Contains Duplicate (LeetCode 217) | Full solution with HashSet explanation | Study Algorithms

[@pavanrajjagdale9369]

Really Great explanation👍.Thank you 😊

[@vineeshreddy9998]

Superb explanation. Thank you

[@viniz0r]

such a great explanation, thank you mate

[@mohammedilyas8824]

Great explanation and amazing content sir

[@nikoo28]

Glad I could help :)

[@SanthoshaK-px8rq]

Another Method that we can sort the array and check its neighbor values. Arrays.sort(nums); for(int i=1;i

[@bhavitvora9364]

Due to sorting, the time complexity will be O(nlogn). And using a HashSet will do it in O(n)

[@vamshisundupalle6141]

Your explanation is so good

[@satyamgupta6030]

thank you so much bhaiya for such a great explaination.

[@nikoo28]

Keep watching

[@pranaym143]

Simple and lucid explaination

[@pentainamerica]

isn't the time complexity n logn because it takes n to traverse through the array and for each element it might take log n to insert in hash set? or am I confusing with hash map?

[@nikoo28]

Hashset insert takes place in O(1)

[@lakshaysaini2070]

i have a doubt. the contains method of hashset also has a time complexity of O(n) and then there is for loop as well so shouldnt the time complexity be O(n^2). Is it the same as brute force approach?

[@nikoo28]

Contains method works in O(1)

[@lakshaysaini2070]

@nikoo28 thanks for replying back. But how is it O(1) cuz it will search through the whole set right?

[@karthik-varma-1579]

GREAT

[@jintodo6550]

very nice

[@abhinavd2]

Super ❤️

[@kadirv]

Sorry, as for the time-complexity - you check for the element in hash-set. How exactly does this happen? Won't all the has set elements be checked against to check if a number is in there or not? then the time complexity will be n square?

[@nikoo28]

Checking in HashSet happens in O(1) time

[@anshumansolanki6775]

Why are we using a HashSet and not a HashMap Nikhil ??

[@nikoo28]

what advantage would a hashmap give you?

[@anshumansolanki6775]

@@nikoo28 you are right we cant say it will improve the algorithm, can we say its one of the solutions? If I find the entry in map I can say there is a duplicate

[@SatishSingh-eb5tq]

We could have avoided the contains check. add() will return false if the number is already present!!

[@rajabhowmick1342]

for (int n : nums) { if(!set.add(n)) { return true; } }

[@SushmaSulthan]

Bro it is tricky in c

[@nikoo28]

what part are you facing a problem with?

[@Daddyboltee]

simply explain ..