Lets devise a generic solution:
Let h(k,m) be the height that can be handled with k eggs and up to m drops.
Let the base at any time be the highest known safe floor. With k eggs and m drops left we can afford to skip h(k-1,m-1) floors and do our next drop at base + h(k-1,m-1) + 1.
When an egg breaks the base remains the same and both k and m decrease by 1.
If the egg doesn’t break, increase the base by h(k-1,m-1) + 1 and then decrease just m by 1.
So lets start with 1 egg and m floors :
h(1,m) = m
start dropping at the lowest floor and work upwards one at a time.So m attempts would be required.
h(2,m) =h(1,m-1)+1 +h(2,m-1)
=(m-1)+1+1+h(1,m-2)+h(2,m-2) and so on
So finally ,
h(2,m)=(m-1)+1 + (m-2)+1 + … + (m-m)+1 = SUM of m, m-1, m-2, …1 = m(m+1) / 2
=1+ (m-2)(m-1)/2 + 1+ h(2,m-2)+h(3,m-2) and so on
So, h(3,m)=(m-1)m/2 +1 + (m-2)(m-1)/2+1 + … = SUM (1/2)j^2 – (1/2)j +1 for j from 1 to m and we get,
(1/12)m(m+1)(2m+1) – (1/4)m(m+1) + m = (1/6)(m^3 -m) + m = (1/6)(m-1)m(m+1) + m.
So , h(3,m)=(1/6)(m-1)m(m+1)+ m
For 120 floor building ,put m=1,2,3…. ans so on until u get a value greater than 120
8 drops is not enough but 9 drops would let us go as high as 120 + 9 which is enough.
Resolving technical problems:
Solve your technical problems instantly
We provide Remote Technical Support from Monday to Sunday, 7:00PM to 1:00 AM
Mail your problem details at firstname.lastname@example.org along with your mobile numberand we will give you a call for further details. We usually attend your problems within 60 minutes and solve it in maximum 2 days.