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C15 - Katrina K #24

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Heaps Practice

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Comprehension Questions

Question Answer
How is a Heap different from a Binary Search Tree? Heaps are defined by min/max and every child follows that rule where BST is best when each parent is roughly halfway through the data if it's balanced.
Could you build a heap with linked nodes? It would be harder for sure but I haven't tried. Why not?
Why is adding a node to a heap an O(log n) operation? In the worst case, the node will need to be heaped up the entire height of the tree until the tree follows the rule again
Were the heap_up & heap_down methods useful? Why? Yes, I used them recursively to reset my tree after adding or removing a node.

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@CheezItMan CheezItMan left a comment

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Nice work Katrina, you hit the learning goals here. Well done.

Comment on lines 3 to 7
def heap_sort(list):
""" This method uses a heap to sort an array.
Time Complexity: ?
Space Complexity: ?
"""

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👍 Time/space complexity?

Comment on lines 21 to 26
def add(self, key, value = None):
""" This method adds a HeapNode instance to the heap
If value == None the new node's value should be set to key
Time Complexity: ?
Space Complexity: ?
Time Complexity: O(log n)
Space Complexity: O(1) *same space is added to the list every time but the behind the scenes buffer might be using more memory than this
"""

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👍 The space complexity is also log n due to the recursive call stack as you note

Comment on lines 30 to 35
def remove(self):
""" This method removes and returns an element from the heap
maintaining the heap structure
Time Complexity: ?
Space Complexity: ?
Time Complexity: O(log n)
Space Complexity: O(1)
"""

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👍 The space complexity is also O(log n)

Comment on lines +62 to +65
if len(self.store) < 1:
return True

return False

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Suggested change
if len(self.store) < 1:
return True
return False
return len(self.store) == 0

if len(self.store) < 1:
return True

return False


def heap_up(self, index):

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👍 Due to the call stack the space complexity is O(log n)

Comment on lines +96 to +97
self.heap_down((index * 2 + 2))
self.heap_down(index)

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This works, but you do potentially end up with an extra swap. However you avoid extra complicated conditionals.

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2 participants