Відео

You made a best video on youtube . I explore all the channel but your channel best😊

Great solution and walkthrough. Thank you

nice explaination.... i upload videos on how to identify what approach like sliding window , two pointer etc.

Well explained

👌

you have written hight instead od right and ur code accepted wow i dont know how it happended but i should give u all right code class Solution { public int smallestDistancePair(int[] nums, int k) { Arrays.sort(nums); int left=0; int right=nums[nums.length-1]-nums[0]; while(left<right){ int mid=(left+right)/2; if(issmallpairs(nums,k,mid)) right=mid; else left=mid+1; } return left; } private boolean issmallpairs(int[] nums,int k,int mid){ int c=0; int left=0; for(int right=1;right<nums.length;right++){ while(nums[right]-nums[left]>mid) left++; c+=right-left; } return (c>=k); } } btw explanation was relly appreciating

NICE SUPER EXCELLENT MOTIVATED

Thank you

sunar explanation

mam you are the goat..!

well explained, thank you

nice

good clean, short explanation.

amazing!!

best explanation, other explanations in youtube should be deleted

NICE SUPER EXCELLENT MOTIVATED

dunno, why 12354 is permutated to 12435 ??? In the description of this task says "ADJACENT" only.. why 12354 > 12435 , how can we swap 3 and 4 in this case? It's not adjacent . 13254 maybee?

Very informative video. I have a question - On teamcity server, I have done soapui setup and reports are generated using ANT and groovy script as well. Can we integrate the generated reports results on teamcity and show the report results on teamcity? please share the details.

i dont understand why sorting?,,,,because ha harmonic subsequence is that in which order remains unchanged

You explained it brilliantly! This is the best explanation I found for this problem

I had to go back to some basic mathematics on Sequences and series to understand the recurrence relations. This is great vid. Thanks for the work and time you put into this !!♥

whats your approach basically? Its confusing little, are you going via the approach of having all permutations of single digits, then 2 digits, then if N itself is 3 digits then all perumtations from 100 till N?

How to run this build xml if you want to run more than 2/3 projects for getting 1 html report

stack.pop().getInteger() how it works ???

wow what amazing expalantion for uick sort

amazing solution

Thank you!

thank you

Amazing Explanation👏👏

Best explanation , so simple.

Excellent, clean cut explanation

but why this is BFS and not dp?

Thanks, very good explanation.

why we dont use a for loop by calcuting the size of queue and exploring all childs once?

I wasn't merely looking for the answer rather I was digging for a clear demonstration. Here it is. Thank you.

Great explanation, subscribed++

Best

if you use loop with in loop time complexitiy is not o(n) anymore

really the logic was good

This solution will not work in C++

great!!

Finally, understandable explanation (focused not on what to do but on why to do). Thank you.

Thanks alot for the solutions, this channel has awsome explanations.

computing isPalindrome for every node in Trie is going to be totally insane and assuming that itself is going to be of order O(n2)

Thank you very much for your explanation! I struggled for several hours to understand the related article in the Leetcode editorial but it was in vain. Then I found your video and got the idea after the first watch. So simple and clear explanation!

can you explain with dry run???

line number 9 m a[] kya hai

You are amazing

jordar

Clear explanation.