Recurrence relation not clear

#1

f(i)=min { f(2i), f(i+1), f(i-1)}

#2

this is not how Dp works
for dp[i] it should be depend on previous calculated values not on future values which we have not solved

so we have modified the relation
correct one
if i is even
dp[i]= min(dp[i/2]+x,dp[i-1]+y} ;

if i is odd
dp[i]=min(dp[i-1]+y,dp[i+1/2]+x+z} ;

you can watch the video again it will definitely help

#3

Sir I didn’t understand why not dp[i+1] will reult to maximum value so not added in even case as done in odd case

#4

this is an observation
if it doesn’t happen you can’t solve it using dp

to get more feel of this
you take 2 -3 examples and try to make dp array manually
you will understand question in better manner and also observe this

#5

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