Java Program for Count ways to reach the n’th stair

There are n stairs, a person standing at the bottom wants to reach the top. The person can climb either 1 stair or 2 stairs at a time. Count the number of ways, the person can reach the top. Consider the example shown in diagram. The value of n is 3. There are 3 ways to reach the top. The diagram is taken from Easier Fibonacci puzzles

Java

class stairs {
    // A simple recursive program to find n'th fibonacci number
    static int fib(int n)
    {
        if (n <= 1)
            return n;
        return fib(n - 1) + fib(n - 2);
    }

    // Returns number of ways to reach s'th stair
    static int countWays(int s)
    {
        return fib(s + 1);
    }

    /* Driver program to test above function */
    public static void main(String args[])
    {
        int s = 4;
        System.out.println("Number of ways = " + countWays(s));
    }
} /* This code is contributed by Rajat Mishra */

Output:

Number of ways = 5

The time complexity of the above implementation is exponential (golden ratio raised to power n). It can be optimized to work in O(Logn) time using the previously discussed Fibonacci function optimizations. 

Java

class stairs {
    // A recursive function used by countWays
    static int countWaysUtil(int n, int m)
    {
        if (n <= 1)
            return n;
        int res = 0;
        for (int i = 1; i <= m && i <= n; i++)
            res += countWaysUtil(n - i, m);
        return res;
    }

    // Returns number of ways to reach s'th stair
    static int countWays(int s, int m)
    {
        return countWaysUtil(s + 1, m);
    }

    /* Driver program to test above function */
    public static void main(String args[])
    {
        int s = 4, m = 2;
        System.out.println("Number of ways = " + countWays(s, m));
    }
} /* This code is contributed by Rajat Mishra */

Output:

Number of ways = 5

Java

// Java program to count number of ways to reach n't stair when
// a person can climb 1, 2, ..m stairs at a time

class GFG {
    // A recursive function used by countWays
    static int countWaysUtil(int n, int m)
    {
        int res[] = new int[n];
        res[0] = 1;
        res[1] = 1;
        for (int i = 2; i < n; i++) {
            res[i] = 0;
            for (int j = 1; j <= m && j <= i; j++)
                res[i] += res[i - j];
        }
        return res[n - 1];
    }

    // Returns number of ways to reach s'th stair
    static int countWays(int s, int m)
    {
        return countWaysUtil(s + 1, m);
    }

    // Driver method
    public static void main(String[] args)
    {
        int s = 4, m = 2;
        System.out.println("Number of ways = " + countWays(s, m));
    }
}

Output:

Number of ways = 5

Time Complexity: O(m*n)

Auxiliary Space: O(n)

Please refer complete article on Count ways to reach the n’th stair for more details!