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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def nsmallest(n, iterable, key=None):
 """Find the n smallest elements in a dataset. Equivalent to: sorted(iterable, key=key)[:n] """


    # Short-cut for n==1 is to use min()
    if n == 1:
        it = iter(iterable)
        sentinel = object()
        if key is None:
            result = min(it, default=sentinel)
        else:
            result = min(it, default=sentinel, key=key)
        return [] if result is sentinel else [result]

    # When n>=size, it's faster to use sorted()
    try:
        size = len(iterable)
    except (TypeError, AttributeError):
        pass
    else:
        if n >= size:
            return sorted(iterable, key=key)[:n]

    # When key is none, use simpler decoration
    if key is None:
        it = iter(iterable)
        # put the range(n) first so that zip() doesn't
        # consume one too many elements from the iterator
        result = [(elem, i) for i, elem in zip(range(n), it)]
        if not result:
            return result
        _heapify_max(result)
        top = result[0][0]
        order = n
        _heapreplace = _heapreplace_max
        for elem in it:
            if elem < top:
                _heapreplace(result, (elem, order))
                top = result[0][0]
                order += 1
        result.sort()
        return [r[0] for r in result]

    # General case, slowest method
    it = iter(iterable)
    result = [(key(elem), i, elem) for i, elem in zip(range(n), it)]
    if not result:
        return result
    _heapify_max(result)
    top = result[0][0]
    order = n
    _heapreplace = _heapreplace_max
    for elem in it:
        k = key(elem)
        if k < top:
            _heapreplace(result, (k, order, elem))
            top = result[0][0]
            order += 1
    result.sort()
    return [r[2] for r in result]