Alf (Asli) - Spruce - C16 #44
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anselrognlie
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anselrognlie
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✨ 💫 Looks good, Alf! I left some comments on your implementation below.
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| Time Complexity: O(n) | ||
| Space Complexity: O(n) |
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✨ Great! By carefully building up the calculations and storing them for later use, we only need to perform O(n) calculations. The storage to keep those calculations is related to n (as is the converted string) giving space complexity of O(n) as well (ignoring a little bit of fiddliness related to the length of larger numbers being longer strings).
| Space Complexity: O(n) | ||
| """ | ||
| pass | ||
| outputs = {} |
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The structure we use to store the memo data can be a list, since we are carefully adding up from the bottom. A dict can be useful if we are storing results for non-integer keys, or if the order that we calculate values is less-predictable.
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| val = outputs[outputs[n-1]] + outputs[n - outputs[n - 1]] | ||
| outputs[n] = val | ||
| res.append(str(val)) |
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Another approach to could be to calculate the set of values using a list as the backing memo structure, then convert the whole structure into strings all at once at the end. It's not really any more efficient, but by focusing on doing one thing in the main calculation (finding the needed numbers) and then doing the conversion separately, we separate the two phases a little which can help with understandability by somewhat separating the concerns.
| Time Complexity: O(n) | ||
| Space Complexity: O(1) |
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✨ Notice how better time complexity this approach achieves over a "naïve" approach of checking for the maximum achievable sum starting from every position and every length. The correctness of this approach might not be apparent, so I definitely encourage reading a bit more about it. This has a fairly good explanation, as well as a description of why this is considered a dynamic programming approach (on the face it might not "feel" like one).
Since like the fibonacci sequence, we are able to maintain a sliding window of recent values to complete our calculation, we can do it with a constant O(1) amount of storage.
| return 0 | ||
| pass | ||
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| max = - 10000000000000000000000000000000000000000000 |
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Another approach would be to initialize the maximum to some value actually found in the list, say nums[0], which we know must at least exist from the guard checks.
| max = - 10000000000000000000000000000000000000000000 | ||
| curr_max = 0 | ||
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| for num in nums: |
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