Find Number of Ways to Reach the K-th Stair

Problem Description

Alice starts at stair 1 with an internal counter jump initialized to 0. In one operation she can either go down one stair (with the extra rule that a down‐move cannot be applied consecutively or when on stair 0) or go up by 2^(jump) stairs (after which jump is incremented by 1). Given a non-negative integer k (where the lowest stair is 0), count the total number of distinct sequences of operations (of any length, even including cycles that re–reach k) that end exactly at stair k.

Key Insights

• The net effect of an up operation is to add 2^(current_jump) to the current stair; the sum of ups (when performed in order) gives a total of 2^n – 1 if there are n ups.
• The down operations subtract exactly 1 per move. Thus, if there are n up moves and d down moves, the final stair is 1 + (2^n – 1) – d = 2^n – d.
• To end on stair k the relation must hold: 2^n – d = k, meaning d = 2^n – k. Not every n produces a valid d because down moves cannot exceed the number of “gaps” available (n + 1) if we require that no two downs occur consecutively.
• For a fixed n (number of up moves), a valid sequence exists if and only if 2^n – k is between 0 and n+1 (inclusive) and 2^n is at least k.
• The number of valid arrangements (given that downs cannot be adjacent) equals choosing which of the n + 1 possible “gaps” to place the d downs, i.e. C(n+1, d).
• It turns out that although Alice can “loop” even after reaching k, every valid sequence is uniquely determined by a choice of n (up count) that satisfies 2^n – k ≤ n+1 and then choosing the positions of the downs.

Space and Time Complexity

Time Complexity: O(N) iterations where N is the maximum number of up moves satisfying 2^N ≤ k + N + 1. Because 2^n grows very fast, N remains small even for k up to 10^9. Space Complexity: O(1) extra space (ignoring the space needed to compute individual binomial coefficients).

Solution

We observe that after n up operations (which add up to 2^n – 1), and d down operations (each subtracting 1), the final stair becomes 2^n – d. To reach stair k, we require d = 2^n – k. Given that down moves cannot be consecutive, they can be thought of as being placed in one of the n + 1 gaps (before the first up and between ups and after the last up); each gap can contain at most one down. Thus, if d is within the range 0 to n + 1, then the number of valid arrangements is binom(n+1, d).

The overall answer is the sum of binom(n+1, 2^n – k) over all n for which:

• 2^n >= k (to ensure d is non-negative)
• 2^n – k <= n+1 (so that the downs can be placed without adjacent downs)

This leads to a simple iterative algorithm where we try increasing values of n until 2^n becomes too large to satisfy 2^n ≤ k + n + 1.

Below are code solutions in Python, JavaScript, C++, and Java with line‐by‐line comments.

Code Solutions