meetforever/GRPO_CL huggingface dataset:
user: Let's define `increasing` numbers as the numbers whose digits, read from left to right, are never less than the previous ones: 234559 is an example of increasing number.

Conversely, `decreasing` numbers have all the digits read from left to right so that no digits is bigger than the previous one: 97732 is an example of decreasing number.

You do not need to be the next Gauss to figure that all numbers with 1 or 2 digits are either increasing or decreasing: 00, 01, 02, ..., 98, 99 are all belonging to one of this categories (if not both, like 22 or 55): 101 is indeed the first number which does NOT fall into either of the categories. Same goes for all the numbers up to 109, while 110 is again a decreasing number.

Now your task is rather easy to declare (a bit less to perform): you have to build a function to return the total occurrences of all the increasing or decreasing numbers *below* 10 raised to the xth power (x will always be >= 0).

To give you a starting point, there are a grand total of increasing and decreasing numbers as shown in the table:

|Total  | Below
|---------------
|1      | 1
|10     | 10
|100    | 100
|475    | 1000
|1675   | 10000
|4954   | 100000
|12952  | 1000000

This means that your function will have to behave like this:
```python
total_inc_dec(0)==1
total_inc_dec(1)==10
total_inc_dec(2)==100
total_inc_dec(3)==475
total_inc_dec(4)==1675
total_inc_dec(5)==4954
total_inc_dec(6)==12952
```

**Tips:** efficiency and trying to figure out how it works are essential: with a brute force approach, some tests with larger numbers may take more than the total computing power currently on Earth to be finished in the short allotted time.

To make it even clearer, the increasing or decreasing numbers between in the range 101-200 are: [110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 122, 123, 124, 125, 126, 127, 128, 129, 133, 134, 135, 136, 137, 138, 139, 144, 145, 146, 147, 148, 149, 155, 156, 157, 158, 159, 166, 167, 168, 169, 177, 178, 179, 188, 189, 199, 200], that is 47 of them. In the following range, 201-300, there are 41 of them and so on, getting rarer and rarer.

**Trivia:** just for the sake of your own curiosity, a number which is neither decreasing of increasing is called a `bouncy` number, like, say, 3848 or 37294; also, usually 0 is not considered being increasing, decreasing or bouncy, but it will be for the purpose of this kata

Answer: {"fn_name": "total_inc_dec", "inputs": [[0], [1], [2], [3], [4], [5], [6], [10], [20], [50]], "outputs": [[1], [10], [100], [475], [1675], [4954], [12952], [277033], [40059819], [87959698326]]}
user: # Task
 You are given a string `s`. Every letter in `s` appears once. 
 
 Consider all strings formed by rearranging the letters in `s`. After ordering these strings in dictionary order, return the middle term. (If the sequence has a even length `n`, define its middle term to be the `(n/2)`th term.)

# Example

 For `s = "abc"`, the result should be `"bac"`.
```
The permutations in order are:
"abc", "acb", "bac", "bca", "cab", "cba"
So, The middle term is "bac".```

# Input/Output


 - `[input]` string `s`

  unique letters (`2 <= length <= 26`)

 - `[output]` a string

  middle permutation.

Answer: {"fn_name": "middle_permutation", "inputs": [["abc"], ["abcd"], ["abcdx"], ["abcdxg"], ["abcdxgz"]], "outputs": [["bac"], ["bdca"], ["cbxda"], ["cxgdba"], ["dczxgba"]]}
user: A product-sum number is a natural number N which can be expressed as both the product and the sum of the same set of numbers.

N = a1 × a2 × ... × ak = a1 + a2 + ... + ak

For example, 6 = 1 × 2 × 3 = 1 + 2 + 3.

For a given set of size, k, we shall call the smallest N with this property a minimal product-sum number. The minimal product-sum numbers for sets of size, k = 2, 3, 4, 5, and 6 are as follows.
```
k=2: 4 = 2 × 2 = 2 + 2
k=3: 6 = 1 × 2 × 3 = 1 + 2 + 3
k=4: 8 = 1 × 1 × 2 × 4 = 1 + 1 + 2 + 4
k=5: 8 = 1 × 1 × 2 × 2 × 2 = 1 + 1 + 2 + 2 + 2
k=6: 12 = 1 × 1 × 1 × 1 × 2 × 6 = 1 + 1 + 1 + 1 + 2 + 6
```

Hence for 2 ≤ k ≤ 6, the sum of all the minimal product-sum numbers is 4+6+8+12 = 30; note that 8 is only counted once in the sum.

Your task is to write an algorithm to compute the sum of all minimal product-sum numbers where 2 ≤ k ≤ n.

Courtesy of ProjectEuler.net

Answer: {"fn_name": "productsum", "inputs": [[3], [6], [12], [2]], "outputs": [[10], [30], [61], [4]]}
user: Motivation
---------

When compressing sequences of symbols, it is useful to have many equal symbols follow each other, because then they can be encoded with a run length encoding. For example, RLE encoding of `"aaaabbbbbbbbbbbcccccc"` would give something like `4a 11b 6c`.

(Look [here](http://www.codewars.com/kata/run-length-encoding/) for learning more about the run-length-encoding.)

Of course, RLE is interesting only if the string contains many identical consecutive characters. But what bout human readable text? Here comes the Burrows-Wheeler-Transformation.

Transformation
-------------
There even exists a transformation, which brings equal symbols closer together, it is called the **Burrows-Wheeler-Transformation**. The forward transformation works as follows: Let's say we have a sequence with length n, first write every shift of that string into a *n x n* matrix:
```
Input: "bananabar"

