Hayden - Edges (C10) - Palindrome Check #29
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shrutivanw
reviewed
Nov 20, 2018
| j = my_phrase.length - 1 | ||
| while i < j | ||
| if my_phrase[i] == " " || my_phrase[j] == " " | ||
| until my_phrase[i] != " " |
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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.
shrutivanw
reviewed
Nov 20, 2018
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| # time complexity: O(n) because it must iterate through each element in the array |
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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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Nice work! Please see some minor comments to make your algorithm further robust. |
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time complexity: O(n) because it must iterate through each element in the array
space complexity: O(1) because it does not require any additional memory space besides the iteration variables