)), where (x, y) = fibfusc(n)

Your job is to produce the code for the `fibfusc` function. In this kata, your function will be tested with large values of n (up to 2000 bits), so you should be concerned about stack overflow and timeouts. Moreover, since the `fibfusc` function grows exponentially, your function should take an extra argument `num_digits` with a default value of `None`. When not `None`, `num_digits` takes a positive `int` value and the returned values shall be truncated to the last `num_digits` digits. 

For example, let `n = 101`. With `num_digits` not set or set to `None`, `fibfusc` should return: 
`(-280571172992510140037611932413038677189525L, 734544867157818093234908902110449296423351L)`. If `num_digits = 12`, the function should return `(-38677189525L, 449296423351L)`. Notice in particular, that for any value of `n` greater than 1, `x` is negative, and the truncated value of `x` should also be negative.

Hint 1: Use modulo `10 ** num_digits` arithmetic to keep all intermediary results small.

Hint 2: Consider using an iterative ["number climber"](http://www.codewars.com/kata/number-climber) to avoid stack overflows.

Answer: {"fn_name": "fibfusc", "inputs": [[0], [1]], "outputs": [[[1, 0]], [[0, 1]]]}
user: To give credit where credit is due: This problem was taken from the ACMICPC-Northwest Regional Programming Contest. Thank you problem writers.

You are helping an archaeologist decipher some runes. He knows that this ancient society used a Base 10 system, and that they never start a number with a leading zero. He's figured out most of the digits as well as a few operators, but he needs your help to figure out the rest.

The professor will give you a simple math expression, of the form

```
[number][op][number]=[number]
```

He has converted all of the runes he knows into digits. The only operators he knows are addition (`+`),subtraction(`-`), and multiplication (`*`), so those are the only ones that will appear. Each number will be in the range from -1000000 to 1000000, and will consist of only the digits 0-9, possibly a leading -, and maybe a few ?s. If there are ?s in an expression, they represent a digit rune that the professor doesn't know (never an operator, and never a leading -). All of the ?s in an expression will represent the same digit (0-9), and it won't be one of the other given digits in the expression. No number will begin with a 0 unless the number itself is  0, therefore 00 would not be a valid number. 

Given an expression, figure out the value of the rune represented by the question mark. If more than one digit works, give the lowest one. If no digit works, well, that's bad news for the professor - it means that he's got some of his runes wrong. output -1 in that case.

Complete the method to solve the expression to find the value of the unknown rune. The method takes a string as a paramater repressenting the expression and will return an int value representing the unknown rune or -1 if no such rune exists.

~~~if:php
**Most of the time, the professor will be able to figure out most of the runes himself, but sometimes, there may be exactly 1 rune present in the expression that the professor cannot figure out (resulting in all question marks where the digits are in the expression) so be careful ;)**
~~~

Answer: {"fn_name": "solve_runes", "inputs": [["123?45*?=?"], ["?*123?45=?"], ["??605*-63=-73???5"], ["123?45+?=123?45"], ["?8?170-1?6256=7?2?14"], ["?38???+595???=833444"], ["123?45-?=123?45"], ["-7715?5--484?00=-28?9?5"], ["50685?--1?5630=652?8?"], ["??+??=??"], ["-?56373--9216=-?47157"]], "outputs": [[0], [0], [1], [0], [9], [2], [0], [6], [4], [-1], [8]]}
user: With your birthday coming up soon, your eccentric friend sent you a message to say "happy birthday":

    hhhappyyyy biirrrrrthddaaaayyyyyyy to youuuu
    hhapppyyyy biirtttthdaaay too youuu
    happy birrrthdayy to youuu
    happpyyyy birrtthdaaay tooooo youu

At first it looks like a song, but upon closer investigation, you realize that your friend hid the phrase "happy birthday" thousands of times inside his message. In fact, it contains it more than 2 million times! To thank him, you'd like to reply with exactly how many times it occurs.

To count all the occurences, the procedure is as follows: look through the paragraph and find a `'h'`; then find an `'a'` later in the paragraph; then find an `'p'` after that, and so on. Now count the number of ways in which you can choose letters in this way to make the full phrase.

