BarbaraWidjon, Edges, palindrome_check #18
Conversation
| end | ||
|
|
||
| def find_start_index(my_phrase, index_val) |
There was a problem hiding this comment.
As due diligence, consider passing my_phrase.length as the third parameter to this method. Then, in the loop on line 37, you can first confirm that new_index is not out of bounds before checking the character at the index.
Because you're coding in Ruby, the algorithm works without it since at index length, the value will be nil and the loop will terminate. However, if you were coding in any other language, your algorithm will need to be more robust and account for index bounds.
| new_index = index_val | ||
|
|
||
| if my_phrase[new_index] == " " | ||
| until my_phrase[new_index] != " " |
There was a problem hiding this comment.
confirm that new_index is greater than or equal to 0 before comparing the value at that index with a white space.
|
Nice work on the algorithm. I added a few comments to make your algorithm further robust. The time and space complexities and your explanation is correct. The only suggestion I have is to always explain what n stands for. In this case n is the length of the input string or the number of characters in the input string. |
**Time complexity: O(n/2) --> O(n). The function will start iterations from index of 0 and index of length-1. Therefore, the function will iterate through half of the string (n/2). Constants are dropped, so the final time complexity is O(n)
**Space complexity: O(1) since additional memory space is not necessary for the execution of this algorithm.