"""Heap queue algorithm (a.k.a. priority queue).
Max Heaps are arrays for which a[k] >= a[2*k+1] and a[k] >= a[2*k+2] for
all k, counting elements from 0.  For the sake of comparison,
non-existing elements are considered to be infinite.  The interesting
property of a max heap is that a[0] is always its largest element.
Usage:
heap = []            # creates an empty heap
heappush(heap, item) # pushes a new item on the heap
item = heappop(heap) # pops the largest item from the heap
item = heap[0]       # largest item on the heap without popping it
heapify(x)           # transforms list into a max heap, in-place, in linear time
item = heapreplace(heap, item) # pops and returns largest item, and adds
                               # new item; the heap size is unchanged
Our API differs from textbook heap algorithms as follows:
- We use 0-based indexing.  This makes the relationship between the
  index for a node and the indexes for its children slightly less
  obvious, but is more suitable since Python uses 0-based indexing.
Our heappop() method returns the largest item. heap[0] is the largest item
and heap.sort() disrupts the heap invariant !
"""

# Original code by Kevin O'Connor, augmented by Tim Peters and Raymond Hettinger, tweaked by Marshel Abraham
__about__ = """Heap queues
[explanation by François Pinard]
Max Heaps are arrays for which a[k] >= a[2*k+1] and a[k] >= a[2*k+2] for
all k, counting elements from 0.  For the sake of comparison,
non-existing elements are considered to be -infinite.  The interesting
property of a max heap is that a[0] is always its largest element.
The strange invariant above is meant to be an efficient memory
representation for a tournament.  The numbers below are `k', not a[k]:
                                   0
                  1                                 2
          3               4                5               6
      7       8       9       10      11      12      13      14
    15 16   17 18   19 20   21 22   23 24   25 26   27 28   29 30


__all__ = ['heappush', 'heappop', 'heapify', 'heapreplace', 'merge',
           'nlargest', 'nsmallest', 'heappushpop']
"""

def heappush(heap, item):
    """Push item onto heap, maintaining the heap invariant."""
    heap.append(item)
    _siftdown(heap, 0, len(heap)-1)

def heappop(heap):
    """Pop the largest item off the heap, maintaining the heap invariant."""
    lastelt = heap.pop()    # raises appropriate IndexError if heap is empty
    if heap:
        returnitem = heap[0]
        heap[0] = lastelt
        _siftup(heap, 0)
        return returnitem
    return lastelt

def heapreplace(heap, item):
    """Pop and return the current largest value, and add the new item.
    This is more efficient than heappop() followed by heappush(), and can be
    more appropriate when using a fixed-size heap.  Note that the value
    returned may be larger than item!  That constrains reasonable uses of
    this routine unless written as part of a conditional replacement:
        if item > heap[0]:
            item = heapreplace(heap, item)
    """
    returnitem = heap[0]    # raises appropriate IndexError if heap is empty
    heap[0] = item
    _siftup(heap, 0)
    return returnitem

def heappushpop(heap, item):
    """Fast version of a heappush followed by a heappop."""
    if heap and heap[0] < item:
        item, heap[0] = heap[0], item
        _siftup(heap, 0)
    return item

def heapify(x):
    """Transform list into a heap, in-place, in O(len(x)) time."""
    n = len(x)
    # Transform bottom-up.  The largest index there's any point to looking at
    # is the largest with a child index in-range, so must have 2*i + 1 < n,
    # or i < (n-1)/2.  If n is even = 2*j, this is (2*j-1)/2 = j-1/2 so
    # j-1 is the largest, which is n//2 - 1.  If n is odd = 2*j+1, this is
    # (2*j+1-1)/2 = j so j-1 is the largest, and that's again n//2-1.
    for i in reversed(range(n//2)):
        _siftup(x, i)

def _heappop_max(heap):
    """Maxheap version of a heappop."""
    lastelt = heap.pop()    # raises appropriate IndexError if heap is empty
    if heap:
        returnitem = heap[0]
        heap[0] = lastelt
        _siftup_max(heap, 0)
        return returnitem
    return lastelt

def _heapreplace_max(heap, item):
    """Maxheap version of a heappop followed by a heappush."""
    returnitem = heap[0]    # raises appropriate IndexError if heap is empty
    heap[0] = item
    _siftup_max(heap, 0)
    return returnitem

def _heapify_max(x):
    """Transform list into a maxheap, in-place, in O(len(x)) time."""
    n = len(x)
    for i in reversed(range(n//2)):
        _siftup_max(x, i)

# 'heap' is a heap at all indices >= startpos, except possibly for pos.  pos
# is the index of a leaf with a possibly out-of-order value.  Restore the
# heap invariant.
def _siftdown(heap, startpos, pos):
    newitem = heap[pos]
    # Follow the path to the root, moving parents down until finding a place
    # newitem fits.
    while pos > startpos:
        parentpos = (pos - 1) >> 1
        parent = heap[parentpos]
        if newitem > parent:
            heap[pos] = parent
            pos = parentpos
            continue
        break
    heap[pos] = newitem

