good bro keep posting more videos

code is not working properly in python3. used pypy3 for this code if __name__ == '__main__': n = int(input()) integer_list =tuple( map(int, input().split())) print(hash(integer_list))

thanks

I think we can also use hashmap

very well explained! for people looking for a reference, here is the law at play: if p = (A mod k) and q = (B mod k) and (A+B) mod k = 0, then p+q is either 0 or k.

Tq it clear

Excellent video thanku

nice explanation👍

Hello

thank you so much please solve insertion sort 1 problem available in hackerrank

probably you should use S1=S1+S.count(s[i:])

happiness=0 n,m=map(int,input().split()) array=list(map(int,input().split())) A=set(map(int,input().split())) B=set(map(int,input().split())) for i in array: if i in A: happiness+=1 elif i in B: happiness-=1 print(happiness)

i need the code to connect to the real device please

Tree playlist: completed✅ Graph playlist: completed✅ Thank you sir hope you will upload more DSA playlist using python

very good explination

Sir plz start uploading this valuable videos.your videos is really helpful for me

Very well taught , visualisation is very important . Thanks a lot bro

Sir itni jaldi khatam ho gi pata ni chala

Clear explanation ever. Keep making more such solutions

Great explanation ever even begineers understands easily👏

Under rated channel, you deserve more subs!

Sir please or bhi DSA se related playlist late rahiye full support h aapko❤

if suppose the cycle consists of duplicate values. This will not work right..

great mic🤣

Thank you

Nice explanation!

your videos are so helpful, thanks for sharing....

Here is the easiest to understand for me logic of all: //Additional impl comments below private static void minimumBribes(List<Integer> q) { int count = 0; //bribesCounter boolean flag = false; for(int i = q.size(); i>=1; i--){ if(q.get(i-1) != i) { //is the last element different than i (5) if((i-2)>=0 && q.get(i-2)==i){ //we have single bribe occurring //swap the values when we have single bribe int swap1 = q.get(i-1); //latest unordered element that is not in the correct position int swap2 = q.get(i-2); //the element next to the latest element also not in the correct position. q.set(i-1, swap2); q.set(i-2, swap1); //bribe counter; count++; } else if ((i-3) >= 0 && q.get(i-3) == i){ //double bribe //i-3 is equal to i and i-2 is not //e.g. 2 1 | 5 (i-3) 3 (i-2) 4 (i-1) //swap operations int swap1 = q.get(i-1); //4 int swap2 = q.get(i-2); //3 int swap3 = q.get(i-3); //5 q.set(i-1, swap3); // 5 q.set(i-2, swap1); // 4 q.set(i-3, swap2); // 3 //bribe counter count+=2; } else { //more than 2 consecutive bribes flag = true; break; } } } System.out.println(flag ? "Too chaotic" : count); }

Thank you so much. You saved my day. Great explanation and examples.

On what grounds we decide that we need to travel in both the directions? i.e. forward as well as backwards? Because for this scenario of {5,4,3,2,1} if we consider then only 9 candies needs to be distributed instead of 15. I am not getting this part! 5,4,3,2,1 (Kids) 1,1,1,1,1 (Initial value of candies) 2,1, (1st iteration) 2,2,1 2,2,2,1, 2,2,2,2,1

Don't have to worry for array out of bound in python? Like in first question loop was running from 0-len(s) and not 0 to len(s) -3

Great playlist, highly recommended!

Good explanation! ❤

Best ❤

Best explained video i have ever seen

Finally i found my perfect channel

is there any updated code?

Great!!!!!! You are really good. Thanks for sharing this knowledgment from yours. Sue.

Hi, I have a doubt with the question in hackerRank. In the function for pre-order they directly have root in parameter and the self keyword for object is missing. Can you please explain how it works?