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Tanya – Pine #25

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17 changes: 13 additions & 4 deletions heaps/heap_sort.py
Original file line number Diff line number Diff line change
@@ -1,8 +1,17 @@

from heaps.min_heap import MinHeap

def heap_sort(list):
""" This method uses a heap to sort an array.
Time Complexity: ?
Space Complexity: ?
Time Complexity: O(log n)

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👀 Time complexity is going to be O(nlogn) as you iterate through n items in list twice, and within each loop perform add or remove which are both O(log n) operations

Space Complexity: O(n)
"""
pass
heap = MinHeap()
for element in list:
heap.add(element)

sorted_list = []
while heap.store:
element = heap.remove()
sorted_list.append(element)

return sorted_list
64 changes: 50 additions & 14 deletions heaps/min_heap.py
Original file line number Diff line number Diff line change
@@ -1,5 +1,5 @@
class HeapNode:

def __init__(self, key, value):
self.key = key
self.value = value
Expand All @@ -19,19 +19,30 @@ def __init__(self):
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)

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✨ Nice however space complexity is going to be O(log n) here because of the recursive call stack of heap_up

"""
pass
if value == None:
value = key

node = HeapNode(key, value)
self.store.append(node)
self.heap_up(len(self.store) -1)

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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✨ However space complexity here will be O(log n) because of the recursive call stack of heap_down

"""
pass
if len(self.store) == 0:
return None

self.swap(0, len(self.store) - 1)
min = self.store.pop()
self.heap_down(0)

return min.value


def __str__(self):
Expand All @@ -44,10 +55,10 @@ def __str__(self):

def empty(self):
""" This method returns true if the heap is empty
Time complexity: ?
Space complexity: ?
Time complexity: O(1)
Space complexity: O(1)

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"""
pass
return len(self.store) == 0


def heap_up(self, index):
Expand All @@ -57,18 +68,43 @@ def heap_up(self, index):
property is reestablished.

This could be **very** helpful for the add method.
Time complexity: ?
Space complexity: ?
Time complexity: O(log n)
Space complexity: O(log n)

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"""
pass
if index == 0:
return

if index % 2 == 0:
# right node parent
parent_index = int((index - 2) / 2)
else:
# left node parent
parent_index = int((index - 1) / 2)

if(parent_index < 0):
return

if self.store[parent_index].key > self.store[index].key:
self.swap(parent_index, index)
self.heap_up(parent_index)

def heap_down(self, index):
""" This helper method takes an index and
moves the corresponding element down the heap if it's
larger than either of its children and continues until
the heap property is reestablished.

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This works, but review my comment below to see how you might improve upon this solution ⬇️

"""
pass
left_child = index * 2 + 1
right_child = index * 2 + 2

if len(self.store) > (left_child) and self.store[index].key > self.store[left_child].key:
self.swap(index, (left_child))
self.heap_down((left_child))
self.heap_down(index)
elif len(self.store) > (right_child) and self.store[index].key > self.store[right_child].key:
self.swap(index, (right_child))
self.heap_down((right_child))
self.heap_down(index)
Comment on lines +100 to +107

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You can reduce your recursive calls to heap_down here. Instead of choosing to swap with the left child automatically if it exists, choose to swap index with whichever child is smaller.

If you swap with the smaller of the two children, that child (as the new parent of index and the other child) will automatically be smaller than both of it's new children and you won't have to call heap_down on it again.



def swap(self, index_1, index_2):
Expand Down