andreyka26_se
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Hey, I'm software engineer at Microsoft, with 7 years of experience. Here we are talking about F(M)AANG big tech interviews: leetcode, system design and corpo life.

YouTube: @andreyka26_se
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So, today we will stream starcraft, it was hard week, I deserved xD
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now see the pain, first pic - my solution, second pic - official solution.. fuck my life😁
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BTW, did it, now 400+😎
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this is TRUE prompt engineering xDD
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Starcraft in 10 mins
Daily is deadly today

I had almost no idea how to solve it, so I just checked the solution after 20 mins. It is too much effort I guess.

If I'm getting this on the interview - I will not join the company 1000%

https://leetcode.com/problems/find-the-minimum-area-to-cover-all-ones-ii/description/?envType=daily-question&envId=2025-08-23

#daily
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Solution

Basically I will explain and draw the solution so at least you can understand it. Understanding the solution is also very useful skill.

We have 6 ways to split the area. Do you remember daily few days ago? Find the minimum area with 1's? So we are going to apply this algo in each of these sections.

Now what we need to do is try out all these sections, and find minimum area in each of them, and then run min(prev, currentlycalculated) over them as result to return
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Visualization of how we split and try to calculate.

Then we rotate the grid and do it again.

In total you will get 6 such combinations of splitting, and just return the min out of them. Fuck it.
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Have a nice day, and remember you need to balance and have your own life apart from leetcode
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Solution

It is sliding window, we should find the biggest contiguous subarray containing only `1`. We must drop at least 1 item.

So whenever we see zero - we drop it, and try to extend the subarray. When we have more than 1 zero in our subarray - we shrink window, until we used only one deletion
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Sorry, was doing system design consultations, so I solved the daily challenge the last 10 mins as well
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Solution

not sure about right solution, mine is cumbersome - but correct.

Just getting all "diagonals" from bottom to up, and reversing them if we need to move them from up to down.
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