b a n a n a b a r
r b a n a n a b a
a r b a n a n a b
b a r b a n a n a
a b a r b a n a n
n a b a r b a n a
a n a b a r b a n
n a n a b a r b a
a n a n a b a r b
```
Then we sort that matrix by its rows. The output of the transformation then is the **last column** and the **row index** in which the original string is in:
```
               .-.
a b a r b a n a n
a n a b a r b a n
a n a n a b a r b
a r b a n a n a b
b a n a n a b a r <- 4
b a r b a n a n a
n a b a r b a n a
n a n a b a r b a
r b a n a n a b a
               '-'

Output: ("nnbbraaaa", 4)
```

```if:java
To handle the two kinds of output data, we will use the preloaded class `BWT`, whose contract is the following:

    public class BWT {
        
        public String s;
        public int n;
        
        public BWT(String s, int n)
        
        @Override public String  toString()
        @Override public boolean equals(Object o)
        @Override public int     hashCode()
    }

```

Of course we want to restore the original input, therefore you get the following hints:

1. The output contains the last matrix column.
2. The first column can be acquired by sorting the last column.
3. **For every row of the table:** Symbols in the first column follow on symbols in the last column, in the same way they do in the input string.
4. You don't need to reconstruct the whole table to get the input back.

Goal
----

The goal of this Kata is to write both, the `encode` and `decode` functions. Together they should work as the identity function on lists. (*Note:* For the empty input, the row number is ignored.)

Further studies
--------------
You may have noticed that symbols are not always consecutive, but just in proximity, after the transformation. If you're interested in how to deal with that, you should have a look at [this Kata](http://www.codewars.com/kata/move-to-front-encoding/).

Answer: {"fn_name": "encode", "inputs": [["bananabar"], ["Humble Bundle"], ["Mellow Yellow"]], "outputs": [[["nnbbraaaa", 4]], [["e emnllbduuHB", 2]], [["ww MYeelllloo", 1]]]}
user: # Longest Palindromic Substring (Linear)

A palindrome is a word, phrase, or sequence that reads the same backward as forward, e.g.,
'madam' or 'racecar'. Even the letter 'x' is considered a palindrome. 

For this Kata, you are given a string ```s```. Write a function that returns the longest _contiguous_ palindromic substring in ```s``` (it could be the entire string). In the event that there are multiple longest palindromic substrings, return the first to occur.

I'm not trying to trick you here:
- You can assume that all inputs are valid strings.
- Only the letters a-z will be used, all lowercase (your solution should, in theory, extend to more than just the letters a-z though). 

**NOTE:** Quadratic asymptotic complexity _(O(N^2))_ or above will __NOT__ work here.

-----

## Examples

### Basic Tests

```
Input: "babad"
Output: "bab"
(Note: "bab" occurs before "aba")
```
```
Input: "abababa"
Output: "abababa"
```
```
Input: "cbbd"
Output: "bb"
```

### Edge Cases

