Find Number of Ways to Reach the K-th Stair

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Dynamic Programming

You are given a non-negative integer k. There exists a staircase with an infinite number of stairs, with the lowest stair numbered 0.

Alice has an integer jump, with an initial value of 0. She starts on stair 1 and wants to reach stair k using any number of operations. If she is on stair i, in one operation she can:

Go down to stair i - 1. This operation cannot be used consecutively or on stair 0.
Go up to stair i + 2^jump. And then, jump becomes jump + 1.

Return the total number of ways Alice can reach stair k.

Note that it is possible that Alice reaches the stair k, and performs some operations to reach the stair k again.

Example 1:

Input: k = 0

Output: 2

Explanation:

The 2 possible ways of reaching stair 0 are:

Alice starts at stair 1.
- Using an operation of the first type, she goes down 1 stair to reach stair 0.
Alice starts at stair 1.
- Using an operation of the first type, she goes down 1 stair to reach stair 0.
- Using an operation of the second type, she goes up 2⁰ stairs to reach stair 1.
- Using an operation of the first type, she goes down 1 stair to reach stair 0.

Example 2:

Input: k = 1

Output: 4

Explanation:

The 4 possible ways of reaching stair 1 are:

Alice starts at stair 1. Alice is at stair 1.
Alice starts at stair 1.
- Using an operation of the first type, she goes down 1 stair to reach stair 0.
- Using an operation of the second type, she goes up 2⁰ stairs to reach stair 1.
Alice starts at stair 1.
- Using an operation of the second type, she goes up 2⁰ stairs to reach stair 2.
- Using an operation of the first type, she goes down 1 stair to reach stair 1.
Alice starts at stair 1.
- Using an operation of the first type, she goes down 1 stair to reach stair 0.
- Using an operation of the second type, she goes up 2⁰ stairs to reach stair 1.
- Using an operation of the first type, she goes down 1 stair to reach stair 0.
- Using an operation of the second type, she goes up 2¹ stairs to reach stair 2.
- Using an operation of the first type, she goes down 1 stair to reach stair 1.

Constraints:

0 <= k <= 10⁹

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible step sizes I can take to reach the next stair, and are they provided as input?
Can the value of K (the target stair) be zero or negative?
If there are no possible ways to reach the K-th stair, what should the function return?
Are the step sizes always positive integers?
Is it possible for K to be a very large number (e.g., exceeding the maximum integer value), and if so, how should I handle potential overflow issues when calculating the number of ways?

Brute Force Solution

Approach

The brute force way to find how many ways to reach the K-th stair is to explore every possible path. We consider taking either one step or two steps at a time, until we either reach the target stair or overshoot it. By trying every single combination of steps, we can count the successful paths.

Here's how the algorithm would work step-by-step:

Start at the bottom of the stairs.
Consider taking one step.
From that new position, again consider taking either one step or two steps.
Keep repeating this process of taking one or two steps from each new position until one of two things happens: either you land exactly on the K-th stair, or you go past it.
If you land exactly on the K-th stair, you've found one way to reach it, so count it.
If you go past the K-th stair, that path doesn't work, so stop.
Go back and try different paths by making different choices of one step versus two steps.
Continue exploring all possible combinations of one-step and two-step moves until you have exhausted all possibilities.
The final count of paths that land exactly on the K-th stair is the answer.

Code Implementation

def find_number_of_ways_to_reach_the_k_th_stair(k_th_stair):
    number_of_ways = 0

    def explore_paths(current_stair):
        nonlocal number_of_ways

        # Base case: We reached the target stair
        if current_stair == k_th_stair:
            number_of_ways += 1
            return

        # Base case: We overshot the target stair
        if current_stair > k_th_stair:
            return

        # Explore taking one step
        explore_paths(current_stair + 1)

        # Explore taking two steps
        explore_paths(current_stair + 2)

    # Initiate exploration from the bottom stair
    explore_paths(0)

    return number_of_ways

Big(O) Analysis

Time Complexity

O(2^k) – The brute force approach explores all possible paths by considering taking either one step or two steps at each stage until reaching the k-th stair or overshooting it. In the worst-case scenario, where we primarily take single steps, the number of possible paths grows exponentially. Specifically, each stair offers a binary choice (one step or two steps), so the number of paths is proportional to 2 raised to the power of k, where k is the target stair. Therefore, the time complexity is O(2^k).

Space Complexity

O(K) – The algorithm uses recursion to explore all possible paths. In the worst-case scenario, the function may call itself `K` times (taking one step each time), creating a call stack of depth `K`. Each recursive call consumes a constant amount of memory for its stack frame. Therefore, the auxiliary space used by the recursion stack is proportional to K, where K is the target stair.

