Sockets - Kelly #8
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CheezItMan
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May 29, 2019
CheezItMan
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CheezItMan
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Check out my comments and let me know what questions you have.
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| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n) |
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This is actually O(n^2) since you make a new string with each recursive call.
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| # Time complexity: ? | ||
| # Space complexity: ? | ||
| def reverse_inplace(s) |
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Think about making a helper which takes the string, a left index and right index. The base case is when left_index >= right_index Each recursive call you swap the characters at left_index or right_index and then make a recursive call with the string, and left_index + 1 and right_index -1.
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