# The child indices of heap index pos are already heaps, and we want to make
# a heap at index pos too.  We do this by bubbling the larger child of
# pos up (and so on with that child's children, etc) until hitting a leaf,
# then using _siftdown to move the oddball originally at index pos into place.
#
# We *could* break out of the loop as soon as we find a pos where newitem >=
# both its children, but turns out that's not a good idea, and despite that
# many books write the algorithm that way.  During a heap pop, the last array
# element is sifted in, and that tends to be large, so that comparing it
# against values starting from the root usually doesn't pay (= usually doesn't
# get us out of the loop early).  See Knuth, Volume 3, where this is
# explained and quantified in an exercise.
#
# Cutting the # of comparisons is important, since these routines have no
# way to extract "the priority" from an array element, so that intelligence
# is likely to be hiding in custom comparison methods, or in array elements
# storing (priority, record) tuples.  Comparisons are thus potentially
# expensive.
#
# On random arrays of length 1000, making this change cut the number of
# comparisons made by heapify() a little, and those made by exhaustive
# heappop() a lot, in accord with theory.  Here are typical results from 3
# runs (3 just to demonstrate how small the variance is):
#
# Compares needed by heapify     Compares needed by 1000 heappops
# --------------------------     --------------------------------
# 1837 cut to 1663               14996 cut to 8680
# 1855 cut to 1659               14966 cut to 8678
# 1847 cut to 1660               15024 cut to 8703
#
# Building the heap by using heappush() 1000 times instead required
# 2198, 2148, and 2219 compares:  heapify() is more efficient, when
# you can use it.
#
# The total compares needed by list.sort() on the same lists were 8627,
# 8627, and 8632 (this should be compared to the sum of heapify() and
# heappop() compares):  list.sort() is (unsurprisingly!) more efficient
# for sorting.

def _siftup(heap, pos):
    endpos = len(heap)
    startpos = pos
    newitem = heap[pos]
    # Bubble up the smaller child until hitting a leaf.
    childpos = 2*pos + 1    # leftmost child position
    while childpos < endpos:
        # Set childpos to index of smaller child.
        rightpos = childpos + 1
        if rightpos < endpos and not heap[childpos] > heap[rightpos]:
            childpos = rightpos
        # Move the smaller child up.
        heap[pos] = heap[childpos]
        pos = childpos
        childpos = 2*pos + 1
    # The leaf at pos is empty now.  Put newitem there, and bubble it up
    # to its final resting place (by sifting its parents down).
    heap[pos] = newitem
    _siftdown(heap, startpos, pos)

def _siftdown_max(heap, startpos, pos):
    'Maxheap variant of _siftdown'
    newitem = heap[pos]
    # Follow the path to the root, moving parents down until finding a place
    # newitem fits.
    while pos > startpos:
        parentpos = (pos - 1) >> 1
        parent = heap[parentpos]
        if parent < newitem:
            heap[pos] = parent
            pos = parentpos
            continue
        break
    heap[pos] = newitem

def _siftup_max(heap, pos):
    'Maxheap variant of _siftup'
    endpos = len(heap)
    startpos = pos
    newitem = heap[pos]
    # Bubble up the larger child until hitting a leaf.
    childpos = 2*pos + 1    # leftmost child position
    while childpos < endpos:
        # Set childpos to index of larger child.
        rightpos = childpos + 1
        if rightpos < endpos and not heap[rightpos] < heap[childpos]:
            childpos = rightpos
        # Move the larger child up.
        heap[pos] = heap[childpos]
        pos = childpos
        childpos = 2*pos + 1
    # The leaf at pos is empty now.  Put newitem there, and bubble it up
    # to its final resting place (by sifting its parents down).
    heap[pos] = newitem
    _siftdown_max(heap, startpos, pos)

def merge(*iterables, key=None, reverse=True):
    '''Merge multiple sorted inputs into a single sorted output.
    Similar to sorted(itertools.chain(*iterables)) but returns a generator,
    does not pull the data into memory all at once, and assumes that each of
    the input streams is already sorted (largest to smallest).
    >>> list(merge([1,3,5,7], [0,2,4,8], [5,10,15,20], [], [25]))
    [0, 1, 2, 3, 4, 5, 5, 7, 8, 10, 15, 20, 25]
    If *key* is not None, applies a key function to each element to determine
    its sort order.
    >>> list(merge(['dog', 'horse'], ['cat', 'fish', 'kangaroo'], key=len))
    ['dog', 'cat', 'fish', 'horse', 'kangaroo']
    '''

    h = []
    h_append = h.append

    if reverse:
        _heapify = _heapify_max
        _heappop = _heappop_max
        _heapreplace = _heapreplace_max
        direction = -1
    else:
        _heapify = heapify
        _heappop = heappop
        _heapreplace = heapreplace
        direction = 1

