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| Original file line number | Diff line number | Diff line change |
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| @@ -1,8 +1,15 @@ | ||
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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(nlogn) | ||
| Space Complexity: O(n) | ||
| """ | ||
| pass | ||
| heap = MinHeap() | ||
| sorted = [] | ||
| for item in list: | ||
| heap.add(item) | ||
| for i in range(len(list)): | ||
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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. In this situation, since we built the heap, we also "know" the number of items in the heap. So it works to iterate using knowledge about the list. But if we were pulling things out of a heap more generally, we would want to make use of the sorted = []
while not heap.empty():
sorted.append(heap.remove())
return sorted |
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| sorted.append(heap.remove()) | ||
| return sorted | ||
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|---|---|---|
| @@ -1,3 +1,5 @@ | ||
| import operator | ||
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| class HeapNode: | ||
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| def __init__(self, key, value): | ||
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@@ -19,18 +21,28 @@ 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(logn) | ||
| Space complexity: O(logn) because of the call stack | ||
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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. ✨ Great. You're exactly right that it's due to the recursive call in |
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| """ | ||
| pass | ||
| self.store.append(HeapNode(key, value or key)) | ||
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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. 👀 To explicitly handle the case where the if value is None:
value = keyUsing |
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| 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(logn) | ||
| Space complexity: O(logn) because of the call stack | ||
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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. ✨ Nice. Just as for |
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| """ | ||
| pass | ||
| if self.empty(): | ||
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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. ✨ Nice use of your own helper method! |
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| return None | ||
| if len(self.store) == 1: | ||
| return self.store.pop().value | ||
| minimum = self.store[0] | ||
| self.store[0] = self.store.pop() | ||
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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. We could avoid this special case by swapping the first element with the last (moves the minimum to the end, and a larger value to the head), then popping from the end. If there were only one thing left, it would be swapped with itself, then removed. |
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| self.heap_down(0) | ||
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| return minimum.value | ||
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@@ -44,10 +56,10 @@ 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) | ||
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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. ✨ |
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| """ | ||
| pass | ||
| return len(self.store) == 0 | ||
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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. Remember that an empty list is falsy return not self.store |
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| def heap_up(self, index): | ||
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@@ -57,18 +69,46 @@ 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(logn) | ||
| Space complexity: O(logn) because of the call stack | ||
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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. ✨ Yes, this function is where the complexity in |
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| """ | ||
| pass | ||
| if index == 0: | ||
| return | ||
| parent = (index-1)//2 | ||
| if self.store[parent].key > self.store[index].key: | ||
| self.swap(parent, index) | ||
| self.heap_up(parent) | ||
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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. | ||
| """ | ||
| pass | ||
| def get_valid_index_or_none(index): | ||
| if index < len(self.store): | ||
| return index | ||
| else: | ||
| return None | ||
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| def get_key_or_none(index): | ||
| if not index: | ||
| return None | ||
| return self.store[index].key | ||
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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. ✨ Nice helpers. They don't really need to be defined locally to the |
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| left_index = get_valid_index_or_none(index * 2 + 1) | ||
| right_index = get_valid_index_or_none(index * 2 + 2) | ||
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| left_key = get_key_or_none(left_index) | ||
| right_key = get_key_or_none(right_index) | ||
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| index_to_swap = min((node for node in [(left_index, left_key), (right_index, right_key)] if node[1] is not None ), key=operator.itemgetter(1), default=None) | ||
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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. This is a really concise set of steps to filter out the invalid children, then select the minimum! However, it's a bit dense, which makes it a little tough to understand. Consider adding a comment about what it does, break it up into several lines with the intermediate products given descriptive names, or move into a helper function with a descriptive name. An example of splitting across multiple lines candidate_children = [(left_index, left_key), (right_index, right_key)] # we could use a tuple rather than a list
valid_children = (node for node in candidate_children if node[1] is not None) # this actually makes a generator function, not a tuple
index_to_swap = min(valid_children, key=operator.itemgetter(1), default=None) |
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| if index_to_swap and self.store[index].key > index_to_swap[1]: | ||
| self.swap(index, index_to_swap[0]) | ||
| self.heap_down(index_to_swap[0]) | ||
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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.
✨ 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).