great explanation

Digit dp

Digit dp

itna hadbad me kyu sir!!! thoda breathe in kariye aur aaram se *reason* k sath aage badhiye. lag raha rat k solution kyu bata rahe ho

Noice 👍

Why is it mp[0] =-1 ? I think there is no need for that

we have to solve this in O(m+n) time complexity . your approach is taking o(m * logn).

It's solving system design using Java DSA

Chutiya explanation bkl

simply sorting and set would work for 1st one

very poor explanation

please explain the time complexity i think its time complexity in worst case o(n^2)

Thats not the better way to explain,i know its tough when you face camera,it could have been better

Even the instructor doesn't know how to solve q2 . Very bad explanation 😢

WRONG EXPLANATION FOR THE 2ND PROBLEM IN THE CODE,IT IS WRITTEN THAT MP[REM]=I; BUT WHILE EXPLANING THE CODE USING DRY RUN YOU ARE TAKING MP[SUM]=I POOR EXPLANATION

But space complexity should be 0(1)

Dear Teacher, this is not a way to teach, can't you improve yourself, ek bhi comment accha nahi hai comments section me, aapko koi pharak nahi padta kya, to pharak padna chahiye, good comments are for motivation and bad comments comments are to learn, agar koi galti bata raha hai to unhe improve kijiye na............ Aise to aap har video ke last me bolti hai ki if you have any doubts, then comment,...aur yaha to aap kisi ka reply bhi nahi kar rahi hai..

nice explaination

Bhai app thora tensed lag rahe hao. Maybe apka shayad 1st time hai. App ache bata rate ho but Thora relax hoke araam se bataiye. Rest all is good. Thanks for your regular post analysis.

poor explanation

You all are correct guys, this teacher explains like this only, not a single video of this teacher is good,....

bro how take x value plase tell bro how to take input for this code.

Great solution mate

Instead of pushing "0" to "255" range into set we just use split (".") function in java and store . Then we apply Integer.parseInt() on each element of that arr@y to check if its there in that range

How is this recursive?

ques2 - 08:10 ques3 - 40:49

for(int i=1;i>1;i++) { cout<<"💯"; }

I'm getting this = Your Code is not producing any output.

7:07 bro you also know you fkd up the explanation at this point..

Best intuitive explanation so far. Thanks.

waste solution

Hi Sidharth @sidhartha166 can you tell me one thing, how do you write with pen tablet on gfg site, because of i am able to use only in whiteboard

bro your explanations is at peaks. thank you so much

I used the first approach it got submitted bro

Solved this one today.

can any one tell what is the difficulty rate for the 2 and 3 question (medium or hard)

I don't think you need to calculate prime numbers beforehand as when you are starting from two and going till sqrt(n) you are already making sure you only divide arr[i] by prime numbers for example you check arr[i] is divisible by 2 and divide the number till it is no more divisible by two so when 4 or 6 comes they will not increase the count same for other prime numbers

Great video

thankyou bro your approach is cleared is there any practice sessons for creating a DB for projects??

i ws not getting the whats the limit of zeros but this video cleared thank you

Great explanation

I did the third question in a similar way with same time complexity but TLE, was it because I did in Python? The code that I did: from typing import List class Solution: def countSubarrays(self, n: int, arr: List[int]) -> int: ans = len(arr) for i in range(n): Min = arr[i] Max = arr[i] for j in range(i + 1, n): Min = min(Min, arr[j]) Max = max(Max, arr[j]) ln = j - i + 1 if (Min + Max) % ln == 0: ans += 1 return ans

bad explaination

Wow it's so simple... And I thought that's I use to the recursive for finding subarr.. to do the product..

best explaination

Nice explanation!

Great Explanation

Sir ques 4 exactly same solution is giving TLE for Java

great video