Reverse Words - Hannah

[hertweckhr1]

No description provided.

[hertweckhr1]

Time Complexity: n^2
Space Complexity: o(n)
-as array increases in size, number of variables stays the same, but new_string variable will increase in length as my_words gets larger)

[shrutivanw]

Since you're updating the characters in the input parameter my_words in place, you don't need to return anything. Line 24 could simply be return.

[shrutivanw]

Ditto with line 42

[shrutivanw]

Good work!

You determined the space complexity correctly for your algorithm. Think through whether you can think of another approach that does not require creating a new string and hence needing O(n) space.

The time complexity of your algorithm is O(n). You go through each character in the input string in the first loop. Once a word boundary is determined, that word gets reversed. Imagine a scenario where there is only one word in the input starting at index 0 and ending at index length-1. In this case, to determine the beginning and end of the word, will need n steps. After that (and hence addition), reversing the whole word would take n/2 steps. Leading to a time complexity of O(n + n/2) = O(n)
If there are only white spaces in the input, then reverse will never get called and the first times loop will run n times. Leading to O(n) time complexity.
In either case, at the end, you copy over each character from the new string back to the input parameter string needing O(n) time.

Take a look at my solution with O(n) time and O(1) space complexity: https://github.com/Ada-C10/reverse_words/blob/solution/lib/reverse_words.rb

Slack me if you have any questions!