Janice H. #20
Janice H. #20
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| # Time Complexity: O(2nlogn)? | ||
| # Space Complexity: O(1) | ||
| # GAH I cannot get those last two numbers | ||
| # sorted properly for some reason and it's | ||
| # time to turn this in. I know they are in | ||
| # the correct order at the end but then flip | ||
| def heapsort(list) |
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It's neat that you were trying to do the in-place heapsort. I'd suggest doing it the simple way first, using the minheap you created in this assignment and adding all the elements of the list to the heap and then extracting them one by one and shoveling them into the resulting list. It's O(n) space complexity but it works. Then try for this one if/when you have time.
| return list | ||
| end | ||
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| def heap_up(list, index) |
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To do this in-place, I would suggest making a Max_heap instead of a min-heap and then as you remove a node adding it to the rear of the array working your way from the last node until index 0. That's a lot easier than using a min-heap like you are doing.
| # Time Complexity: O(nlogn) | ||
| # Space Complexity: O(n) | ||
| def add(key, value = key) |
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This works, but why would adding one value to the heap be O(n log n) time complexity and why would it take O(n) space complexity?
I suggest it's O(log n) for space (the recursive stack) and O(log n) for time.
| # Time Complexity: O(nlogn) | ||
| # Space Complexity: O(1) | ||
| def remove() |
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See my notes above about space/time complexity.
| def heap_up(index) | ||
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| # for right children | ||
| if index % 2 == 0 |
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These if statements could be condensed a bit.
| def heap_down(index) | ||
| raise NotImplementedError, "Method not implemented yet..." | ||
| if !@store[index+1].nil? && @store[index].key > @store[index+1].key | ||
| swap(index, index+1) | ||
| heap_down(index+1) | ||
| elsif !@store[index+2].nil? && @store[index].key > @store[index+2].key | ||
| swap(index, index+2) | ||
| heap_down(index+2) | ||
| end | ||
| end |
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This method is NOT doing head_down. Instead it's comparing a node to the nodes next to it. Remember the formula to find a node's left child is:
left_child_index = index * 2 + 1
and the right child
right_child_index = index * 2 + 2
Because you're not doing the correct heap_down method this method is just using a sorted array which is O(n)
| # Time complexity: O(nlogn) | ||
| # Space complexity: O(1) | ||
| def heap_up(index) |
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How is the time complexity to heap up 1 element O(n log n), you don't, or shouldn't look at every node in the list for this method.
Co-Authored-By: Chris M <[email protected]>
Heaps Practice
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Comprehension Questions
heap_up&heap_downmethods useful? Why?