Related to MrZizoScream's Product Array kata. You might want to solve that one first :)

```if:javascript
**Note:** Node 10 has now been enabled, and you can now use its BigInt capabilities if you wish, though your resulting array must still contain strings (e.g. "99999999999", not 9999999999n)

Pre-node 10: You will need to use the BigNumber.js library! Please use `.toFixed(0)` or `.toPrecision()` to round instead of `.toString(10)`, as the latter is _very_ slow
```

This is an adaptation of a problem I came across on LeetCode. 

Given an array of numbers, your task is to return a new array where each index (`new_array[i]`) is equal to the product of the original array, except for the number at that index (`array[i]`).

**Two things to keep in mind:**

* Zeroes will be making their way into some of the arrays you are given
* O(n^2) solutions will not pass.

Examples:

**Note**: All inputs will be valid arrays of nonzero length.

Have fun! Please upvote if you enjoyed :)

```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 = 'from functools import reduce\n# Since it must be less thatn O(n^2) the solution may not involve mul(Array[:]) for each\n# element of Array since mul would be O(n-1) and you would do O(n) calls to mul.\n    \n# Since it must be less thatn O(n^2) the solution may not involve mul(Array[:]) for each\n# element of Array since mul would be O(n-1) and you would do O(n) calls to mul.\n    \n# The only way that occurs to me is calling mul just once [that is O(n)] and then mapping / on the\n# result Array [that is another O(n)], but since the calls are concatenated the final complexity\n# is O(2n)\n\n# Also, since \n\ndef product_sans_n(nums):\n    # The case for two or more zeros is trivial and can be handled separately:\n    # Just a list of N zeroes where N is the length of nums\n    if len([ x for x in nums if x == 0]) > 1:\n        return [0] * len(nums)\n    \n    # Now, if there is only one zero this must be stripped from the product P\n    # The product can be computed by the functool reduce: this is O(n)\n    from operator import mul\n    from functools import reduce\n    onezero = 0 in nums\n    product = reduce(mul, nums if not onezero else nums[:nums.index(0)]+nums[nums.index(0)+1:])\n    \n    # Now, we iterate through the original list dividing the product P by each element\n    # The case of one zero is handled separately\n    # There is only one list traversing with map: this is O(n)\n    if onezero:\n        return list(map((lambda x: 0 if x != 0 else product), nums))\n    else:\n        return list(map((lambda x: product / x), nums))\n    \n    # In the end, the algorithm is O(2*n)\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

_target = None
_fn_name = 'product_sans_n'
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: 6016