```
Input: "ab"
Output: "a"
```
```
Input: ""
Output: ""
```
-----

## Testing

Along with the example tests given:

 - There are **500** tests using strings of length in range [1 - 1,000]
 - There are **50** tests using strings of length in range [1,000 - 10,000]
 - There are **5** tests using strings of length in range [10,000 - 100,000]
 
All test cases can be passed within 10 seconds, but non-linear solutions will time out every time. _Linear performance is essential_. 

## Good Luck!

-----

This problem was inspired by [this](https://leetcode.com/problems/longest-palindromic-substring/) challenge on LeetCode. Except this is the performance version :^)

Answer: {"fn_name": "longest_palindrome", "inputs": [["babad"], ["madam"], ["dde"], ["ababbab"], ["abababa"], ["banana"], ["abba"], ["cbbd"], ["zz"], ["dddd"], [""], ["abcdefghijklmnopqrstuvwxyz"], ["ttaaftffftfaafatf"], ["bbaaacc"], ["m"]], "outputs": [["bab"], ["madam"], ["dd"], ["babbab"], ["abababa"], ["anana"], ["abba"], ["bb"], ["zz"], ["dddd"], [""], ["a"], ["aaftffftfaa"], ["aaa"], ["m"]]}
user: Given two strings s1 and s2, we want to visualize how different the two strings are.
We will only take into account the *lowercase* letters (a to z).
First let us count the frequency of each *lowercase* letters in s1 and s2.

`s1 = "A aaaa bb c"`

`s2 = "& aaa bbb c d"`

`s1 has 4 'a', 2 'b', 1 'c'`

`s2 has 3 'a', 3 'b', 1 'c', 1 'd'`

So the maximum for 'a' in s1 and s2 is 4 from s1; the maximum for 'b' is 3 from s2.
In the following we will not consider letters when the maximum of their occurrences
is less than or equal to 1.

We can resume the differences between s1 and s2 in the following string:
`"1:aaaa/2:bbb"`
where `1` in `1:aaaa` stands for string s1 and `aaaa` because the maximum for `a` is 4.
In the same manner `2:bbb` stands for string s2 and `bbb` because the maximum for `b` is 3.

The task is to produce a string in which each *lowercase* letters of s1 or s2 appears as many times as
its maximum if this maximum is *strictly greater than 1*; these letters will be prefixed by the 
number of the string where they appear with their maximum value and `:`. 
If the maximum is in s1 as well as in s2 the prefix is `=:`.

In the result, substrings (a substring is for example 2:nnnnn or 1:hhh; it contains the prefix) will be in decreasing order of their length and when they have the same length sorted in ascending lexicographic order (letters and digits - more precisely sorted by codepoint); the different groups will be separated by '/'. See examples and "Example Tests".

Hopefully other examples can make this clearer.

```
s1 = "my&friend&Paul has heavy hats! &"
s2 = "my friend John has many many friends &"
mix(s1, s2) --> "2:nnnnn/1:aaaa/1:hhh/2:mmm/2:yyy/2:dd/2:ff/2:ii/2:rr/=:ee/=:ss"

s1 = "mmmmm m nnnnn y&friend&Paul has heavy hats! &"
s2 = "my frie n d Joh n has ma n y ma n y frie n ds n&"
mix(s1, s2) --> "1:mmmmmm/=:nnnnnn/1:aaaa/1:hhh/2:yyy/2:dd/2:ff/2:ii/2:rr/=:ee/=:ss"

s1="Are the kids at home? aaaaa fffff"
s2="Yes they are here! aaaaa fffff"
mix(s1, s2) --> "=:aaaaaa/2:eeeee/=:fffff/1:tt/2:rr/=:hh"
```
# Note for Swift, R, PowerShell
The prefix `=:` is replaced by `E:`
``` 
s1 = "mmmmm m nnnnn y&friend&Paul has heavy hats! &"
s2 = "my frie n d Joh n has ma n y ma n y frie n ds n&"
mix(s1, s2) --> "1:mmmmmm/E:nnnnnn/1:aaaa/1:hhh/2:yyy/2:dd/2:ff/2:ii/2:rr/E:ee/E:ss"
```

Answer: {"fn_name": "mix", "inputs": [["Are they here", "yes, they are here"], ["looping is fun but dangerous", "less dangerous than coding"], [" In many languages", " there's a pair of functions"], ["Lords of the Fallen", "gamekult"], ["codewars", "codewars"], ["A generation must confront the looming ", "codewarrs"]], "outputs": [["2:eeeee/2:yy/=:hh/=:rr"], ["1:ooo/1:uuu/2:sss/=:nnn/1:ii/2:aa/2:dd/2:ee/=:gg"], ["1:aaa/1:nnn/1:gg/2:ee/2:ff/2:ii/2:oo/2:rr/2:ss/2:tt"], ["1:ee/1:ll/1:oo"], [""], ["1:nnnnn/1:ooooo/1:tttt/1:eee/1:gg/1:ii/1:mm/=:rr"]]}
user: ## Task

Given a positive integer, `n`, return the number of possible ways such that `k` positive integers multiply to `n`. Order matters.

**Examples**
```
n = 24
k = 2
(1, 24), (2, 12), (3, 8), (4, 6), (6, 4), (8, 3), (12, 2), (24, 1) -> 8

