Sigrid Benezra - Edges - Palindrome check #16
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sdbenezra
changed the title
shrutivanw
reviewed
Nov 20, 2018
| end_index = my_phrase.length - 1 | ||
| while beg_index < end_index | ||
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| while my_phrase[beg_index] == " " |
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In the two nested inner loops, continue to check and confirm that beg_index is less than end_index before comparing the character at the index beg_index (or end_index) with a white space.
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Nice work! I added a minor comment to make your algorithm even more robust. Slack me if you have any questions if you'd like to discuss further. |
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Time complexity is O(n) because characters from each end of the string are compared in turn and that means you will make about n/2 comparisons depending on how many blank spaces there are. Dropping constants, you end up with O(n), linear time.
Space complexity is O(1) because there is only space needed to store a few variables for comparison, and that doesn't change no matter how long the string is.