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by ColinWright
3835 days ago
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OK, now I don't understand what you are trying to say. There are n^2 results, each result takes O(n) to compute, so the naive algorithm is O(n^3). That's what you asked - why does the naive algorithm take O(n^3) - and that seems to answer your question. Yes, clearly there is an overlap of some partial sums, and that's why a less naive algorithm is sub-O(n^3). So, what are you asking, or what additional point are you making? |
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