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Tanya – Pine #25
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Tanya – Pine #25
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Original file line number | Diff line number | Diff line change |
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@@ -1,8 +1,17 @@ | ||
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from heaps.min_heap import MinHeap | ||
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def heap_sort(list): | ||
""" This method uses a heap to sort an array. | ||
Time Complexity: ? | ||
Space Complexity: ? | ||
Time Complexity: O(log n) | ||
Space Complexity: O(n) | ||
""" | ||
pass | ||
heap = MinHeap() | ||
for element in list: | ||
heap.add(element) | ||
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sorted_list = [] | ||
while heap.store: | ||
element = heap.remove() | ||
sorted_list.append(element) | ||
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return sorted_list |
Original file line number | Diff line number | Diff line change |
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@@ -1,5 +1,5 @@ | ||
class HeapNode: | ||
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def __init__(self, key, value): | ||
self.key = key | ||
self.value = value | ||
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@@ -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) | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ✨ Nice however space complexity is going to be O(log n) here because of the recursive call stack of |
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""" | ||
pass | ||
if value == None: | ||
value = key | ||
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node = HeapNode(key, value) | ||
self.store.append(node) | ||
self.heap_up(len(self.store) -1) | ||
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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) | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ✨ However space complexity here will be O(log n) because of the recursive call stack of |
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""" | ||
pass | ||
if len(self.store) == 0: | ||
return None | ||
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self.swap(0, len(self.store) - 1) | ||
min = self.store.pop() | ||
self.heap_down(0) | ||
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return min.value | ||
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def __str__(self): | ||
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@@ -44,10 +55,10 @@ def __str__(self): | |
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def empty(self): | ||
""" This method returns true if the heap is empty | ||
Time complexity: ? | ||
Space complexity: ? | ||
Time complexity: O(1) | ||
Space complexity: O(1) | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ✨ |
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""" | ||
pass | ||
return len(self.store) == 0 | ||
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def heap_up(self, index): | ||
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@@ -57,18 +68,43 @@ def heap_up(self, index): | |
property is reestablished. | ||
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This could be **very** helpful for the add method. | ||
Time complexity: ? | ||
Space complexity: ? | ||
Time complexity: O(log n) | ||
Space complexity: O(log n) | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ✨ |
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""" | ||
pass | ||
if index == 0: | ||
return | ||
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if index % 2 == 0: | ||
# right node parent | ||
parent_index = int((index - 2) / 2) | ||
else: | ||
# left node parent | ||
parent_index = int((index - 1) / 2) | ||
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if(parent_index < 0): | ||
return | ||
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if self.store[parent_index].key > self.store[index].key: | ||
self.swap(parent_index, index) | ||
self.heap_up(parent_index) | ||
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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. | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. This works, but review my comment below to see how you might improve upon this solution ⬇️ |
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""" | ||
pass | ||
left_child = index * 2 + 1 | ||
right_child = index * 2 + 2 | ||
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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
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. You can reduce your recursive calls to If you swap with the smaller of the two children, that child (as the new parent of |
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def swap(self, index_1, index_2): | ||
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There was a problem hiding this comment.
Choose a reason for hiding this comment
The reason will be displayed to describe this comment to others. Learn more.
👀 Time complexity is going to be O(nlogn) as you iterate through n items in
list
twice, and within each loop performadd
orremove
which are both O(log n) operations