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anselrognlie
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anselrognlie
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✨ 💫 Looks good, Vange! I left some comments on your implementation below.
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| elif num == 1: | ||
| return '1' | ||
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| memo = [0,1,1] |
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✨ Nice use of a buffer slot to account for the 1-based calculation.
| memo = [0,1,1] | ||
| count = 3 | ||
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| while count <= num: |
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👀 Prefer a for loop, since we know exactly how many times this will loop.
| NCnums = [] | ||
| for num in memo: | ||
| NCnums.append(str(num)) |
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Consider using a comprehension or map to apply the str transformation to each numeric value. And avoid using identifiers that start with capital letters for things that aren't classes.
nc_nums = [str(num) for num in memo]or
nc_nums = map(str, memo)| Time Complexity: O(n), iterates through relative to the size of the list given. | ||
| Space Complexity: ? O(1), we will only ever make 2 |
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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.
| max_this_subarray = nums[0] | ||
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| for i in range(1, len(nums)): # O(n) | ||
| max_this_subarray = max(max_this_subarray + nums[i], nums[i]) |
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✨ This is a really nice way to represent this calculation, which captures the underlying invariant that makes Kadane's algorithm work.
You could also refer back to the pseudocode presented in the project description for a slightly different (though equivalent) approach. Again, to me it's not obvious why Kadane's algorithm is correct, so do check out that linked article.
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I relied heavily on the class recording for this assignment. I think it'd be a good idea for me to do more algorithm problems independentlyl.