More precisely, given a text string, you are to determine how many times the search string appears as a sub-sequence of that string.

Write a function called `countSubsequences` that takes two arguments: `needle`, the string to be search for and `haystack`, the string to search in. In our example, `"happy birthday"` is the needle and the birthday message is the haystack. The function should return the number of times `needle` occurs as a sub-sequence of `haystack`.  Spaces are also considered part of the needle.

Since the answers can be very large, return only the last 8 digits of the answer in case it exceeds 8 digits. The answers to the test cases will all be shorter than 8 digits.

Answer: {"fn_name": "count_subsequences", "inputs": [["happy birthday", "appyh appy birth day"], ["happy birthday", "hhaappyy bbiirrtthhddaayy"], ["happy birthday", "happy holidays"], ["happy birthday", "happy birthday"], ["happy", "hhhappy"], ["happy birthday", "hhhappyyyy biirrrrrthddaaaayyyyyyy to youuuu"]], "outputs": [[1], [2048], [0], [1], [3], [7680]]}
user: This kata is  blatantly copied from  inspired by This Kata
Welcome

this is the second in the series of the string iterations kata!

Here we go!

---------------------------------------------------------------------------------

We have a string s

Let's say you start with this: "String"

The first thing you do is reverse it: "gnirtS"

Then you will take the string from the 1st position and reverse it again: "gStrin"

Then you will take the string from the 2nd position and reverse it again: "gSnirt"

Then you will take the string from the 3rd position and reverse it again: "gSntri"

Continue this pattern until you have done every single position, and then you will return the string you have created. For this particular string, you would return: 
"gSntir"

now,

The Task:

In this kata, we also have a number x

take that reversal function, and apply it to the string x times.


return the result of the string after applying the reversal function to it x times.

    example where s = "String" and x = 3:

    after 0 iteration  s = "String"
    after 1 iteration  s = "gSntir"
    after 2 iterations s = "rgiStn"
    after 3 iterations s = "nrtgSi"
    
    so you would return "nrtgSi".

 Note 

String lengths may exceed 2 million


x exceeds a billion


be read to optimize



if this is too hard, go here https://www.codewars.com/kata/string-%3E-n-iterations-%3E-string/java

Answer: {"fn_name": "string_func", "inputs": [["This is a string exemplification!", 0], ["String for test: incommensurability", 1], ["Ohh Man God Damn", 7], ["Ohh Man God Damnn", 19], ["I like it!", 1234], ["codingisfornerdsyounerd", 10101010], ["this_test_will_hurt_you", 12345678987654321]], "outputs": [["This is a string exemplification!"], ["ySttirliinbga rfuosrn etmemsotc:n i"], [" nGOnmohaadhMD  "], ["haG mnad MhO noDn"], ["iitkIl !e "], ["fonroisreinrddgdneyscou"], ["tt_rt_swuhyeihiotl_su_l"]]}
user: The aim is to calculate `exponential(x)` (written `exp(x)` in most math libraries) as an irreducible fraction, the numerator of this fraction having a given number of digits.

We call this function `expand`, it takes two parameters, `x` of which we want to evaluate the exponential, `digits` which is the required number of digits for the numerator. 

The function `expand` will return an array of the form `[numerator, denominator]`; we stop the loop in the Taylor expansion (see references below) when the numerator has a number of digits `>=` the required number of digits

