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Your largest subarray works, but isn't using dynamic programming and has a pretty bad runtime. See my comments and let me know if you have questions or need help working through it.
We can review it if needed.
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| while i < nums.length do | ||
| max_ending_here = max_ending_here + nums[i] | ||
| if (max_so_far < max_ending_here) | ||
| max_so_far = max_ending_here | ||
| end | ||
| i+=1 | ||
| end |
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You can do this without the nested loop. Think about how you could use same saved values, like the largest subarray so far and the largest ending at the current index and how that could work.
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| # Time Complexity: O(n!)? You only go throught the each with index n times, but the inner loop completes decrementing number of loops... | ||
| # Space Complexity: O(1), no additional data structures created outside of 2 variables each times | ||
| def max_sub_array(nums) |
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| # Time complexity: O(n) | ||
| # Space Complexity: O(n), always creating 1 array num + 1 length long and 1 string proprtionate to n | ||
| # P(1) = 1 | ||
| # P(2) = 1 | ||
| # for all n > 2 | ||
| # P(n) = P(P(n - 1)) + P(n - P(n - 1)) | ||
| # 1 1 2 2 3 4 4 4 5 6 7 7…. | ||
| def newman_conway(num) |
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