MINIMUM DELETIONS TO MAKE CHARACTER FREQUENCIES UNIQUE | PYTHON | LEETCODE # 1647

[@Justice4x]

really awesome job man! was stuck on this question for a couple hours. searched around youtube and your explanation was the best one. period.

[@crackfaang]

Thank you for the kind words! If there’s any videos you’d like to see, let me know and I can make the videos

[@izzuddinsyamil1479]

same! this is the best solution to understand and follow thanks

[@xRoshtack]

Great video!

[@crackfaang]

Glad you enjoyed the video! Make sure to subscribe and let me know if there’s any problems you’d like me to make videos for

[@xRoshtack]

@@crackfaang Can you do Leetcode 2060?

[@subee128]

Thanks

[@symbol767]

Wow this solution is epic and really great explaination, you're super underrated dude. Liked! We gotta get you more popular, this video is too good

[@ThisIsntmyrealnameGoogle]

awesome explanation! Subscribed!

[@abdullahgarra8477]

I don't really get why it is O(N)🤔

[@TheQuancy]

Just because there is a nested loop, doesnt necessarily mean that its N^2 The for loop is iterating through the frequency dictionary, while the while loop is iterating through the hashset. If the while loop was also iterating through the dictionary, then yes, it would be N^2, but its 2 separate data structures

[@ebenezeracquah478]

The reason is that the for loops are not nested. There are two separate for loops. The first one is in Counter(s) which traverse s and stores it in key-value pairs in a dictionary. And the second for loop traverses the key-value pairs to delete the keys with equal counts. So as he said the time complexity is O(n) + O(n) which is ultimately O(n).