# Examples:
```
expand(1, 2) --> 65/24 (we will write this [65, 24] or (65, 24) in Haskell; 
65/24 ~ 2.708...)

expand(2, 5) --> [20947, 2835]

expand(3, 10) --> [7205850259, 358758400]

expand(1.5, 10) --> [36185315027,8074035200]
```
**Note**
```expand(1,5) = [109601, 40320]``` is the same as ```expand(1, 6)```

#Method:
As said above the way here is to use `Taylor expansion` of the exponential function though it is not always the best approximation by a rational.

#References:
https://en.wikipedia.org/wiki/Exponential_function#Formal_definition

http://www.efunda.com/math/taylor_series/exponential.cfm

Answer: {"fn_name": "expand", "inputs": [[1, 2], [1, 5], [1, 6], [10, 3], [10, 4], [10, 40], [10, 39], [10, 38], [1.5, 10], [1.6, 10], [1.7, 10], [1.7, 12], [1.7, 15], [1.8, 20], [2.0, 20], [1.95, 60]], "outputs": [[[65, 24]], [[109601, 40320]], [[109601, 40320]], [[683, 3]], [[1933, 3]], [[1239343290542858204293897268755807211243, 56266098342252742161373730812700463]], [[1239343290542858204293897268755807211243, 56266098342252742161373730812700463]], [[1239343290542858204293897268755807211243, 56266098342252742161373730812700463]], [[36185315027, 8074035200]], [[27425286391, 5537109375]], [[3933851989, 720000000]], [[22069373989681, 4032000000000]], [[1986362246947787, 362880000000000]], [[13246837419514181887, 2189687500000000000]], [[355236915407681299637, 48076088562799171875]], [[12382830214749612215651955187527316512224932853910667762866497, 1761755672416929554471321600000000000000000000000000000000000]]]}
user: # How many ways can you make the sum of a number?

From wikipedia: https://en.wikipedia.org/wiki/Partition_(number_theory)#

>In number theory and combinatorics, a partition of a positive integer *n*, also called an *integer partition*, is a way of writing n as a sum of positive integers. Two sums that differ only in the order of their summands are considered the same partition. If order matters, the sum becomes a composition. For example, 4 can be partitioned in five distinct ways:
```
4
3 + 1
2 + 2
2 + 1 + 1
1 + 1 + 1 + 1
```

## Examples

### Basic

```python
exp_sum(1) # 1
exp_sum(2) # 2  -> 1+1 , 2
exp_sum(3) # 3 -> 1+1+1, 1+2, 3
exp_sum(4) # 5 -> 1+1+1+1, 1+1+2, 1+3, 2+2, 4
exp_sum(5) # 7 -> 1+1+1+1+1, 1+1+1+2, 1+1+3, 1+2+2, 1+4, 5, 2+3

exp_sum(10) # 42
```

### Explosive

```python
exp_sum(50) # 204226
exp_sum(80) # 15796476
exp_sum(100) # 190569292
```

See [here](http://www.numericana.com/data/partition.htm) for more examples.

Answer: {"fn_name": "exp_sum", "inputs": [[1], [2], [3], [4], [5], [20], [30], [40], [43], [60], [70], [90], [200], [275]], "outputs": [[1], [2], [3], [5], [7], [627], [5604], [37338], [63261], [966467], [4087968], [56634173], [3972999029388], [1520980492851175]]}
user: When we attended middle school were asked to simplify mathematical expressions like "3x-yx+2xy-x" (or usually bigger), and that was easy-peasy ("2x+xy"). But tell that to your pc and we'll see! 

Write a function: `simplify`, that takes a string in input, representing a *multilinear non-constant polynomial in integers coefficients* (like `"3x-zx+2xy-x"`), and returns another string as output where the same expression has been simplified in the following way ( `->` means application of `simplify`):

- All possible sums and subtraction of equivalent monomials ("xy==yx") has been done, e.g.: `"cb+cba" -> "bc+abc"`, `"2xy-yx" -> "xy"`, `"-a+5ab+3a-c-2a" -> "-c+5ab"`

- All monomials appears in order of increasing number of variables, e.g.: `"-abc+3a+2ac" -> "3a+2ac-abc"`, `"xyz-xz" -> "-xz+xyz"`
 
- If two monomials have the same number of variables, they appears in lexicographic order, e.g.: `"a+ca-ab" -> "a-ab+ac"`, `"xzy+zby" ->"byz+xyz"`
  
- There is no leading `+` sign if the first coefficient is positive, e.g.: `"-y+x" -> "x-y"`, but no restrictions for `-`:  `"y-x" ->"-x+y"`

---

__N.B.__ to keep it simplest, the string in input is restricted to represent only *multilinear non-constant polynomials*, so you won't find something like `-3+yx^2'. **Multilinear** means in this context: **of degree 1 on each variable**.

**Warning**: the string in input can contain arbitrary variables represented by lowercase characters in the english alphabet.

__Good Work :)__

Answer: {"fn_name": "simplify", "inputs": [["dc+dcba"], ["2xy-yx"], ["-a+5ab+3a-c-2a"], ["-abc+3a+2ac"], ["xyz-xz"], ["a+ca-ab"], ["xzy+zby"], ["-y+x"], ["y-x"], ["3a+b+4ac+bc-ab+3a-cb-a-a"], ["+n-5hn+7tjhn-4nh-3n-6hnjt+2jhn+9hn"], ["-8fk+5kv-4yk+7kf-qk+yqv-3vqy+4ky+4kf+yvqkf"]], "outputs": [["cd+abcd"], ["xy"], ["-c+5ab"], ["3a+2ac-abc"], ["-xz+xyz"], ["a-ab+ac"], ["byz+xyz"], ["x-y"], ["-x+y"], ["4a+b-ab+4ac"], ["-2n+2hjn+hjnt"], ["3fk-kq+5kv-2qvy+fkqvy"]]}
user: You are given a binary tree:

