Karis - Edges - Palindrom Check #30
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| @@ -1,5 +1,43 @@ | |||
| # A method to check if the input string is a palindrome. | |||
| # Return true if the string is a palindrome. Return false otherwise. | |||
| def palindrome_check(my_phrase) | |||
| raise NotImplementedError | |||
| return false if !my_phrase | |||
| return true if my_phrase == "" | |||
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You could combine lines 5 and 6 to be: return true if my_phrase.length < 2
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| until min > max | ||
| # unless my_phrase[min] == " " || my_phrase[max] == " " | ||
| until my_phrase[min] != " " |
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In the two nested inner loops, continue to check and confirm that min is less than max before comparing the character at the index min (or max) with a white space.
| # require 'pry'; binding.pry | ||
| end | ||
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| return check |
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If the code reaches this point, you'll always be returning true. So you could get rid of the variable check and simply return true here. Accordingly, you could simplify lines 30 through 34.
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Nice work! I added some comments inline to make your algorithm even more robust and to simplify your code. I didn't see any comments on the time and space complexity of your algorithm. Give it some thought. Friendly reminder: Always explain what n stands for while explaining your time and space complexity. That along with your explanation for reasoning behind your answer, is the complete answer for the time and space complexity. In this case, n will be the number of characters in the input string parameter. |
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