To find out the possible combination to reach the last step is a problem of combination. We are using the Dynamic programming approach to solve it with less time complexity.

To solve this problem, we will follow these steps,

1. We need an array of size equals the number of steps N.
2. We initialize the first step with the value one.
3. Initialize the second step with two because you can take two steps to reach the second step or you can directly move to the second step. Therefore, in the second step there in two possibilities.
4. For the steps those are greater than the 2nd there is two possibilities,
   1. To reach the ith step, we can go from the (i-1) step.
   2. To reach the ith step, we can go from the (i-2) step.

Let f(i)= no. Of possible combination to reach the ith step.
f(i)= f(i-1) +f(i-2)

C++ Implementation:

#include <bits/stdc++.h>
using namespace std;

int count(int num)
{
    int arr[num];
    arr[0] = 1;
    arr[1] = 2;
    for (int i = 2; i < num; i++) {
        arr[i] = arr[i - 1] + arr[i - 2];
    }
    return arr[num - 1];
}

int main()
{
    int t;
    
    cout << "Testcase : ";
    cin >> t;
    
    while (t--) {
        int num;
    
        cout << "Enter the Steps : ";
        cin >> num;
    
        cout << "Step Count : " << count(num) << endl;
    }
    
    return 0;
}

Output

Testcase : 3
Enter the Steps : 5
Step Count : 8
Enter the Steps : 10
Step Count : 89
Enter the Steps : 7
Step Count : 21