Cell metosis hackerblock

#1

sir in the even case , we did not consider f(i+1) in our calculation because you have said it would take more time.
sir please explain me

#2

hello @mzk1994

say if i is even then

f(i) = min(f(i-1) , f(i+1) , f(i/2) )

now becuase i is even i+1 will be odd so here we cannot perform division step , only option we are left with is either add one to it or subtract one to it.
so
i+1 will become -> (i+2) on adding or i (on subtracting ,discard this as we reached fthe same from where we started)

f(i) = min(cost of subtracting + f(i-1), cost of div + f(i/2) , cost of adding 2 ones + f(i+2) )

f(i+2 /2 ) -> cost of div + f(i/2)
here u can see we have extra cost for div and 2 cost of adding.
thats why we discarded this state.

refer this for clarity

#3

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