# Java Challenge - Stairmaster

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The formula for this function is f(n) = f(n-1) + f(n-2) + f(n-3).
f(1) = 1
f(2) = 1 + f(1), and
f(3) = 1 + f(2) + f(1).

Thus,
f(4) = f(3) + f(2) + f(1) = f(3) + f(2) + 1 = f(3) + 2 + 1 = 4 + 2 + 1 = 7.

I chose to solve this recursively by having a function call itself 3 times inside itself, each representing the three possible ways of taking steps.

``````import java.util.*;
import java.util.concurrent.*;

public class Stairmaster {

public static int ways = 0;

public static int stairmaster(int n) {
ways = 0;

if (n == 0) {
return 0;
}

int toReturn = stairmasterRecursive(n);
}

public static int stairmasterRecursive(int n) {
if (n == 0) {
return ++ways;
}
else if (n < 0) {
return 0;
}
else {
stairmasterRecursive(n-1);
stairmasterRecursive(n-2);
stairmasterRecursive(n-3);
}

return ways;
}

public static void main(String args[]) {
System.out.println(stairmaster(0));
}
}
``````