Ruby one-liner, beats 100%/100%
```ruby []
# @param {Integer} n
# @return {String}
def base_neg2(n)
  Enumerator.produce([n, '']) { [(_1[0] - _1) / 2, _2 << _1[0].to_s] }.find { _1[0] == 0 }[1].reverse.ljust(1, '0')
end
```
['Ruby']