We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsDisclaimersPrivacyTermsContactAbout Us

Find Number of Ways to Reach the K-th Stair

Difficulty: Hard

Problem Description

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 stair k, and performs some operations to reach stair k again.

Key Insights

  • Alice can either go down one stair or jump up by increasing powers of two.
  • The jumps increase exponentially, which can lead to reaching higher stairs quickly.
  • The key challenge is to account for both upward and downward movements while ensuring that downward moves do not occur consecutively.

Space and Time Complexity

Time Complexity: O(log k)
Space Complexity: O(1)

Solution

To solve this problem, we can use a combinatorial approach to count the ways to reach stair k. The upward jumps increase exponentially, which means we can reach stair k using a combination of jumps and downward moves. We can represent the stair reaching process as a tree, where each node represents a stair and branches represent possible operations.

  1. Each jump to stair i + 2^jump can be represented as increasing the jump count, which allows us to reach further stairs.
  2. We can use a recursive approach or memoization to store previously computed results to avoid recalculating the number of ways to reach a given stair.
  3. Special care must be taken to ensure that downward moves are not repeated consecutively.

Code Solutions

def findWays(k):
    # Base case for when k is 0
    if k == 0:
        return 2  # Two ways: stay at 1 or go down to 0 and come back
    
    # Initialize the number of ways to reach each stair
    ways = [0] * (k + 1)
    
    # There's one way to stay on stair 1
    ways[1] = 1
    
    # Calculate ways for each stair up to k
    for i in range(2, k + 1):
        # Count ways by jumping up and going down
        ways[i] = ways[i - 1] + (ways[i - (1 << (i - 1))] if i - (1 << (i - 1)) >= 1 else 0)
    
    return ways[k]

# Example usage
print(findWays(0))  # Output: 2
print(findWays(1))  # Output: 4
← Back to All Questions