Sockets - Sarah #1
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CheezItMan
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May 29, 2019
CheezItMan
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CheezItMan
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Nice work, you hit the learning goals here. See my comments regarding space and time as you missed the cost of some of the built-in ruby methods.
Overall, outstanding work
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| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n), where n is the length of s, reverse will get called n/2 times |
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This is actually O(n^2) since s[1...-1] creates a new string of length n-2 each iteration. This is true for both space & time.
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| # Time complexity: O(n), where n is the length of s, reverse will get called n/2 times | ||
| # Space complexity: O(n) where n is the length of s, n/2 number of stack frames will be used. Unless you are using a language that utilizes tail recursion, then the space complexity would be constant | ||
| def r_reverse_inplace(s, i) |
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n), where n is the length of s, nested will get called n/2 times | ||
| # Space complexity: O(n) where n is the length of s. A new stack frame will be used each call, and by slicing into the array a new array is created during each call, which i think would make it 0(2n) |
| # Space complexity: ? | ||
| # Time complexity: O(n) | ||
| # Space complexity: O(n), | ||
| def search(array, value) |
| # Space complexity: ? | ||
| # Time complexity: O(n), where n is the length of s, nested will get called n/2 times | ||
| # Space complexity: O(n) where n is the length of s. A new stack frame will be used each call | ||
| def is_palindrome(s) |
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O(n^2) similar to reverse.
Could you have used a technique similar to the one you used with reverse_in_place?
| return is_palindrome(s.slice!(1...-1)) | ||
| end | ||
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| # Time complexity: O(log10 n) where n is the smaller number between n and m, each call to r_didget_match divids both numbres by 10 and terminates when wither becomes 0. |
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