-
Notifications
You must be signed in to change notification settings - Fork 51
Cedar - Kristin L #30
New issue
Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.
By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.
Already on GitHub? Sign in to your account
base: master
Are you sure you want to change the base?
Changes from all commits
File filter
Filter by extension
Conversations
Jump to
Diff view
Diff view
There are no files selected for viewing
| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -1,8 +1,41 @@ | ||
| from heapq import heappush, heappop | ||
| from heaps.min_heap import MinHeap | ||
|
|
||
|
|
||
| def heap_sort(list): | ||
| def heap_sort(list): # O(n) | ||
| """ This method uses a heap to sort an array. | ||
| Time Complexity: ? | ||
| Space Complexity: ? | ||
| Time Complexity: O(n log n) | ||
| Space Complexity: O(n) | ||
| """ | ||
| pass | ||
| heap = MinHeap() | ||
| for num in list: | ||
| # has heap_up inside and sorts it in heap everytime new node is added | ||
| heap.add(num) | ||
|
|
||
| # while the heap is not empty, it removes the most minimal value which is always the root | ||
| index = 0 | ||
| while not heap.empty(): | ||
| # puts sorted nums back in list using index | ||
| # remove is O(log n) | ||
| list[index] = heap.remove() | ||
| index += 1 | ||
|
|
||
| return list | ||
|
Comment on lines
+15
to
+22
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. Note the since this isn't a fully in-place solution (the MinHeap has a O(n) internal store), we don't necessarily need to modify the passed in list. The tests are written to check the return value, so we could unpack the heap into a new result list to avoid mutating the input. result = []
while not heap.empty():
result.append(heap.remove())
return result |
||
|
|
||
| # from class | ||
| # Time: n log n | ||
| # space: n | ||
| def heapsort(unsorted): | ||
| heap = [] | ||
| for item in unsorted: | ||
| heappush(heap, item) | ||
| # heapq is putting all of this together. append will treat it like any other list and does order it | ||
|
|
||
| ordered = [] | ||
| while len(heap) > 0: | ||
| value = heappop(heap) | ||
| ordered.append(value) | ||
|
|
||
| # returns ordered elements | ||
| return ordered | ||
|
|
||
| print(heapsort([4,7,1,33,0,-8])) | ||
| Original file line number | Diff line number | Diff line change |
|---|---|---|
|
|
@@ -19,18 +19,29 @@ 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) | ||
|
Comment on lines
+22
to
+23
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. 👀 In the worst case, the new value we're inserting is the new root of the heap, meaning it would need to move up the full height of the heap (which is log n levels deep) leading to O(log n) time complexity. Your implementation of the |
||
| """ | ||
| pass | ||
| if value == None: | ||
|
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. 👀 Prefer using |
||
| 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) | ||
|
Comment on lines
+35
to
+36
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. 👀 In the worst case, the value that got swapped to the top will need to move all the way back down to a leaf, meaning it would need to move down the full height of the heap (which is log n levels deep) leading to O(log n) time complexity. Your implementation of the |
||
| """ | ||
| pass | ||
| if self.empty(): | ||
|
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 use of your own helper method! |
||
| return None | ||
| self.swap(0, len(self.store) - 1) | ||
| min = self.store.pop() | ||
| self.heap_down(0) | ||
|
|
||
| return min.value | ||
|
|
||
|
|
||
|
|
||
|
|
@@ -44,10 +55,11 @@ def __str__(self): | |
|
|
||
| def empty(self): | ||
| """ This method returns true if the heap is empty | ||
| Time complexity: ? | ||
| Space complexity: ? | ||
| Time complexity: O(n) | ||
| Space complexity: O(1) | ||
|
Comment on lines
+58
to
+59
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. 👀 Getting the length of a list is a constant time operation, so the overall time complexity will be constant time as well. |
||
| """ | ||
| pass | ||
| if len(self.store) == 0: | ||
| return True | ||
|
Comment on lines
+61
to
+62
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 will return Python would fall off the end of the function, with the default if len(self.store) == 0:
return True
else:
return Falsewhich simplifies to return len(self.store) == 0We could also remember that empty lists are falsy, and write this as return not self.store |
||
|
|
||
|
|
||
| def heap_up(self, index): | ||
|
|
@@ -57,18 +69,44 @@ 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(1) | ||
|
Comment on lines
+72
to
+73
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 function is the main source of time and space complexity for |
||
|
|
||
| check my value vs parent's value | ||
| if parent is larger | ||
| swap with parent | ||
| recursively call heap_up from new location | ||
|
Comment on lines
+75
to
+78
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 explanation of the approach |
||
| """ | ||
| pass | ||
| if index == 0: | ||
| return | ||
|
|
||
| parent = (index - 1) // 2 | ||
| store = self.store | ||
| if store[parent].key > store[index].key: | ||
| self.swap(parent, index) | ||
| self.heap_up(parent) | ||
|
|
||
| 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. | ||
| """ | ||
| pass | ||
| left_child = index * 2 + 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. 👀 This is identical to the instructor solution and has no additions that would indicate to me that time was spent time understanding this approach. |
||
| right_child = index * 2 + 2 | ||
| store = self.store | ||
| if left_child < len(self.store): | ||
| if right_child < len(self.store): | ||
| if store[left_child].key < store[right_child].key: | ||
| smaller = left_child | ||
| else: | ||
| smaller = right_child | ||
| else: | ||
| smaller = left_child | ||
|
|
||
| if store[index].key > store[smaller].key: | ||
| self.swap(index, smaller) | ||
| self.heap_down(smaller) | ||
|
|
||
|
|
||
| def swap(self, index_1, index_2): | ||
|
|
||
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.
✨ Great. Since sorting using a heap reduces down to building up a heap of n items one-by-one (each taking O(log n)), then pulling them back out again (again taking O(log n) for each of n items), we end up with a time complexity of O(2n log n) → O(n log n). While for the space, we do need to worry about the O(log n) space consume during each add and remove, but they aren't cumulative (each is consumed only during the call to add or remove). However, the internal store for the
MinHeapdoes grow with the size of the input list. So the maximum space would be O(n + log n) → O(n), since n is a larger term than log n.Note that a fully in-place solution (O(1) space complexity) would require both avoiding the recursive calls, as well as working directly with the originally provided list (no internal store).