n = 100
k = 1
100 -> 1

n = 20
k = 3
(1, 1, 20), (1, 2, 10), (1, 4, 5), (1, 5, 4), (1, 10, 2), (1, 20, 1),
(2, 1, 10), (2, 2, 5), (2, 5, 2), (2, 10, 1), (4, 1, 5), (4, 5, 1),
(5, 1, 4), (5, 2, 2), (5, 4, 1), (10, 1, 2), (10, 2, 1), (20, 1, 1) -> 18
```
**Constraints**
`1 <= n <= 500_000_000`
and `1 <= k <= 1000`

Answer: {"fn_name": "multiply", "inputs": [[24, 2], [100, 1], [20, 3], [1, 2], [1000000, 3], [10, 2], [36, 4]], "outputs": [[8], [1], [18], [1], [784], [4], [100]]}
user: You are given 2 numbers is `n` and `k`. You need to find the number of integers between 1 and n (inclusive) that contains exactly `k` non-zero digit.

Example1 

`
almost_everywhere_zero(100, 1) return 19`

by following condition we have 19 numbers that have k = 1 digits( not count zero ) 
` [1,2,3,4,5,6,7,8,9,10,20,30,40,50,60,70,80,90,100]`


Example2  

`
almost_everywhere_zero(11, 2) return 1`

we have only `11` that has 2 digits(ten not count because zero is not count) 
` 11`





constrains


`1≤n<pow(10,100)`


`1≤k≤100`

Answer: {"fn_name": "almost_everywhere_zero", "inputs": [[100, 1], [11, 2], [20, 2], [101, 2], [10001, 2], [10001000, 2], [500309160, 2], [10000000000000000000000, 3], [10000000000000000000000, 21], [1203, 4]], "outputs": [[19], [1], [9], [82], [487], [1729], [2604], [1122660], [2407217760893271902598], [81]]}
user: [BasE91](http://base91.sourceforge.net/) is a method for encoding binary as ASCII characters. It is more efficient than Base64 and needs 91 characters to represent the encoded data.

The following ASCII charakters are used:

    'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789'
    '!#$%&()*+,./:;<=>?@[]^_`{|}~"'
    
Create two functions that encode strings to basE91 string and decodes the other way round. 

    b91encode('test') = 'fPNKd'
    b91decode('fPNKd') = 'test'
    
    b91decode('>OwJh>Io0Tv!8PE') = 'Hello World!'
    b91encode('Hello World!') = '>OwJh>Io0Tv!8PE'
    
Input strings are valid.

Answer: {"fn_name": "b91decode", "inputs": [["fPNKd"], [">OwJh>Io0Tv!8PE"]], "outputs": [["test"], ["Hello World!"]]}
user: The task is simply stated. Given an integer n (3 < n < 10^(9)), find the length of the smallest list of [*perfect squares*](https://en.wikipedia.org/wiki/Square_number) which add up to n. Come up with the best algorithm you can; you'll need it!

Examples:

sum_of_squares(17) = 2  17 = 16 + 1 (4 and 1 are perfect squares).
sum_of_squares(15) = 4  15 = 9 + 4 + 1 + 1. There is no way to represent 15 as the sum of three perfect squares.
sum_of_squares(16) = 1  16 itself is a perfect square.

Time constraints:

5 easy (sample) test cases: n < 20

5 harder test cases: 1000 < n < 15000

5 maximally hard test cases: 5 * 1e8 < n < 1e9