    if key is None:
        for order, it in enumerate(map(iter, iterables)):
            try:
                next = it.__next__
                h_append([next(), order * direction, next])
            except StopIteration:
                pass
        _heapify(h)
        while len(h) > 1:
            try:
                while True:
                    value, order, next = s = h[0]
                    yield value
                    s[0] = next()           # raises StopIteration when exhausted
                    _heapreplace(h, s)      # restore heap condition
            except StopIteration:
                _heappop(h)                 # remove empty iterator
        if h:
            # fast case when only a single iterator remains
            value, order, next = h[0]
            yield value
            yield from next.__self__
        return

    for order, it in enumerate(map(iter, iterables)):
        try:
            next = it.__next__
            value = next()
            h_append([key(value), order * direction, value, next])
        except StopIteration:
            pass
    _heapify(h)
    while len(h) > 1:
        try:
            while True:
                key_value, order, value, next = s = h[0]
                yield value
                value = next()
                s[0] = key(value)
                s[2] = value
                _heapreplace(h, s)
        except StopIteration:
            _heappop(h)
    if h:
        key_value, order, value, next = h[0]
        yield value
        yield from next.__self__


# Algorithm notes for nlargest() and nsmallest()
# ==============================================
#
# Make a single pass over the data while keeping the k most extreme values
# in a heap.  Memory consumption is limited to keeping k values in a list.
#
# Measured performance for random inputs:
#
#                                   number of comparisons
#    n inputs     k-extreme values  (average of 5 trials)   % more than min()
# -------------   ----------------  ---------------------   -----------------
#      1,000           100                  3,317               231.7%
#     10,000           100                 14,046                40.5%
#    100,000           100                105,749                 5.7%
#  1,000,000           100              1,007,751                 0.8%
# 10,000,000           100             10,009,401                 0.1%
#
# Theoretical number of comparisons for k smallest of n random inputs:
#
# Step   Comparisons                  Action
# ----   --------------------------   ---------------------------
#  1     1.66 * k                     heapify the first k-inputs
#  2     n - k                        compare remaining elements to top of heap
#  3     k * (1 + lg2(k)) * ln(n/k)   replace the topmost value on the heap
#  4     k * lg2(k) - (k/2)           final sort of the k most extreme values
#
# Combining and simplifying for a rough estimate gives:
#
#        comparisons = n + k * (log(k, 2) * log(n/k) + log(k, 2) + log(n/k))
#
# Computing the number of comparisons for step 3:
# -----------------------------------------------
# * For the i-th new value from the iterable, the probability of being in the
#   k most extreme values is k/i.  For example, the probability of the 101st
#   value seen being in the 100 most extreme values is 100/101.
# * If the value is a new extreme value, the cost of inserting it into the
#   heap is 1 + log(k, 2).
# * The probability times the cost gives:
#            (k/i) * (1 + log(k, 2))
# * Summing across the remaining n-k elements gives:
#            sum((k/i) * (1 + log(k, 2)) for i in range(k+1, n+1))
# * This reduces to:
#            (H(n) - H(k)) * k * (1 + log(k, 2))
# * Where H(n) is the n-th harmonic number estimated by:
#            gamma = 0.5772156649
#            H(n) = log(n, e) + gamma + 1 / (2 * n)
#   http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Rate_of_divergence
# * Substituting the H(n) formula:
#            comparisons = k * (1 + log(k, 2)) * (log(n/k, e) + (1/n - 1/k) / 2)
#
# Worst-case for step 3:
# ----------------------
# In the worst case, the input data is reversed sorted so that every new element
# must be inserted in the heap:
#
#             comparisons = 1.66 * k + log(k, 2) * (n - k)
#
# Alternative Algorithms
# ----------------------
# Other algorithms were not used because they:
# 1) Took much more auxiliary memory,
# 2) Made multiple passes over the data.
# 3) Made more comparisons in common cases (small k, large n, semi-random input).
# See the more detailed comparison of approach at:
# http://code.activestate.com/recipes/577573-compare-algorithms-for-heapqsmallest

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()
        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, _order = result[0]
                order += 1
        result.sort()
        return [elem for (elem, order) 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, _order, _elem = result[0]
            order += 1
    result.sort()
    return [elem for (k, order, elem) in result]

def nlargest(n, iterable, key=None):
    """Find the n largest elements in a dataset.
    Equivalent to:  sorted(iterable, key=key, reverse=True)[:n]

    Less optimal compared to original implementation.
    """

    # Short-cut for n==1 is to use max()
    if n == 1:
        it = iter(iterable)
        sentinel = object()
        result = max(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, reverse=True)[:n]

    result = [heappop(iterable) for i in range(n)]
    return result

if __name__ == "__main__":

    import doctest # pragma: no cover
    print(doctest.testmod()) # pragma: no cover

    h = []
    heappush(h, 1)
    heappush(h, 2)
    heappush(h, 3)
    print(h)
    print(nlargest(2, h))


