Great explanation ! It would be even more interesting to continue on with determining a formula to calculate n directly from N (100 in this example). And a little quadratic equations magic would yield the following : n = ceil(√(2N + 1/4) - 1/2)
+Just PK then your first egg is broken and you enter the second phase of the algorithm which is a linear search from 0 to 13 using the second egg, this is O(n). The first egg enjoys an O(logN) runtime. So if the egg broke on the 14th floor (first iteration of the first algorithm) then you've identified the worst case scenario and the work done between the two eggs is the most unbalanced.
I think in this case you would divide each interval another K-1 times? For example 100 separates to |14 | |+13 | |+12 |... then you divide the |14| section to smaller sections such as n(n-1)/2>=14 and so on...
i still do not understand, why as the floor number increases, we want to decrease the number of times we use egg 2 (as well as the # of floors going up smaller)? at video 9:34. can someone please help? :(
This is a great explanation for someone who already has some idea of what's going on, but not so much for a complete noobie I don't think.
Could you please explain why you have considered the difference as 1? why not N, N-2, N-4,... or N, N-3, N-6....
Naveen I'm guessing you'll be skipping a floors
Great explanation ! It would be even more interesting to continue on with determining a formula to calculate n directly from N (100 in this example). And a little quadratic equations magic would yield the following :
n = ceil(√(2N + 1/4) - 1/2)
why don't we decrease the separators by 2
eg for the total 10 floors
first go 6 floors high
next go 4 floors high
yolob 21 only have 2 eggs you'll be skipping floors
Thank you. Way to use your knowledge to help the world.
best explanation finally...Thanks for all your efforts
Considering 10 floors & 2 eggs, what if it breaks at 4th floor, the next check would be at 3rd floor and if it breaks there as well?
You are going from bottom to top
If you only have 2 eggs. How can you possible ever find out where the threshold is on 2 tries
your explanation was perfect for anyone to understand but please make it brief
Thanks to you! Finally understood this weird puzzle
what if u failed on the first try i.e, on 14th floor the egg broke?
+Just PK then your first egg is broken and you enter the second phase of the algorithm which is a linear search from 0 to 13 using the second egg, this is O(n). The first egg enjoys an O(logN) runtime. So if the egg broke on the 14th floor (first iteration of the first algorithm) then you've identified the worst case scenario and the work done between the two eggs is the most unbalanced.
what about 3 eggs instead of 2
Great explanation. Thank you!
finally understood the problem.. keep up the good work (y)
Great video!
What if you have K eggs?
I think in this case you would divide each interval another K-1 times?
For example 100 separates to |14 | |+13 | |+12 |... then you divide the |14| section to smaller sections such as n(n-1)/2>=14 and so on...
skip to @10:00
the way he belabored the explanation man hahahaha, it's like trying to defocate with a big constipated poop
Real good explanation !!
what if there are n eggs and n floors???
i still do not understand, why as the floor number increases, we want to decrease the number of times we use egg 2 (as well as the # of floors going up smaller)? at video 9:34. can someone please help? :(
yeah
u can mail me at procoder9973@gmail.com
Roushan Singh wtf
amazing explanation ... I seldom write comments under a video . :) keep it up .
eggazing!
Nice explanation...
nice explanation.
would be better if short videos.
Your right! I will try to correct that mistake In the future to increase my retention time and not be a bore.
Awesome explanation , do more (y)
great explanation
nyc explanation.
Good explanation but little boring