Scale Engineer

Find Number of Ways to Reach the K-th Stair

Hard
#2802Time: Exponential, roughly O(2^N) where N is related to the number of moves. It explores a vast number of redundant paths and is too slow for the given constraints.Space: O(D), where D is the maximum recursion depth. This can be very large.
Patterns
MathDynamic ProgrammingBit ManipulationCombinatoricsMemoization

Prompt

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

Approaches

3 approaches with complexity analysis and trade-offs.

A straightforward recursive approach that explores all possible paths. The state of the recursion is defined by (current_stair, current_jump, can_go_down). This method directly models the operations available to Alice and explores the state graph without any optimization.

Algorithm

  • Define a recursive function, say countWays(stair, jump, canGoDown).
  • The function explores two possible moves from the current state:
    • Go Up: Move to stair + 2^jump, increment jump to jump + 1, and allow the next move to be a "go down" move. Recursively call countWays(stair + 2^jump, jump + 1, true).
    • Go Down: If stair > 0 and canGoDown is true, move to stair - 1. The jump value remains the same, and the next move cannot be a "go down" move. Recursively call countWays(stair - 1, jump, false).
  • The function's return value is the sum of ways from the subsequent recursive calls. A way is counted if the path eventually lands on stair k.
  • The base case for the recursion is when the number of up-moves (jump) or the current stair exceeds a practical limit, beyond which it's impossible to reach k. If stair == k, we add 1 to our count for the current path.
  • The initial call would be countWays(1, 0, true).

Walkthrough

This approach directly translates the problem statement into a recursive function. We define a state by (stair, jump, canGoDown). stair is the current stair, jump is the power for the next upward jump, and canGoDown is a boolean flag to enforce the rule that "go down" operations cannot be consecutive. The function countWays(stair, jump, canGoDown) calculates the number of ways to reach k from this state. It sums the results from the recursive calls for the "go up" and "go down" moves. A crucial issue is defining the termination/base cases correctly to avoid infinite recursion and count paths properly. Due to the massive state space (k up to 10^9), this approach is not practical and will time out.

// This is a conceptual illustration. A direct implementation is not feasible// for the given constraints due to the massive state space and termination issues.public int waysToReachK_conceptual(int k) {    // A practical implementation would need bounds for stair and jump to terminate.    return solve(1, 0, true, k);} private int solve(long stair, int jump, boolean canGoDown, int k) {    // Pruning is essential but difficult to get right.    // If we are too far above k, it might be impossible to return.    if (stair > k + 2 && (1L << jump) > stair - k) { // Heuristic pruning        return 0;    }    if (jump > 35) { // Heuristic pruning        return (stair == k) ? 1 : 0;    }     int count = 0;    if (stair == k) {        count = 1;    }     // Explore Go Up move    count += solve(stair + (1L << jump), jump + 1, true, k);     // Explore Go Down move    if (canGoDown && stair > 0) {        count += solve(stair - 1, jump, false, k);    }     return count;}

Complexity

Time

Exponential, roughly O(2^N) where N is related to the number of moves. It explores a vast number of redundant paths and is too slow for the given constraints.

Space

O(D), where D is the maximum recursion depth. This can be very large.

Trade-offs

Pros

  • Simple to conceptualize as it directly follows the problem description.

Cons

  • Extremely inefficient and not feasible for the given constraints due to an enormous state space.

  • Prone to stack overflow for deep recursion paths.

  • Defining correct and effective termination/pruning conditions is very difficult.

Solutions

class Solution {private  Map<Long, Integer> f = new HashMap<>();private  int k;public  int waysToReachStair(int k) {    this.k = k;    return dfs(1, 0, 0);  }private  int dfs(int i, int j, int jump) {    if (i > k + 1) {      return 0;    }    long key = ((long)i << 32) | jump << 1 | j;    if (f.containsKey(key)) {      return f.get(key);    }    int ans = i == k ? 1 : 0;    if (i > 0 && j == 0) {      ans += dfs(i - 1, 1, jump);    }    ans += dfs(i + (1 << jump), 0, jump + 1);    f.put(key, ans);    return ans;  }}

Video walkthrough

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