Utkarsh solution

project_14_UT.R

single.call <- function(limit) { # Another cute function that returns the vector that contains the number of iterations for each number.

        memo <- rep(-1, limit)   
        memo[1] <- 0

        for(i in c(2:limit)) {  
                l <- 0
                n <- i
                while(n >= i) {  # Check only so long as "n > i" and not "1" this is basically the optimization we wanted. 
                        l <- l + 1
                        if(n %% 2 == 0) {
                                n <- n / 2
                        } else {
                                n <- 3 * n + 1
                        }
                } # This while loop makes sure that the number is iterated only till the time it is greater than one of the previously computed number. 
# In our case, (where the number is 13) this loop will run till the value after iteration reaches 10 (which has been previously computed and is stored in memo[10] think why?)
                memo[i] <- memo[n] + l  # This is where the magic takes place. You count the steps that took while iterating 13 -> 40 -> 20 -> 10 (that is 4, this is "l") and then just add it to memo[10] (which contains the number of iteration that are needed to go from 10 to 1) 
        }

        return(memo)
}

system.time(temp <- single.call(1000000)) # Now do this for 1,000,000 (instead of 13)
which(temp == max(temp)) # Which number has the maximum iterations