Question

#8 Write an iterative method that calculates the SUM of all integers between 1 and a given integer N (input into the method).  Write a corresponding recursive solution. (return answer, don’t print)

#8 Solutions:

6      publicstaticintIterative(intn){
7    intsum = 0;
8    for(inti = 1 ; i<n; i++){
9        sum = sum + i;
10    }
11    returnsum;
12  }

Recursive Solution

6      publicstaticintIterative(intn){
7
8    if(n == 0) return0;
9    return1 + Iterative(n-1);
10 }

Please answer below question:
Looking at your solution(s) in #8 above – it’s easy to see that as n gets larger, the solution is takes longer to solve (or increases the chance of “overflow”).  Is this the best we can do? (is this the most efficient solution)? If so, explain why.  If not, give your solution (and make sure you explain why it is more efficient).

Answer 1

publicstaticintIterative(intn){

if(n == 0) return0;
return n + Iterative(n-1);
}

The solution is efficient, reason being there are two ways of finding sum recursively. Either add from 1 to n or from n to 1 in the first case we pass n so it's not possible to generate the stop condition as we wouldn't know where to stop. So the only solution is add from n to 0.

Only change in the code is return n + Iterative(n-1) instead of 1. If 1 is written it'll add 1 n times so doesn't serve our purpose. This is the only efficient solution.