Optimal Solution

Approach

The most efficient way to solve this problem is to recognize it can be broken down into smaller, overlapping subproblems. We use a 'remembering' trick to avoid recalculating the same things multiple times. This drastically speeds up the process.

Here's how the algorithm would work step-by-step:

Think about reaching the K-th stair. You can only get there from the (K-1)-th stair or the (K-2)-th stair.
The total number of ways to reach the K-th stair is the sum of the number of ways to reach the (K-1)-th stair and the number of ways to reach the (K-2)-th stair.
Now, imagine you already know the number of ways to reach each stair up to the (K-1)-th and (K-2)-th stair. You can just add those values together to get the answer for the K-th stair.
To avoid recomputing these values, store the number of ways to reach each stair as you calculate them. This 'memory' makes the whole process much faster.
Start with the base cases: There's one way to reach the first stair, and one way to reach the second stair (assuming you can only take steps of size one or two).
Then, systematically calculate the number of ways to reach each subsequent stair, using the 'remembered' values from the stairs before.
Keep going until you've calculated the number of ways to reach the K-th stair. The value you've stored is your answer.

Code Implementation

def find_number_of_ways_to_climb(k_th_stair):    ways_to_reach_stair = [0] * (k_th_stair + 1)
    # There is one way to reach the first stair.
    ways_to_reach_stair[1] = 1
    if k_th_stair > 1:
      # There is one way to reach the second stair.
      ways_to_reach_stair[2] = 1

    for stair_number in range(3, k_th_stair + 1):
      # Sum ways to reach previous two stairs.
      ways_to_reach_stair[stair_number] = ways_to_reach_stair[stair_number - 1] + ways_to_reach_stair[stair_number - 2]

    # Return the number of ways to reach the K-th stair.
    return ways_to_reach_stair[k_th_stair]

Big(O) Analysis

Time Complexity

O(k) – The provided solution uses dynamic programming to calculate the number of ways to reach the k-th stair. The algorithm iterates from the base cases (stair 1 and stair 2) up to the k-th stair, calculating the number of ways to reach each stair based on the two preceding stairs. Since the algorithm iterates through each stair from 1 to k once, the time complexity is directly proportional to k. Therefore, the time complexity is O(k).

Space Complexity

O(K) – The algorithm uses a memory to store the number of ways to reach each stair up to the K-th stair. This memory is an array (or similar data structure) of size K to store these intermediate results. Therefore, the auxiliary space required is proportional to K, the number of stairs. This results in a space complexity of O(K).

Edge Cases

Case	How to Handle
K is zero or negative	Return 1 if K is 0 (base case: already at the destination), and 0 if K is negative (invalid destination).
K is a very large number, potentially leading to integer overflow	Use a data type that can accommodate larger numbers (e.g., long in Java/C++, arbitrary-precision integers in Python).
The step sizes array is empty	If the step sizes array is empty, and K > 0, there are no ways to reach the K-th stair, so return 0.
The step sizes array contains zero	If the step sizes array contains zero, then zero is a valid step, so the solution should still work, though potentially causing infinte ways to reach k if k > 0 - therefore the code must check if zero is the only step.
The step sizes array contains negative numbers	If steps are only allowed to be of positive value, negative step sizes mean that the problem is impossible to solve.
Only one stair and a step size of 1	Should return 1, since it's directly reachable.
K is 1, and step sizes array does not contain 1	If no step size of 1 exists, return 0 because it is impossible to reach the first stair.
K is small, but the allowed steps are all much larger than K.	Return 0, as it's impossible to reach K with the given steps.

K is zero or negative

How to Handle:

Return 1 if K is 0 (base case: already at the destination), and 0 if K is negative (invalid destination).

K is a very large number, potentially leading to integer overflow

How to Handle:

Use a data type that can accommodate larger numbers (e.g., long in Java/C++, arbitrary-precision integers in Python).

The step sizes array is empty

How to Handle:

If the step sizes array is empty, and K > 0, there are no ways to reach the K-th stair, so return 0.

The step sizes array contains zero

How to Handle:

If the step sizes array contains zero, then zero is a valid step, so the solution should still work, though potentially causing infinte ways to reach k if k > 0 - therefore the code must check if zero is the only step.

The step sizes array contains negative numbers

How to Handle:

If steps are only allowed to be of positive value, negative step sizes mean that the problem is impossible to solve.

Only one stair and a step size of 1

How to Handle:

Should return 1, since it's directly reachable.

K is 1, and step sizes array does not contain 1

How to Handle:

If no step size of 1 exists, return 0 because it is impossible to reach the first stair.

K is small, but the allowed steps are all much larger than K.

How to Handle:

Return 0, as it's impossible to reach K with the given steps.