He checks if character arr[R](as a key) exists in the map and returns the number of its repetitions, or 0 if it doesn't.So when hm.put is executed it either puts the key arr[R] for the first time with 1 as the number of its repetitions, or incremented value of the previous number of repetitions.
![[Java] Leetcode 3. Longest Substring Without Repeating Characters [Sliding Windows #3]](http://i.ytimg.com/vi/-QPgQJl_CDg/mqdefault.jpg)
![[Java] Leetcode 3. Longest Substring Without Repeating Characters [Sliding Windows #3]](/img/tr.png)
Nice! Other solutions feel like magic, when they update the left pointer to the least recently seen element
so much simplified explanation. thanks!
Thanks a lot for the clear and calm explanation, I complete misunderstood it.
Glad it helped!
🐱👤Wow! this is actually a premium question on leetcode. Thank you for sharing your knowledge! 👏
My pleasure!
Nice this one was a little confusing so i gotta restudy it
Practice makes perfect and there are a couple more videos that are similar to this, so worth trying as well
@@EricProgramming thx i learned it after going thru it a couple times
Sir nice explanation 🙏🙏
Thanks for liking
Good video. What's the time and space complexity of the solution?
thank you very much❤
Thank you!
can someone explain the second part of the HashMap...why inside of the hm.put its hm.getOrDefault(arr[R], 0) + 1), like what is that saying exactly?
He checks if character arr[R](as a key) exists in the map and returns the number of its repetitions, or 0 if it doesn't.So when hm.put is executed it either puts the key arr[R] for the first time with 1 as the number of its repetitions, or incremented value of the previous number of repetitions.
If we know that K is two why can't we just do hm. remove(arr[L]) ? Instead of decrementing it first?
for this example what python libraries in basics we have to study...and is this kind of examples are asked in FAANG interviews?
huhh??