Number of ways in which you can climb a staircase with 1, 2 or 3 steps - memoization

Question

I have to calculate the number of ways one can climb a staircase taking 1, 2, or 3 steps at a time. I know of ways to do this, for example, f(n-1) + f(n-2) + f(n-3) but I would like to know why in my implementation (which is different from the above) I do not get the correct answer. I'm using a for loop instead. The problem is every time a value is returned that is non-zero but that value includes a pre-calculated answer, and hence is not used by my code. What changes do I need to make?

#include <iostream>
#include <map>

using namespace std;
int nWays = 0;

int stepPerms(int n, map<int, int> &memo) {

    if (memo.find(n) != memo.end())
        return memo[n];
    if (n == 0)
        return 0;
    if (n < 0)
        return -1;
    for (int i = 1; i <= 3; i++) {
        int retVal = stepPerms(n - i, memo);
        if (retVal == 0)
            memo[n] = ++nWays;
    }
    return memo[n];
}

int main() {
    map<int, int> memo;
    cout << stepPerms(5, memo);//cout << stepPerms(3) << endl;
}

Answer 1

I solved it like this :

#include <iostream>
#include <map>

using namespace std;
map<int, int> memo;

int stepPerms(int n) {
    if (memo.find(n) != memo.end())
        return memo[n];
    if (n == 0)
        return 0;
    if (n == 1)
        return 1;
    if (n == 2)
        return 2;
    if (n == 3)
        return 4;
    for (int i = 1; i <= 3; i++)
        memo[n] += stepPerms(n - i);
    return memo[n];
}

int main() {
    cout << stepPerms(27);
}