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Hayden - Edges (C10) - Palindrome Check #29
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Original file line number | Diff line number | Diff line change |
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@@ -1,5 +1,28 @@ | ||
# 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 | ||
if my_phrase == nil | ||
return false | ||
end | ||
i = 0 | ||
j = my_phrase.length - 1 | ||
while i < j | ||
if my_phrase[i] == " " || my_phrase[j] == " " | ||
until my_phrase[i] != " " | ||
i += 1 | ||
end | ||
until my_phrase[j] != " " | ||
j -= 1 | ||
end | ||
elsif my_phrase[i] != my_phrase[j] | ||
return false | ||
end | ||
i += 1 | ||
j -= 1 | ||
end | ||
return true | ||
end | ||
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# time complexity: O(n) because it must iterate through each element in the array | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Always explain what n stands for while explaining your time and space complexity. In this case, it would be the number of characters in the input string or the size of the input string. |
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# space complexity: O(1) because it does not require any additional memory space besides the iteration variables |
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In the two nested inner loops, continue to check and confirm that i is less than j before comparing the character at the index i (or j) with a white space.