In bottom up approach what are we exactly doing in line no 31 . arr[j] + dp[i-j] . how are we breaking down the problem . suppose n = 3 so, we need to get ans from n=0 , n=1 , n=2. and the add the arr[3] in it and find the maximum.
why are we not using arr[i]+dp[i-j] .
hello @sinhadivyam19
let say we have rod of length i .
and we have cut length j from that rod so how we will represent the total cost.
it will be a[j] -> profit from j length rod
dp[i-j] -> profit from remaining rod.
so in total it will be a[j] + dp[i-j]
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