# Solving Leetcode Interviews in Seconds with AI: Find Number of Ways to Reach the K-th Stair


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3154" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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 + 2jump. 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  

	# Explanation
	Here's the approach and code:

*   **Dynamic Programming with Memoization:** The problem lends itself to dynamic programming because the number of ways to reach a stair depends on the number of ways to reach previous stairs. Memoization is crucial to avoid recomputing the same subproblems.
*   **State Definition:** The state of the DP will be defined by the current stair `i` and the current jump value `j`.  This captures all the necessary information to make decisions about future moves.
*   **Base Cases and Transitions:**  The base case is when `i` reaches `k`. The transitions involve either going down (if allowed) or going up based on the current jump value.

*   **Runtime Complexity:** O(k * sqrt(k)), **Storage Complexity:** O(k * sqrt(k))

	
	# Code
	```python
	def solve():
    k = int(input())
    memo = {}

    def count_ways(i, jump, last_down):
        if (i, jump, last_down) in memo:
            return memo[(i, jump, last_down)]

        if i == k:
            return 1

        if i < 0:
            return 0

        if i > k + 200 :
            return 0
        
        ways = 0

        # Option 1: Go down
        if i > 0 and not last_down:
            ways += count_ways(i - 1, jump, True)

        # Option 2: Go up
        ways += count_ways(i + 2 * jump, jump + 1, False)

        memo[(i, jump, last_down)] = ways
        return ways

    result = count_ways(1, 0, False)
    print(result)

solve()
	```