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CheezItMan
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Overall nice work, you hit the learning goals here. Well done.
| # Hint, you may want a recursive helper method | ||
| # Time Complexity - O(n) | ||
| # Space Complexity - O(n) | ||
| def fibonacci(n) |
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This works, but the solution keeps an array of all the fibonacci numbers from 0 to n for the entire length of the recursive calls. Instead you only need to keep the last 2 fibonacci numbers.
| # Space Complexity - ? | ||
| # Time Complexity - O(n) | ||
| # Space Complexity - O(n) | ||
| def super_digit(n) |
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👍
Your time complexity is right, if n is the number of digits.
| # Space Complexity - ? | ||
| # Time Complexity - Is this O(2^n) (exponential), because the method gets called twice each time? | ||
| # Space Complexity - Is this O(2^n) (exponential), because the method gets called twice each time? | ||
| def refined_super_digit(n, k) |
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| # Time Complexity - Is this O(2^n) (exponential), because the method gets called twice each time? | ||
| # Space Complexity - Is this O(2^n) (exponential), because the method gets called twice each time? |
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This isn't exponential since the number of digits goes down by 1 with each iteration.
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