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Number of comparisons in build_heap
def parent(i):
return i/2
def left(i):
return 2*i
def right(i):
return 2*i+1
def heapify_cost(n, i):
most = 0
if left(i) <= n:
most = 1 + heapify_cost(n, left(i))
if right(i) <= n:
most = 1 + max(most, heapify_cost(n, right(i)))
return most
def build_heap_cost(n):
return sum(heapify_cost(n, i) for i in xrange(n/2, 1, -1))
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