how much time did it took to complete the playlist and how many videos were you watching everyday

Bro there is competative programming sheet also, if you have finished that do a deep dive in stl if c++ or collections if in Java, then do math a lot, remember no tutorial can prep you for competitive programming it is just you and you have to go deep dive in it and eventually you will learn ,from some who is 2000 on Codeforces. Best of luck

WILL THE BUCKET WORK IF THE NO,S ARE NOT IN CLUSTERS

@@ArkojyotiDeyOfficial Yes, Works for every possible arrangment of the elements.

both are correct because that line is used to execute the point "num[i] will go to twos if it is in ones" . lets consider the the array be arr={2,2,2,3} step 1: ones=2 and twos=0 step 2: num[i]==2 and ones is also 2 which means we have to delete 2 from ones. now ones==0 if it the first occurrence of 2 then it should be present in ones but it is not (because ones==0) which means it is the second occurrence of 2 which means 2 is now added in twos that why we use the condition that ***it should not be present in ones*** by ~ones.

Ya similar doubt, I think the array should be sorted....

No it works for unsorted array as well

Same Thought Bro

i didnt understand this concept can you explain

Same vro😢

it will give desired output after the complete traversal of array may be you done some error in your calculation #include using namespace std; int main() { vector arr={1,2,3,2,3,2,3}; int ones=0,twos=0; int n=arr.size(); for(int i=0;i

this is not wrong the answer will be (6^4)&(~two). ones will change here but if you completely traverse whole array the ones will contain 6

yes

that's why u are here

not update yest..

Congratulations bro🎉🎉🎉

kl rashtrapati bhavan aa jana prize lene

@@modiji8706 😂😂😂😂😂😂

bro i am also searching for it 🤣🤣

as elements are not in even number if we xor 1 1 1 2 then 1^2 will be left at the end

@@artifice_abhi ok ok got it... thank you for the explanation

loop will not go ahead after finding 2....2 will be returned

no it only return ans==2 for every index we are checking we will use if(arr[i] != arr[i-1]) return arr[n-1]; it will return n-1 index value

i think we add the number and also delete it if it is duplicate thats why only unique element will remain in ones

it will give desired output after the complete traversal of array may be you done some error in your calculation. #include using namespace std; int main() { vector arr={1,2,3,2,3,2,3}; int ones=0,twos=0; int n=arr.size(); for(int i=0;i

suggestion for you:: you can also try to get answer from chatgpt . most of the time it will find error in your code . It will save your lot of time.

no, the way in the video is correct. we want to count the number of 1's in each bit position. so we need to iterate through each bit position first, and then each number from the array to count how many 1's there are. hopefully that makes sense

U need to take all elements thrice except one element. If u change 6 to 4.. you'll get the correct output.

😂😂 ha bhai sach m kabhi kabhi samjh m aana band ho jata h ,,bhai ki english to nxt lvl hai