Amazing :) just one question: if I use a proper hash function(one that doesn't map different elements to the same value), would the performance still degrade to O(n) even if the hash table is completely full? would collision/clustering still happen in that case?
Thank you for your great explanation i followyou from Jordan ❤❤
U r the best than my teacher
Thanks a lot ❤
Thank you our great doctor❤️
thanks Dr.. :)
Amazing :)
just one question:
if I use a proper hash function(one that doesn't map different elements to the same value), would the performance still degrade to O(n) even if the hash table is completely full?
would collision/clustering still happen in that case?
When the table is completely full the performance degrade to O(n) when the collision occurs
h(x)=h(y) iffx=y
طيب ما ممكن x!=y ويبقي ليهم نفس ال hash عشان ليهم نفس module بالنسبة لل size
@@يوسفعبدالقادر-غ3ث ممكن طبعا ما ده collision
مفروض hash الكويس يبقى صعب فيه collision