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18 changes: 15 additions & 3 deletions heaps/heap_sort.py
Original file line number Diff line number Diff line change
@@ -1,8 +1,20 @@
from heapq import heappush, heappop


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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✨ Would love to see how you could write heap_sort with the MinHeap class you implemented below as well. Additionally, time complexity is O(n logn) because you are doing heappush/heappop which is a log(n) operation n times.

Space Complexity: O(n)
"""
pass
heap = []

for item in list:
heappush(heap, item)

ordered = []

while len(heap) > 0:
value = heappop(heap)
ordered.append(value)

return ordered
80 changes: 59 additions & 21 deletions heaps/min_heap.py
Original file line number Diff line number Diff line change
@@ -1,5 +1,8 @@
from turtle import right


class HeapNode:

def __init__(self, key, value):
self.key = key
self.value = value
Expand All @@ -10,67 +13,102 @@ def __str__(self):
def __repr__(self):
return str(self.value)


class MinHeap:

def __init__(self):
self.store = []


def add(self, key, value = None):
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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"""
pass
if value is None:
value = key

new_node = HeapNode(key, value)
self.store.append(new_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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✨ Space complexity is O(log n) here because of the recursive call stack of heap_down

"""
pass

if self.empty():
return None

first_node = 0
last_node = len(self.store)-1

self.swap(first_node, last_node)
result = self.store.pop()

self.heap_down(first_node)

return result.value


def __str__(self):
""" This method lets you print the heap, when you're testing your app.
"""
if len(self.store) == 0:
return "[]"
return f"[{', '.join([str(element) for element in self.store])}]"


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):
""" This helper method takes an index and
moves the corresponding element up the heap, if
it is less than it's parent node until the Heap
property is reestablished.

This could be **very** helpful for the add method.
Time complexity: ?
Space complexity: ?
Time complexity: O(log n)
Space complexity: O(1)
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✨ Nice iterative solution! The only thing I would say is style wise, I might stick to recursion or iteration across heap_up and heap_down for consistency.


(index - 1) // 2 -> to find the parent

"""
pass
parent = (index-1)//2
temp = index
while parent >= 0 and self.store[parent].key > self.store[temp].key:
self.swap(temp, parent)
temp = parent
parent = (temp - 1) // 2

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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pass
right_child = (index * 2) + 2
left_child = (index * 2) + 1

if left_child < len(self.store):
if right_child < len(self.store):
if self.store[left_child].key < self.store[right_child].key:
child = left_child
else:
child = right_child
else:
child = left_child

if self.store[child].key < self.store[index].key:
self.swap(index, child)
self.heap_down(child)


def swap(self, index_1, index_2):
""" Swaps two elements in self.store
at index_1 and index_2
Expand Down