Climbing Stairs with Quick Power Calculations

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Topic description

Suppose you are climbing stairs. You need  n steps to get to the top.

You can climb  one 1 or  2 more steps at a time. How many different ways can you get to the top of a building?

Thought analysis

There are many algorithms for climbing stairs, from the recursion learned at the beginning, and the constant storage of the pre value through two variables (this idea has a shadow of dynamic programming, which hides a state transition equation, but does not need to apply for additional O( n) level of memory space).

However, recursion requires a deep stack, and the time complexity of dynamic programming is also O(n). Can the time complexity be further reduced?

See the following formula

$$\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} f(n) \\ f(n - 1) \end{bmatrix} = \begin{bmatrix} f(n - 1) + f(n) \\ f(n) \end{bmatrix} = \begin{bmatrix} f(n + 1) \\ f(n) \end{bmatrix}$$

Therefore

$$\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n \begin{bmatrix} f(1) \\ f(0) \end{bmatrix} = \begin{bmatrix} f(n + 1) \\ f(n) \end{bmatrix}$$

Then the time complexity of this problem simplifies to the solution

$$\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n$$

time complexity of

To solve this n-th power, we can use the fast power. Change the time complexity of O(n) to O( $log_2n$)

Code

 public int[][] pow(int[][] a, int n) {
        int[][] ret = {{1, 0}, {0, 1}};
        while (n > 0) {
            if ((n & 1) == 1) {
                ret = multiply(ret, a);
            }
            n >>= 1;
            a = multiply(a, a);
        }
        return ret;
    }
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Summarize

In fact, I personally think that it is enough to use dynamic programming for climbing stairs, but occasionally I do encounter some people who pursue time complexity or competitions that require fast power to answer.

The difficulty of fast power is not its implementation, but how to abstract a problem into a problem that can be solved using fast power.