```python
class Node:
    def __init__(self, L, R, n):
        self.left = L
        self.right = R
        self.value = n
```

Your task is to return the list with elements from tree sorted by levels, which means the root element goes first, then root children (from left to right) are second and third, and so on.
```if:ruby
Return empty array if root is `nil`.
```
```if:haskell
Return empty list if root is `Nothing`.
```
```if:python
Return empty list if root is `None`.
```
```if:csharp
Return empty list if root is 'null'.
```
```if:java
Return empty list is root is 'null'.
```

Example 1 - following tree:

                     2
                8        9
              1  3     4   5

Should return following list:

    [2,8,9,1,3,4,5]

Example 2 - following tree:

                     1
                8        4
                  3        5
                             7
Should return following list:

    [1,8,4,3,5,7]

Answer: {"fn_name": "tree_by_levels", "inputs": [[null]], "outputs": [[[]]]}
user: In this Kata we focus on finding a sum S(n) which is the total number of divisors taken for all natural numbers less or equal to n. More formally, we investigate the sum of n components denoted by d(1) + d(2) + ... + d(n) in which for any i starting from 1 up to n the value of d(i) tells us how many distinct numbers divide i without a remainder. 

Your solution should work for possibly large values of n without a timeout.
Assume n to be greater than zero and not greater than 999 999 999 999 999.
Brute force approaches will not be feasible options in such cases. It is fairly simple to conclude that for every n>1 there holds a recurrence S(n) = S(n-1) + d(n) with initial case S(1) = 1.

For example:
S(1) = 1
S(2) = 3
S(3) = 5
S(4) = 8
S(5) = 10

But is the fact useful anyway? If you find it is rather not, maybe this will help: 

Try to convince yourself that for any natural k, the number S(k) is the same as the number of pairs (m,n) that solve the inequality mn <= k in natural numbers.
Once it becomes clear, we can think of a partition of all the solutions into classes just by saying that a pair (m,n) belongs to the class indexed by n.
The question now arises if it is possible to count solutions of n-th class. If f(n) stands for the number of solutions that belong to n-th class, it means that S(k) = f(1) + f(2) + f(3) + ...

The reasoning presented above leads us to some kind of a formula for S(k), however not necessarily the most efficient one. Can you imagine that all the solutions to inequality mn <= k can be split using sqrt(k) as pivotal item?

Answer: {"fn_name": "count_divisors", "inputs": [[5], [10], [20], [59], [105], [785], [1001], [8009], [9999999999999], [9999999999998], [9999999999995], [9999999949950]], "outputs": [[10], [27], [66], [249], [510], [5364], [7077], [73241], [300880375389561], [300880375389537], [300880375389493], [300880373832097]]}
user: Complete the function/method (depending on the language) to return `true`/`True` when its argument is an array that has the same nesting structures and same corresponding length of nested arrays as the first array.

For example:

```python
# should return True
same_structure_as([ 1, 1, 1 ], [ 2, 2, 2 ] )
same_structure_as([ 1, [ 1, 1 ] ], [ 2, [ 2, 2 ] ] )

