Leaves - Dominique #23
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CheezItMan
reviewed
Mar 18, 2020
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| while i > 0 do | ||
| heapify(root_array, i, size) | ||
| i -= 1 | ||
| end | ||
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| while size > 1 do | ||
| root_array[1], root_array[size] = root_array[size], root_array[1] | ||
| size -= 1 | ||
| heapify(root_array, 1, size) | ||
| end |
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| # O(n) | ||
| # Space Complexity: O(1) | ||
| def heap_sort(array) |
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👍 This is a great implementation of the in-place heapsort. Nice!
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| # Time Complexity: (logn) because in the worst case an element inserted at the bottom has to be swapped at | ||
| # every level from bottom to top up to the root node to maintain the heap property. So essentially you are only traversing half the number of the heap. | ||
| # Space Complexity: O(1) because there will always be one space of memory allocated for a new heap node | ||
| def add(key, value = key) |
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| # Time Complexity: (logn) because after the swapping of the root node and the last node, the last node may have to traverse the tree height (heap down) | ||
| # to satisfy the heap | ||
| # Space Complexity: O(1) because only one space in memory will be created as the algorithm scales. | ||
| def remove() |
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| # Time complexity: O(1) because the algorithm is only checking to see if there are elements in the array. This | ||
| # will stay as the problem scales | ||
| # Space complexity: O(1) because no new memory will be created as the algorithm scales | ||
| def empty? |
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| if @store[0] | ||
| return | ||
| else | ||
| return false | ||
| end |
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Suggested change
| if @store[0] | |
| return | |
| else | |
| return false | |
| end | |
| return @store.empty? |
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| # Time complexity: ? | ||
| # Space complexity: ? | ||
| def heap_up(index) | ||
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| def heap_up(current_index) |
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👍 , but no estimates in time/space complexity?
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| while @store[left_child_index] || @store[right_child_index] | ||
| # having trouble figuring out when to end the while loop | ||
| if @store[left_child_index].key < @store[right_child_index].key |
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What if @store[right_child_index] is nil?
| end | ||
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| # This helper method takes an index and | ||
| # moves it up the heap if it's smaller | ||
| # than it's parent node. | ||
| def heap_down(index) | ||
| raise NotImplementedError, "Method not implemented yet..." | ||
| def heap_down(current_index) |
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| # heap.add(3, "Pasta") | ||
| # print heap.add(2, "Soup") | ||
| heap.add(3, "Pasta") | ||
| heap.add(6, "Soup") | ||
| heap.add(1, "Pizza") | ||
| heap.add(0, "Donuts") | ||
| heap.add(16, "Cookies") | ||
| heap.add(57, "Cake") | ||
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| print heap |
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Suggested change
| heap = MinHeap.new | |
| # heap.add(3, "Pasta") | |
| # print heap.add(2, "Soup") | |
| heap.add(3, "Pasta") | |
| heap.add(6, "Soup") | |
| heap.add(1, "Pizza") | |
| heap.add(0, "Donuts") | |
| heap.add(16, "Cookies") | |
| heap.add(57, "Cake") | |
| print heap |
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Heaps Practice
Congratulations! You're submitting your assignment!
Comprehension Questions
heap_up&heap_downmethods useful? Why?