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CheezItMan
reviewed
Mar 18, 2020
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| list.size.times do |node| | ||
| last_node = heap.remove() | ||
| if sorted_list.empty? | ||
| sorted_list.push(last_node) | ||
| elsif last_node < sorted_list[-1] | ||
| temp = sorted_list[-1] | ||
| sorted_list[-1] = last_node | ||
| sorted_list.push(temp) | ||
| else | ||
| sorted_list.push(last_node) | ||
| end | ||
| end |
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This is a bit overly complicated. Remember you always remove the smallest element of a heap.
Suggested change
| list.size.times do |node| | |
| last_node = heap.remove() | |
| if sorted_list.empty? | |
| sorted_list.push(last_node) | |
| elsif last_node < sorted_list[-1] | |
| temp = sorted_list[-1] | |
| sorted_list[-1] = last_node | |
| sorted_list.push(temp) | |
| else | |
| sorted_list.push(last_node) | |
| end | |
| end | |
| until heap.empty? | |
| sorted_list << heap.remove() | |
| end |
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| # Time Complexity: O(log n) | ||
| # Space Complexity: O(1) | ||
| def add(key, value = key) |
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| # Time Complexity: O(log n) | ||
| # Space Complexity: O(1) | ||
| def remove() |
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| # Time complexity: O(n) | ||
| # Space complexity: O(1) | ||
| def empty? |
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It works, but how is this O(n) time complexity? Why not O(1)?
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| # Time complexity: O(log n) | ||
| # Space complexity: O(1) | ||
| def heap_up(index) |
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Note, you're doing recursion so this is O(log n) space complexity. You gotta remember the system stack!
| right_child = 2*index+2 | ||
|
|
||
| # if either is nil, can use, so return | ||
| return if @store[left_child].nil? || @store[right_child].nil? |
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What if the left child is not nil, but it IS smaller than the parent? Small edge-case the test missed here.
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Heaps Practice
Congratulations! You're submitting your assignment!
Comprehension Questions
heap_up&heap_downmethods useful? Why?