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optimal ka matlab kuch v samjha dete ho, alluwa understanding video h

thanks bro

I am glad that I discovered your channel today. Your explanation is so much better than the leetcode editorial. Your video is very concise. I hit the "Subscribe" button.

I discovered your channel today and this is the second video I watched. Thank you for making this awesome video. I only solved it with O(n) space complexity in the past. Now I know the O(1) space complexity solution.

For those who didn't understand the intuition behind this solution. Imagine 2 supports s1 and s2. s1 starts from 0 and s2 starts from sum of array. Now each rod will have 3 choice 1. Add to S1 2. Add to S2 3. Remove from S2. Notice the third choice is removing from s2 not adding into Null Set. So comparing with the solution provided in the video. op1 is Choice 1 op2 is Choice 3 op3 is Choice 2, as rod is already added into s2 thats why no change. In recursive solution op3 is Choice of adding rod into a null set. which is different in memoized answer.

How it will pass all the test cases ?. It will give tle for sure. You are traversing some tree in every dfs call and you are saying it is o(n)?

hello 10:48 video me aapne run kiya vo kese kiya?

Nice Explanation !

I think question is fairly simple, language is also easy to understand

GIVING TLE

Sometime it cant able to solve palindrome questions

@15:59 how can the min heap have <7,2> it should be <1,9> right ? we first popped <1,2> and then inserted {i,j+1} thats <1,9> and its lesser than <7,2> someone pls make me understand

Very good explanation :)

Nice explanation

How to install axioms?

great explanatiob bro

Bro can you provide the brute force solution for this in java if possible or c++ will be fine

good explanation

Thank you! Great explanation

I didn’t find anywhere this way solution. Thank you so much

this is O(nlogn) solution but we can do this with o(n) time complexity

Thank you sir 🎉

Brilliant explanation

Hello can you create a video on how to connect the node application with

Thanks for the video , nicely explained.

I really appreciate the hard work you have put into this video. Thanks for making this video.

it really helped a lot thankyou😊😊

will this work on negative numbers ?

The explanation was easy to understand, thanks

while swapping, why do we take (2*min) ?

simple and neat solution bro

The explanation was really awesome. Thanks a lot❤❤❤

volume is very low

Liked the solution. I made a small improvement with memoization. class Solution: def __init__(self): self.memo = {} def minOperations(self, n: int) -> int: if n in self.memo: return self.memo[n] v = int(math.log(n, 2)) if n == pow(2, v): return 1 low = pow(2, v) high = pow(2, v+1) d1 = n - low d2 = high - n self.memo[n] = 1 + min(self.minOperations(d1) , self.minOperations(d2)) return self.memo[n]

hy your approach is very efficient. But I did not understood the intuition of summing up ALL elements of a diagonal rather than only 2 for each corresponding element of a matrix. Like in the intuition, you explained if both (right and bottom element) sum is greater than 1, then there surely exists a path to dest even after flipping. But whats the logic with summing up ALL diagonal elements?

The apprpach 1 won't work, it depends on the order in which you remove the edge

Bhaiya can you please suggest me some platform where i can practice question asked in OA

Thank you sir, Very nice explanation!

Could you please explain why you took ans as long?

best explanation thanks ! : )

Nice explanation

What a technique! Just Awesome!

So good solution my friend

I really didn't understood anything

nice bro!!

bro, I'm so happy with this explaination

the best explaination i have ever seen

Good Explanation Bro everything connects

NIce explanation

The second if() inside check() should be a XOR not a OR.