3139. Minimum Cost to Equalize Array | Cases | Reduce values in Pairs Trick | Greedy | Math
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- Опубліковано 26 сер 2024
- In this video, I'll talk about how to solve Leetcode 3139. Minimum Cost to Equalize Array | Cases | Reduce values in Pairs Trick | Greedy | Math
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I am Aryan Mittal - A Software Engineer in Goldman Sachs, Speaker, Creator & Educator. During my free time, I create programming education content on this channel & also how to use that to grow :)
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Big salute for your efforts in explaining Hard prblms🙇♂If you're reading this comment Pleaseeee like this video so he'll post more such videos as mentioned 0:40 :))
Please UPLOAD THIS KIND OF VIDEOS!!!!!!!!! WE WILL WATCH EACH OF THEM and we're so greatful that you're giving a lot of efforts for us!
bhaiya you explain very very good please don't close
you are really different
meh chaah ke bhi apko striver ya auro se compare nhi kr skta
jabhi tough question aata hai
bas mann me aata hai aryan bhaiya ki video hogi toh pakka aa hii jayega
I am repeatedly visiting this channel since the contest ending to check his explanation of this problem. Thanks for your effort anna ❤
++
full support for these videos. we need this!
Please dont stop these kind of videos.. its amazingggg.. i recently came to know about your channel and i have been visiting it everyday since then because of how much it helps..
Big salute to your efforts. Your video is truly helpful!!!
Keep making long videos. I am also getting helped by these explanations certainly and i watch till end each of them.
Big salute for your efforts
One feedback :
You tend to explain so much & elongate the video so much that sometimes the context of the video starts to diminish.
explained it beautifylly
Karte raho aryaan bhaiya.❤
Can you also share notes that you make?, they are really helpfull
holyshiet u deserve more views brother, so insightful
bro dont stop these long vids
we will support you
we are highly dependent on you for all these contests
Likes Will Not Justify Your Content Man! Superbly Awesom
In the last case, when you have 7 gaps of length 1 and 1 gap of length 5 and when you are done pairing, no gaps will be left, because both are odd. We only increment once more in the case when, say, instead of a gap size of 5, you have a gap size of 4.
Thank u❤
Can someone pls explain the n-2 part again?
Thanks for the explanation
Thanks for the explanation...
Thanks
How are you sure that if the sum of remaining gaps is more than the greatest gap, then only 0 or 1 operation 1 is needed?
I appreciate your effort. ❤
Thanks for the efforts ✨ Really helpful
can we get the ipad notes of these all videos u post ?? plz it'll be great help
Thanks bhaiya
170 likes
Bro giving his 5 hours, so others can save their (700++ * 4 ) hours.