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Jane's palindrome_check #32
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Original file line number | Diff line number | Diff line change |
---|---|---|
@@ -1,5 +1,45 @@ | ||
# A method to check if the input string is a palindrome. | ||
# Return true if the string is a palindrome. Return false otherwise. | ||
# Time Complexity: O(n) | ||
# Space Complexity: O(n) | ||
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||
def string_reverse(my_string) | ||
i = 0 | ||
j = my_string.length - 1 | ||
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||
while i < j | ||
placeholder = my_string[i] | ||
my_string[i] = my_string[j] | ||
my_string[j] = placeholder | ||
i += 1 | ||
j -= 1 | ||
end | ||
return my_string | ||
end | ||
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def palindrome_check(my_phrase) | ||
raise NotImplementedError | ||
return false if my_phrase.nil? | ||
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my_phrase_original= [] | ||
my_phrase_reversed = [] | ||
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i = 0 | ||
while i < my_phrase.length | ||
if my_phrase[i] != " " | ||
my_phrase_original << my_phrase[i] | ||
my_phrase_reversed << my_phrase[i] | ||
end | ||
i += 1 | ||
end | ||
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string_reverse(my_phrase_reversed) | ||
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i = 0 | ||
while i < my_phrase_original.length | ||
return false if my_phrase_original[i] != my_phrase_reversed[i] | ||
i += 1 | ||
end | ||
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return true | ||
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end |
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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 is the number of characters in the input string parameter.