(Last time I checked) indexOf uses linear search to check for an element. So it's like doing a for loop and checking every element until it's found. It doesn't benefit from the fast look-up time of a hashtable.
I learned the benefits and use cases of different data-structures by reading a book called Grokking Algorithms and watching Introduction to Algorithms on the MIT website (MIT Course Number 6.006). Textbooks can help as well as programming puzzle websites like hackerrank and leetcode where you can test the performance of your code and read peoples' discussions of solutions. Looking up (and learning) what 'Big O' is helps too!
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I know I'm late to the party, but thanks for this reply, as you provide some links that make me understand why it "should be around O(1)" ^
If you reach this reply, how do you know/learn how to use the Set instead of doing a nested loop ?
Edit : what if in the first exemple; instead of the nested tab, we use jewels.indexOf(S[i]) !== -1 ? Exactly like using jewels.has(S[i]) ?
(Last time I checked)
indexOfuses linear search to check for an element. So it's like doing a for loop and checking every element until it's found. It doesn't benefit from the fast look-up time of a hashtable.I learned the benefits and use cases of different data-structures by reading a book called Grokking Algorithms and watching Introduction to Algorithms on the MIT website (MIT Course Number 6.006). Textbooks can help as well as programming puzzle websites like hackerrank and leetcode where you can test the performance of your code and read peoples' discussions of solutions. Looking up (and learning) what 'Big O' is helps too!