คุณกำลังพยายามหาข้อมูลเกี่ยวกับหัวข้อการหาเงินออนไลน์หรือไม่? คุณกำลังมองหาหัวข้อที่เหมาะสม Google Coding Interview With A Competitive Programmer หรือไม่? ถ้าเป็นเช่นนั้นโปรดอ่านบทความนี้ทันที.
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Google Coding Interview With A Competitive Programmer | เฉลย programming.
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รูปภาพที่เกี่ยวข้องกับหัวข้อ เฉลย programming.
Google Coding Interview With A Competitive Programmer
ความรู้ทั่วไปเกี่ยวกับการลงทุน, ดูรายละเอียดที่นี่ : https://brokengroundgame.com/business-investment/.
ข้อมูลที่เกี่ยวข้องกับหัวข้อ เฉลย programming.
ในวิดีโอนี้ ฉันทำการสัมภาษณ์เกี่ยวกับการเขียนโปรแกรม Google จำลองกับ Errichto ซึ่งเป็นโปรแกรมเมอร์ที่มีการแข่งขันสูง ในฐานะวิศวกรซอฟต์แวร์ของ Google ฉันได้สัมภาษณ์ผู้สมัครหลายสิบคน นี่คือการสัมภาษณ์แบบเข้ารหัสที่คุณจะได้รับจาก Google หรือบริษัทเทคโนโลยีรายใหญ่อื่นๆ ดูวิดีโอที่เราทำในช่องของ Errichto: กำลังเตรียมบทสัมภาษณ์เกี่ยวกับการเข้ารหัสหรือไม่ ฝึกฝนด้วยวิดีโอคำอธิบาย 77 รายการของคำถามสัมภาษณ์ยอดนิยมและพื้นที่ทำงานเขียนโค้ดเต็มรูปแบบบน AlgoExpert: (ใช้รหัสโปรโมชัน “clem” เพื่อรับส่วนลด!) ..
https://brokengroundgame.com/ หวังว่าข้อมูลในบทความนี้จะมีคุณค่ามากสำหรับคุณ. ขอแสดงความนับถือ.
การค้นหาที่เกี่ยวข้องกับหัวข้อGoogle Coding Interview With A Competitive Programmer.
เฉลย programming
Google Coding Interview With A Competitive Programmer
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Hope you all found this coding interview insightful! And check out the video we made on Errichto's channel where we chatted about coding interviews, Math in Software Engineering, and more: https://www.youtube.com/watch?v=Y8VeyLG3NhY
If Errichto is not hired by Google yet, I think I will have to reconsider my career choices. He is too intelligent that his brain works faster than his speech.
Pls do one with Kevin Sun (ksun48)
He actually completed the given question on 11:56 then Clement realize oh it's been done.
Damn in 30:20 the solution would've been as simple as changing a few arguments in the first algorithm. Instead of creating a hash map of middle points, the hash map would consist of the middle points and the length, and the same way they counted all possible rectangles in a row this would work as well.
There is a simple way to reduce the second problem to the first. Considering pair a, b apply rotation from the origin to an angle such that vector ab becomes paralel to oX then store the angle in the hashmap too.
A fucking disgraceful to candidate by google.
Im currently finishing my Master degree in Systems Engineering and I will have to look for a job after next year, and seeing this video makes me super anxious because I think I dont know anything valuable enough to get a job atm
Don't compare your level 1 with someone's level 1000. LOL
10:20 an interviewer with C++ knowledge asks what is make_pair?
@Errichto This is a legendary solution. It gets as beautiful as it can get.
For the second one I like the idea of creating a hash table and throwing line segments with the same slope all under one key. Then we can check perpendicular slopes. Two lines are perpendicular if one is the negative reciprocal of the other, like n and -1/n. Maybe there’s a way to do this using the hash table without going too far past O(n^2) time.
I think with solution 2 you should get a complexity of N^2+N.
Finding the center of diagonal is smart as all points of the rectangle will be equidistant.
From A and C you can guess all possible B and D.
Thank you so much!
gives 2 axis…
I am good with 2 axis…well its easy until 10 axis…11 will be hard
no shitt man !!!!
Why are they calling squares rectangles?
That's what a real-life Gilfoyle from Silicon Valley looks like
I software engineering Y2 near Y3
And I feel I should end my studies and go to countryside and start farming after watching this
Gotta say, this looks intimidating
For the diamond problem would interating over a matrix of 3×3 then determining that only 1 diamond could be formed in such an area be useful as I could see that would be a linear time complexity. As the area can only contain 1 diamond meaning for every 3×3 area 1 result is the case. If more points were added for each y row to the left or right as soon as 3 more were added another diamond would exist allowing for 2 diamonds in a 3×4 area.
I'm not really a programmer just an engineer, so correct me if I'm wrong.
I feel there is a very beautiful solution for the general case, similar to his 2nd proposed solution. You always get a rectangle if the middle points of the diagonals meet at the same point and the diagonals have the same length. So can't I just do the following? Iterate through all pairs of points and find their distances (distances squared should also suffice, so no roots would need to be computed) and middle points. Then count those number of triples (xm, ym, dist[or dist squared]) or rather the number of combinations of triples, similar how he did it for the "easy" version.
Would this work? What would the time complexity? Isn't it also N*N*log(N)?
Really slow.
rotating 90º is exchange x by y and put one negative. if you double de "vector" size it will rotate 45º (I say vector because you now you must put the rectangle on the origin)
25:12 When you solve all the problems in interview and there is time🤣🤣
Interviewer- let me start the 45 min counter.
Erichto – 8 min later next problem please.
Wont the first solution until 20 mins which counts vertical lines fail if we have 6 points, ie. add two points to the right?
We would have 3 small rectangles with (1234), (3456), (5678) one huge rectangle (1278) and two other rectangles (1256) and (3478) which gives us a total of 6 rectangles.
But the first solution just counts vertical lines with same y and y_above coordinates which would give the answer as 4?
I was hoping that he start with a truth table before jumping into coding the states.
For me its just two people gone crazy
How many of you dead now
What just happened here? I am still struggling to understand the solution.
oj kamil pozaginal klementa
Gutka thookde bhai
The diagonal problem I think can be solved in O(n^2) by keeping track of both the middle point and length between each unique point pair in a hash map. Then you can count similarly to the solution of the first problem (don't count first occurrence).
i just realize im dumb.
Diagonals of a rectangle, besides bisecting each other, should have the same length as well. Maybe a more complex form of the first part could be a solution to the second part?
Dividing the result that you overcounted is my new life motto.
so janitor it is…
Click bait