Emily - Leaves #47
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Emily - Leaves #47
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CheezItMan
reviewed
Sep 19, 2019
| (1...length).each do |k| | ||
| if list[k] == list[k-1] | ||
| i = k | ||
| until i > length || list[i-1] == nil do |
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I would say until i >= length || list[i-1] == nil do
| @@ -1,13 +1,45 @@ | |||
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| # Time Complexity: ? | |||
| # Space Complexity: ? | |||
| # Time Complexity: O(n^2) | |||
| end | ||
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| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: O(n), where n is total length of all strings combined |
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I would say O(n * m) where n is the number of strings and m is the number of letters in the average string. Unless the strings are small in which case you could say O(n)
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