# should return False 
same_structure_as([ 1, [ 1, 1 ] ], [ [ 2, 2 ], 2 ] )
same_structure_as([ 1, [ 1, 1 ] ], [ [ 2 ], 2 ] )

# should return True
same_structure_as([ [ [ ], [ ] ] ], [ [ [ ], [ ] ] ] )

# should return False
same_structure_as([ [ [ ], [ ] ] ], [ [ 1, 1 ] ] )
```

~~~if:javascript
For your convenience, there is already a function 'isArray(o)' declared and defined that returns true if its argument is an array, false otherwise.
~~~

~~~if:php
You may assume that all arrays passed in will be non-associative.
~~~

Answer: {"fn_name": "same_structure_as", "inputs": [[[1, [1, 1]], [[2, 2], 2]], [[1, [1, 1]], [2, [2]]], [[[[], []]], [[[], []]]], [[[[], []]], [[1, 1]]], [[1, [[[1]]]], [2, [[[2]]]]], [[], 1], [[], {}], [[1, "[", "]"], ["[", "]", 1]]], "outputs": [[false], [false], [true], [false], [true], [false], [false], [true]]}
user: ## Description

Given an array X of positive integers, its elements are to be transformed by running the following operation on them as many times as required:

```if X[i] > X[j] then X[i] = X[i] - X[j]```

When no more transformations are possible, return its sum ("smallest possible sum").

For instance, the successive transformation of the elements of input X = [6, 9, 21] is detailed below:
```
X_1 = [6, 9, 12] # -> X_1[2] = X[2] - X[1] = 21 - 9
X_2 = [6, 9, 6]  # -> X_2[2] = X_1[2] - X_1[0] = 12 - 6
X_3 = [6, 3, 6]  # -> X_3[1] = X_2[1] - X_2[0] = 9 - 6
X_4 = [6, 3, 3]  # -> X_4[2] = X_3[2] - X_3[1] = 6 - 3
X_5 = [3, 3, 3]  # -> X_5[1] = X_4[0] - X_4[1] = 6 - 3
```
The returning output is the sum of the final transformation (here 9).

## Example

## Solution steps:

## Additional notes:

There are performance tests consisted of very big numbers and arrays of size at least 30000. Please write an efficient algorithm to prevent timeout.

Answer: {"fn_name": "solution", "inputs": [[[6, 9, 21]], [[9]], [[30, 12]], [[11, 22]], [[1, 21, 55]], [[4, 16, 24]], [[3, 13, 23, 7, 83]], [[60, 12, 96, 48, 60, 24, 72, 36, 72, 72, 48]], [[71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71]]], "outputs": [[9], [9], [12], [22], [3], [12], [5], [132], [923]]}
user: # Unflatten a list (Harder than easy)

This is the harder version of Unflatten a list (Easy)

So you have again to build a method, that creates new arrays, that can be flattened!

# Shorter: You have to unflatten a list/an array.

You get an array of integers and have to unflatten it by these rules:
```
- You have to do several runs. The depth is the number of runs, you have to do.
- In every run you have to switch the direction. First run from left, next run from right. Next left...
Every run has these rules:
- You start at the first number (from the direction).
- Take for every number x the remainder of the division by the number of still available elements (from 
  this position!) to have the number for the next decision.
- If the remainder-value is smaller than 3, take this number x (NOT the remainder-Value) direct
  for the new array and continue with the next number.
- If the remainder-value (e.g. 3) is greater than 2, take the next remainder-value-number (e.g. 3)
  elements/numbers (inclusive the number x, NOT the remainder-value) as a sub-array in the new array.
  Continue with the next number/element AFTER this taken elements/numbers.
- Every sub-array in the array is independent and is only one element for the progress on the array. 
  For every sub-array you have to follow the same rules for unflatten it.
  The direction is always the same as the actual run.
```