```if:java
300 random maximally hard test cases: 1e8 < n < 1e9
```
```if:c#
350 random maximally hard test cases: 1e8 < n < 1e9
```
```if:python
15 random maximally hard test cases: 1e8 < n < 1e9
```
```if:ruby
25  random maximally hard test cases: 1e8 < n < 1e9
```
```if:javascript
100 random maximally hard test cases: 1e8 < n < 1e9
```
```if:crystal
250 random maximally hard test cases: 1e8 < n < 1e9
```
```if:cpp
Random maximally hard test cases: 1e8 < n < 1e9
```

Answer: {"fn_name": "sum_of_squares", "inputs": [[15], [16], [17], [18], [19], [2017], [1008], [3456], [4000], [12321], [661915703], [999887641], [999950886], [999951173], [999998999]], "outputs": [[4], [1], [2], [2], [3], [2], [4], [3], [2], [1], [4], [1], [3], [2], [4]]}
user: This is a very simply formulated task. Let's call an integer number `N` 'green' if `N²` ends with all of the digits of `N`. Some examples:

`5` is green, because `5² = 25` and `25` ends with `5`.

`11` is not green, because `11² = 121` and `121` does not end with `11`.

`376` is green, because `376² = 141376` and `141376` ends with `376`.

Your task is to write a function `green` that returns `n`th green number, starting with `1` - `green (1) == 1`

---

## Data range

```if:haskell
`n <= 4000` for Haskell
```
```if:java
`n <= 5000` for Java
```
```if:python
`n <= 5000` for Python
```
```if:javascript
`n <= 3000` for JavaScript

Return values should be `String`s, and should be exact. A BigNum library is recommended.
```

Answer: {"fn_name": "green", "inputs": [[1], [2], [3], [4], [12], [13], [100], [110]], "outputs": [[1], [5], [6], [25], [2890625], [7109376], [6188999442576576769103890995893380022607743740081787109376], [9580863811000557423423230896109004106619977392256259918212890625]]}
user: Create a function that takes a positive integer and returns the next bigger number that can be formed by rearranging its digits. For example:

```
12 ==> 21
513 ==> 531
2017 ==> 2071
```

If the digits can't be rearranged to form a bigger number, return `-1` (or `nil` in Swift):

```
9 ==> -1
111 ==> -1
531 ==> -1
```

Answer: {"fn_name": "next_bigger", "inputs": [[12], [513], [2017], [414], [144], [123456789], [1234567890], [9876543210], [9999999999], [59884848459853]], "outputs": [[21], [531], [2071], [441], [414], [123456798], [1234567908], [-1], [-1], [59884848483559]]}
user: This is the performance version of [this kata](https://www.codewars.com/kata/59afff65f1c8274f270020f5).

---

Imagine two rings with numbers on them. The inner ring spins clockwise and the outer ring spins anti-clockwise. We start with both rings aligned on 0 at the top, and on each move we spin each ring by 1. How many moves will it take before both rings show the same number at the top again?

The inner ring has integers from 0 to innerMax and the outer ring has integers from 0 to outerMax, where innerMax and outerMax are integers >= 1.

```
e.g. if innerMax is 2 and outerMax is 3 then after
1 move: inner = 2, outer = 1
2 moves: inner = 1, outer = 2
3 moves: inner = 0, outer = 3
4 moves: inner = 2, outer = 0
5 moves: inner = 1, outer = 1
Therefore it takes 5 moves for the two rings to reach the same number
Therefore spinningRings(2, 3) = 5
```
```
e.g. if innerMax is 3 and outerMax is 2 then after
1 move: inner = 3, outer = 1
2 moves: inner = 2, outer = 2
Therefore it takes 2 moves for the two rings to reach the same number
spinningRings(3, 2) = 2
```

---

Test input range:

- `100` tests with `1 <= innerMax, outerMax <= 10000`
- `400` tests with `1 <= innerMax, outerMax <= 2^48`

Answer: {"fn_name": "spinning_rings", "inputs": [[5, 5], [2, 10], [10, 2], [7, 9], [1, 1], [16777216, 14348907]], "outputs": [[3], [13], [10], [4], [1], [23951671]]}
user: We want to generate all the numbers of three digits where:

- the sum of their digits is equal to 10.

- their digits are in increasing order (the numbers may have two or more equal contiguous digits)

The numbers that fulfill the two above constraints are: ```118, 127, 136, 145, 226, 235, 244, 334```

Make a function that receives two arguments:

- the sum of digits value 

- the desired number of digits for the numbers

The function should output an array with three values: \[1,2,3\]

1 - the total number of possible numbers

2 - the minimum number

3 - the maximum number

The example given above should be:

```python
find_all(10, 3) == [8, 118, 334]
```

If we have only one possible number as a solution, it should output a result like the one below:

```python
find_all(27, 3) == [1, 999, 999]
```

If there are no possible numbers, the function should output the empty array.

```python
find_all(84, 4) == []
```

The number of solutions climbs up when the number of digits increases.

