1

I would like to know how to determine the time complexity for the following code, using top-down dynamic programming approach with memoization (note that i, k, d can be any integer value):

solve(i, k, d) {
    if (k >= d) {
        A[i][k] = 0;
        return 0;
    }

    if (i == 0) {
        A[i][k] = 0;
        return 0;
    }

    if (A[i][k] == null) {
        for (int j = 0; j < k + 1; j++) {
            A[i][k] = solve(i - 1, (k + 1 - j), d);
        }
    }

    return A[i][k];
}
stackyyflow
  • 767
  • 3
  • 11
  • 30
  • Well. You're missing a return statement. But why don't you start with the first branch when `j=0` and trace what happens there -- it terminates after `min(i, d-k)` steps. – MFisherKDX Nov 04 '17 at 04:04

1 Answers1

1

Issue :

solve(i - 1, (k + 1 - j), d)

will give out of bound error when j = 0 as A index should go from 0 to K [K being largest index]

ANSWER : The function will give O(n * n) complexity.

Intuition :

(it is clear that the worst case is the one where there are no solutions, so I focus on that)

Since the recursive call is made before putting values in the memoization-cache, the last (shortest) suffixes will get cached first. This is because the function is first called with a array of length N, which then calls itself with a array of length N-1, which then .... , with a array of len 0, which is cached and returns, then the array of length 1 is cached and returns, ..., length N is cached and returns.

Explanation :

Assuming size of matrix I x K and Considering worst case,

[Note] In worst case, Function call will start from right bottom of matrix

A initialization happens in two cases :

  • k >= d
  • i == 0

[Note] In worst case, k is always less that d. 

For loop for `(I, K)`
- Recursion call `for (i-1, k:[K to 0])` 
- Update value for `A[i, k]`

[Note] i = 0 is base case when function initializes A and returns 0.

akshay dhule
  • 167
  • 3
  • 17