You are given 2 eggs.
-> You have access to a 100-storey building.
-> Eggs can be very hard or very fragile means it may break if dropped from the first floor or may not even break if dropped from 100 th floor.Both eggs are identical.
-> You need to figure out the highest floor of a 100-storey building an egg can be dropped without breaking.
-> Now the question is how many drops you need to make. You are allowed to break 2 eggs in the process
Plz Explain in easy way

• First thought which comes to our mind is to use binary search, we first drop Egg#1 from 50th floor, if it does not break, then try the middle of second half, if breaks then we have to try each floor in first half. But this will give worst case number of drops as 50(if it brakes on 50th floor, then we have to try from 1 to 49 floors sequentially).
  
  Second thought is to try xth floor then 2xth floor till 100th, in this case worst case time will be (100/x)+(x-1). worst case will be when Egg #1 breaks at 100th floor then we have to try Egg #2 from (100-x)th to 99th floors. In (100/x) + (x-1) equation, with increase in x, 100/x decreases while (x-1) increases, thus we can minimize it when 100/x = (x-1), this gives x ~10, which gives worst case number as 19 drops.
  
  But increasing the floor every time by x is not a very nice idea, as with each new increase in Egg #1 drop, we should decrease Egg #2 drops to minimize worst case number. so if we drop Egg #1 from xth floor initially, then in next turn we should try x + (x-1)th floor(to keep the worst case number same).
  Thus we can say X + (x-1) + (x-2)…1 = 100
  X(X+1)/2 = 100 => X=14.
  
  So we should drop Egg #1 from 14th, then 27th, then 39th and so on.
• 9 years agoHelpfull: Yes(0) No(4)