```python
find_all(35, 6) == [123, 116999, 566666]
```

Features of the random tests:

* Number of tests: `112`
* Sum of digits value between `20` and `65`
* Amount of digits between `2` and `17`

Answer: {"fn_name": "find_all", "inputs": [[10, 3], [27, 3], [84, 4], [35, 6]], "outputs": [[[8, 118, 334]], [[1, 999, 999]], [[]], [[123, 116999, 566666]]]}
user: A format for expressing an ordered list of integers is to use a comma separated list of either

* individual integers
* or a range of integers denoted by the starting integer separated from the end integer in the range by a dash, '-'. The range includes all integers in the interval including both endpoints.  It is not considered a range unless it spans at least 3 numbers. For example "12,13,15-17"

Complete the solution  so that it takes a list of integers in increasing order and returns a correctly formatted string in the range format. 

**Example:**

```python
solution([-6, -3, -2, -1, 0, 1, 3, 4, 5, 7, 8, 9, 10, 11, 14, 15, 17, 18, 19, 20])
# returns "-6,-3-1,3-5,7-11,14,15,17-20"
```

```C#
RangeExtraction.Extract(new[] {-6, -3, -2, -1, 0, 1, 3, 4, 5, 7, 8, 9, 10, 11, 14, 15, 17, 18, 19, 20});
# returns "-6,-3-1,3-5,7-11,14,15,17-20"
```

*Courtesy of rosettacode.org*

Answer: {"fn_name": "solution", "inputs": [[[-6, -3, -2, -1, 0, 1, 3, 4, 5, 7, 8, 9, 10, 11, 14, 15, 17, 18, 19, 20]], [[-3, -2, -1, 2, 10, 15, 16, 18, 19, 20]], [[1, 2, 3, 4, 5]]], "outputs": [["-6,-3-1,3-5,7-11,14,15,17-20"], ["-3--1,2,10,15,16,18-20"], ["1-5"]]}
user: Your task in order to complete this Kata is to write a function which formats a duration, given as a number of seconds, in a human-friendly way.

The function must accept a non-negative integer. If it is zero, it just returns `"now"`. Otherwise,  the duration is expressed as a combination of `years`, `days`, `hours`, `minutes` and `seconds`.

It is much easier to understand with an example:

```Fortran
formatDuration (62)    // returns "1 minute and 2 seconds"
formatDuration (3662)  // returns "1 hour, 1 minute and 2 seconds"
```
```python
format_duration(62)    # returns "1 minute and 2 seconds"
format_duration(3662)  # returns "1 hour, 1 minute and 2 seconds"
```

**For the purpose of this Kata, a year is 365 days and a day is 24 hours.**

Note that spaces are important.

### Detailed rules

The resulting expression is made of components like `4 seconds`, `1 year`, etc.  In general, a positive integer and one of the valid units of time, separated by a space. The unit of time is used in plural if the integer is greater than 1.

The components are separated by a comma and a space (`", "`). Except the last component, which is separated by `" and "`, just like it would be written in English. 

A more significant units of time will occur before than a least significant one. Therefore, `1 second and 1 year` is not correct, but `1 year and 1 second` is.

Different components have different unit of times. So there is not repeated units like in `5 seconds and 1 second`.

A component will not appear at all if its value happens to be zero.  Hence, `1 minute and 0 seconds` is not valid, but it should be just `1 minute`.

 A unit of time must be used "as much as possible". It means that the function should not return `61 seconds`, but `1 minute and 1 second` instead.  Formally, the duration specified by  of a component must not be greater than any valid more significant unit of time.

Answer: {"fn_name": "format_duration", "inputs": [[0], [1], [62], [120], [3600], [3662], [15731080], [132030240], [205851834], [253374061], [242062374], [101956166], [33243586]], "outputs": [["now"], ["1 second"], ["1 minute and 2 seconds"], ["2 minutes"], ["1 hour"], ["1 hour, 1 minute and 2 seconds"], ["182 days, 1 hour, 44 minutes and 40 seconds"], ["4 years, 68 days, 3 hours and 4 minutes"], ["6 years, 192 days, 13 hours, 3 minutes and 54 seconds"], ["8 years, 12 days, 13 hours, 41 minutes and 1 second"], ["7 years, 246 days, 15 hours, 32 minutes and 54 seconds"], ["3 years, 85 days, 1 hour, 9 minutes and 26 seconds"], ["1 year, 19 days, 18 hours, 19 minutes and 46 seconds"]]}