### self_converge
`Goal`: Given a number (with a minimum of 3 digits), return the number of iterations it takes to arrive at a derived number that converges on to itself, as per the following [Kaprekar][Kaprekar] routine. As a learning exercise, come up with a solution that uses recursion. The
following steps would be applicable to a number that originally had
exactly 4 digits.

0. Initialize a counter to count the number of iterations
1. Take any four-digit number `n`, using at least two different digits.
2. Arrange the digits in descending and then in ascending order to get two four-digit numbers, adding leading zeros if necessary. 
    - Add as many zeroes so that the width of the original number is maintained.
3. Subtract the smaller number from the bigger number. Let us call this `nseq`.
4. Check if `nseq` (the remainder) from Step 4 equals the previous value of `n`. If not, increment the iteration counter and go back to step 2 and perform it on the `nseq`. 

If the number of digits to start with was more than 4, convergence occurs on to a `cycle` of numbers. Therefore in Step 5, detect this cycle by comparing to not only the previous value, but to all previous values of `n`.
  - If there is a match, then return the count of iterations
  - If the sequence_number collapses to zero, then return -1 

[Kaprekar]: http://mathworld.wolfram.com/KaprekarRoutine.html

### Converge values  
While 3-digit numbers converge to the same unique number `k` which is also 3 digits long, all 4-digit numbers also converge to the same unique value `k1` which is 4 digits long. However, 5 digit numbers converge to any one of the following values: `53955, 59994, 61974, 62964, 63954, 71973, 74943, 75933, 82962, 83952`.

### Example
`1234`  ->   `4`
```
1. 4321 - 1234 =  3087 /   
2. 8730 - 378  =  8352 /   
3. 8532 - 2358 =  6174 /   
4. 7641 - 1467 =  6174 // same as previous
```
`414` -> `5`
```
1. 441 - 144 =  297 /   
2. 972 - 279 =  693 /   
3. 963 - 369 =  594 /
4. 954 - 459 =  495 /
5. 954 - 459 =  495 // same as previous
```
`50000` -> `4`
```
1. 50000 - 5      =  49995 /   
2. 99954 - 45999  =  53955 /  # first 
3. 95553 - 35559  =  59994 /
4. 99954 - 45999  =  53955 /  # 2nd time
```

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

def _disabled_input(*args, **kwargs):
    raise RuntimeError("input() is disabled in callable mode")
builtins.input = _disabled_input

ns = {"__name__": "__judge__"}  # prevent if __name__ == "__main__" blocks
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = '\ndef self_converge(number):\n    def sub_converge(n, l, min_length):\n        digits = [d for d in str(n)]\n        while len(digits) < min_length:\n            digits.append(\'0\')\n        asc = sorted(digits)\n        desc = asc[::-1]\n        if asc == desc:\n            return -1\n        next = int("".join(desc)) - int("".join(asc))\n        if next in l:\n            return len(l) + 1\n        else:\n            return sub_converge(next, l + [next], min_length)\n    return sub_converge(number, [], len(str(number)))'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

_target = None
_fn_name = 'self_converge'
if _fn_name:
    if "Solution" in ns:
        try:
            _obj = ns["Solution"]()
            if hasattr(_obj, _fn_name):
                _target = getattr(_obj, _fn_name)
        except Exception:
            _target = None
    if _target is None and _fn_name in ns and callable(ns[_fn_name]):
        _target = ns[_fn_name]
# Fallback: if there's exactly one user-defined function, use it
if _target is None:
    _cands = [v for k, v in ns.items() if callable(v) and getattr(v, "__module__", "") in ("__judge__", "__main__")]
    _user_funcs = [c for c in _cands if getattr(c, "__name__", "").startswith("_") is False]
    if len(_user_funcs) == 1:
        _target = _user_funcs[0]
if _target is None:
    raise RuntimeError(f"Could not resolve callable '{_fn_name}'")

_raw = sys.stdin.read().strip()
if _raw:
    _parsed = json.loads(_raw)
    if isinstance(_parsed, dict):
        _kwargs = _parsed.get("kwargs", {})
        _args = _parsed.get("args", [])
    else:
        _args, _kwargs = _parsed, {}
else:
    _args, _kwargs = [], {}
if not isinstance(_args, (list, tuple)):
    _args = [_args]
_res = _target(*_args, **_kwargs)
try:
    sys.stdout.write(repr(_res) + "\n")
except Exception:
    pass

```

Measured memory_bytes: 8127