Sockets - Kirsten #9
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CheezItMan
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May 29, 2019
CheezItMan
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CheezItMan
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Very nicely done. See my inline comments, but this is outstanding work. Check a little bit on your big-O as you missed some of the overhead of some of the built-in functions, something most people do.
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| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n), where n is the length of the string |
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This is not in place since you create a new string.
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.
| return reverse_helper(s, i, j) | ||
| end | ||
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| def reverse_helper(s, i, j) |
| # Space complexity: ? | ||
| # Time complexity: O(n), where n is the input number | ||
| # Space complexity: O(n), where n is the input number | ||
| def bunny(n) |
| # Space complexity: ? | ||
| # Time complexity: O(n), where n is the size of the input string | ||
| # Space complexity: O(n), where n is the size of the input string | ||
| def nested(s) |
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| end | ||
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| # Time complexity: O(n), where n is the length of the input array |
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This is actually O(n^2) since you create a new array with each recursive call.
| # Space complexity: ? | ||
| # Time complexity: O(n), where n is the length of the string | ||
| # Space complexity: O(n), where n is the length of the string | ||
| def is_palindrome(s) |
| # Space complexity: ? | ||
| # Time complexity: O(log10(n)), where n is the smallest input number | ||
| # Space complexity: O(log10(n)), where n is the smallest input number | ||
| def digit_match(n, m) |
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