n steps with 1, 2 or 3 steps taken. How many ways to get to the top?

Question

If we have n steps and we can go up 1 or 2 steps at a time, there is a Fibonacci relation between the number of steps and the ways to climb them. IF and ONLY if we do not co

Answer 1

Here is an O(Nk) Java implementation using dynamic programming:

public class Sample {
    public static void main(String[] args) {
        System.out.println(combos(new int[]{4,3,2,1}, 100));
    }

    public static int combos(int[] steps, int stairs) {
        int[][] table = new int[stairs+1][steps.length];
        for (int i = 0; i < steps.length; i++) {

            for (int n = 1; n <= stairs; n++ ) {
                int count = 0;
                if (n % steps[i] == 0){
                    if (i == 0)
                        count++;
                    else {
                        if (n <= steps[i])
                            count++;
                    }
                }

                if (i > 0 && n > steps[i]) {
                    count += table[n - steps[i]][i];
                }

                if (i > 0)
                    count += table[n][i-1];

                table[n][i] = count;
            }
        }

        for (int n = 1; n < stairs; n++) {
            System.out.print(n + "\t");
            for (int i = 0; i < steps.length; i++) {
                System.out.print(table[n][i] + "\t");
            }
            System.out.println();
        }
        return table[stairs][steps.length-1];
    }
}

The idea is to fill the following table 1 column at a time from left to right:

N    (4)    (4,3)  (4,3,2) (4,3,2,1)
1   0   0   0   1   
2   0   0   1   2   
3   0   1   1   3   
4   1   1   2   5   
5   0   0   1   6   
6   0   1   3   9   
7   0   1   2   11  
8   1   1   4   15  
9   0   1   3   18  
10  0   1   5   23  
11  0   1   4   27  
12  1   2   7   34  
13  0   1   5   39  
..
..
99  0   9   217 7803
100             8037