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CheezItMan
reviewed
Mar 18, 2020
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| // Time Complexity: O(n log n) - first iterate over list (O(n)) | ||
| // and run .add() for each item (O(log n)) | ||
| // Space Complexity: O(1) - uses original list param to store sorted list | ||
| function heapsort(list) { |
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👍 , except that you're building a heap, so the space complexity is O(n).
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| // Time Complexity: O(log n) | ||
| // Space Complexity: O(log n) | ||
| add(key, value = key) { |
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The method works, and you have the space complexity right because you're doing recursion.
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| // Time complexity: O(1) | ||
| // Space complexity: O(1) | ||
| isEmpty() { |
| const leftChild = 2 * index + 1; | ||
| const rightChild = 2 * index + 2; | ||
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| if (leftChild >= this.store.length || rightChild >= this.store.length) return; |
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What if the left child is smaller than the length and the left child is smaller than it's parent, and at the same time the right child is equal to the length?
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@CheezItMan can you give me an example? I'm having a hard time thinking of how a child index would be smaller than its parent index.
| if (this.store[parent].key > this.store[index].key) { | ||
| this.swap(parent, index); | ||
| return this.heapUp(parent); | ||
| } | ||
| } | ||
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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. | ||
| heapDown(index) { |
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This method works in most cases, but I think you're missing an edge-case here.
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Heaps Practice
Congratulations! You're submitting your assignment!
Comprehension Questions
heap_up&heap_downmethods useful? Why?