Sounds complicated? Yeah, thats why, this is the harder version...
Maybe an example will help.

```
 Array: [4, 5, 1, 7, 1] Depth: 2 -> [[ 4, [ 5, 1, 7 ] ], 1]
 
Steps: 
First run: (start from left side!)
1. The first number is 4. The number is smaller than the number of remaining elements, so it is the remainder-value (4 / 5 -> remainder 4).
   So 4 numbers (4, 5, 1, 7) are added as sub-array in the new array.
2. The next number is 1. It is smaller than 3, so the 1 is added direct to the new array.
Now we have --> [[4, 5, 1, 7], 1]

Second run: (start from right side!)
1. The last number (first from other side) is 1. So the 1 is added direct to the new array.
2. The next element is the sub-array. So we use the rules for this.
2a.The last number is 7. There are 4 elements in the array. So for the next decision you have to
   take the remainder from 7 / 4 -> 3. So 3 numbers (5, 1, 7) are added as sub-array in the 
   new array.
2b.Now there is the 4 and only one element last in this array. 4 / 1 -> remainder 0. It is smaller
   than 3. So the 4 is added direct to the new array.
Now we have --> [[ 4, [ 5, 1, 7 ] ], 1]
```

The given array will always contain numbers. There will only be numbers > 0.


Have fun coding it and please don't forget to vote and rank this kata! :-) 

I have created other katas. Have a look if you like coding and challenges.

Answer: {"fn_name": "unflatten", "inputs": [[[4, 5, 1, 7, 1], 2], [[12, 1, 5, 3, 1, 55, 2, 3, 7, 8, 1], 3]], "outputs": [[[[4, [5, 1, 7]], 1]], [[12, 1, [5, [3, 1, 55], 2], [3, 7, 8], 1]]]}
user: Write a function that takes a positive integer and returns the next smaller positive integer containing the same digits.

For example:

```python
next_smaller(21) == 12
next_smaller(531) == 513
next_smaller(2071) == 2017
```

Return -1 (for `Haskell`: return `Nothing`, for `Rust`: return `None`), when there is no smaller number that contains the same digits. Also return -1 when the next smaller number with the same digits would require the leading digit to be zero.

```python 
next_smaller(9) == -1
next_smaller(135) == -1
next_smaller(1027) == -1  # 0721 is out since we don't write numbers with leading zeros
```
```ruby 
next_smaller(9) == -1
next_smaller(135) == -1
next_smaller(1027) == -1  # 0721 is out since we don't write numbers with leading zeros
```

 * some tests will include very large numbers.
 * test data only employs positive integers.

*The function you write for this challenge is the inverse of this kata: "[Next bigger number with the same digits](http://www.codewars.com/kata/next-bigger-number-with-the-same-digits)."*

Answer: {"fn_name": "next_smaller", "inputs": [[21], [907], [531], [1027], [441], [123456798], [513], [351], [315], [153], [135], [100], [2071], [1207], [414], [123456789], [29009], [1234567908], [9999999999], [59884848483559], [1023456789], [51226262651257], [202233445566], [506789]], "outputs": [[12], [790], [513], [-1], [414], [123456789], [351], [315], [153], [135], [-1], [-1], [2017], [1072], [144], [-1], [20990], [1234567890], [-1], [59884848459853], [-1], [51226262627551], [-1], [-1]]}
user: Alice has a hand of cards, given as an array of integers.
Now she wants to rearrange the cards into groups so that each group is size W, and consists of W consecutive cards.
Return true if and only if she can.
 


Example 1:
Input: hand = [1,2,3,6,2,3,4,7,8], W = 3
Output: true
Explanation: Alice's hand can be rearranged as [1,2,3],[2,3,4],[6,7,8].
Example 2:
Input: hand = [1,2,3,4,5], W = 4
Output: false
Explanation: Alice's hand can't be rearranged into groups of 4.

 
Constraints:

1 <= hand.length <= 10000
0 <= hand[i] <= 10^9
1 <= W <= hand.length

Note: This question is the same as 1296: https://leetcode.com/problems/divide-array-in-sets-of-k-consecutive-numbers/

Answer: {"fn_name": "isNStraightHand", "inputs": [[[1, 2, 3, 6, 2, 3, 4, 7, 8], 3]], "outputs": [true]}