Find Number of Ways to Reach the K-th Stair

• 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).

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.

1// This is a conceptual illustration. A direct implementation is not feasible2// for the given constraints due to the massive state space and termination issues.3public int waysToReachK_conceptual(int k) {4    // A practical implementation would need bounds for stair and jump to terminate.5    return solve(1, 0, true, k);6}7 8private int solve(long stair, int jump, boolean canGoDown, int k) {9    // Pruning is essential but difficult to get right.10    // If we are too far above k, it might be impossible to return.11    if (stair > k + 2 && (1L << jump) > stair - k) { // Heuristic pruning12        return 0;13    }14    if (jump > 35) { // Heuristic pruning15        return (stair == k) ? 1 : 0;16    }17 18    int count = 0;19    if (stair == k) {20        count = 1;21    }22 23    // Explore Go Up move24    count += solve(stair + (1L << jump), jump + 1, true, k);25 26    // Explore Go Down move27    if (canGoDown && stair > 0) {28        count += solve(stair - 1, jump, false, k);29    }30 31    return count;32}

• Define a recursive function solve(jump, diff, canGoDown) with a memoization table (e.g., a HashMap).
• The state is defined by jump (the next jump power), diff (the difference current_stair - k), and canGoDown (boolean flag).
• The base case: if diff == 0, we are at stair k, so we add 1 to our count.
• Pruning: If jump is too large (e.g., > 32) or diff makes it impossible to reach k, return 0. For example, if diff > jump + 1, we need more "down" steps than available slots between "up" moves.
• Check the memoization table. If the result for the current state (jump, diff, canGoDown) is already computed, return it.
• Recursively call for the two moves:
  • Go Up: solve(jump + 1, diff + 2^jump, true).
  • Go Down: If canGoDown and k + diff > 0, call solve(jump, diff - 1, false).
• Store the sum of the results in the memoization table and return it.
• The initial call is solve(0, 1 - k, true).

This method refines the recursive approach by avoiding recomputation of the same subproblems using dynamic programming. We define a state relative to the target k to try and reduce the state space. The state is (jump, diff, canGoDown), where diff = current_stair - k. The function solve(jump, diff, canGoDown) computes the number of ways to reach a state with diff = 0. The state transitions are:

• Go Up: From (jump, diff), we move to (jump + 1, diff + 2^jump).
• Go Down: From (jump, diff), we move to (jump, diff - 1). The main challenge remains the size of the state space for diff. diff can range from -k up to a positive value. For k=10^9, this is too large. This approach is only practical for small k.

1// This approach is only feasible for small k.2// For k <= 10^9, the 'diff' can be very large and negative, making the memoization map huge.3class Solution {4    Map<String, Integer> memo;5    int targetK;6 7    public int waysToReachStair(int k) {8        this.targetK = k;9        // Using a HashMap for memoization. The key represents the state.10        this.memo = new HashMap<>();11        return solve(1, 0, true);12    }13 14    private int solve(long stair, int jump, boolean canGoDown) {15        // Pruning: if we are too far, it's impossible to come back.16        // The number of down moves needed is stair - k. The max we can do is jump + 1.17        if (stair - targetK > jump + 1) {18            return 0;19        }20        // Further pruning21        if (jump > 31) {22            return (stair == targetK) ? 1 : 0;23        }24 25        String key = stair + "," + jump + "," + canGoDown;26        if (memo.containsKey(key)) {27            return memo.get(key);28        }29 30        int count = (stair == targetK) ? 1 : 0;31 32        // Go Up33        count += solve(stair + (1L << jump), jump + 1, true);34 35        // Go Down36        if (canGoDown && stair > 0) {37            count += solve(stair - 1, jump, false);38        }39 40        memo.put(key, count);41        return count;42    }43}

• Analyze the net displacement. If we perform j "go up" moves, the total upward displacement is 2^0 + 2^1 + ... + 2^(j-1) = 2^j - 1. Starting from stair 1, we would reach stair 1 + (2^j - 1) = 2^j.
• To reach stair k, we must perform d "go down" moves. The final position will be 2^j - d. Setting this to k gives d = 2^j - k.
• Since d must be non-negative, we must have 2^j >= k.
• The "no consecutive down-moves" rule implies that a down-move must be followed by an up-move (unless it's the final move). This means we can place at most one down-move between any two consecutive up-moves, one before the first up-move, and one after the last up-move.
• This gives j+1 available "slots" to place the d down-moves.
• The problem reduces to choosing d slots out of j+1. The number of ways is the binomial coefficient C(j+1, d). This also implies d <= j+1.
• Iterate through the number of up-moves j from 0 upwards (up to ~31, as 2^31 is > 10^9).
• For each j, calculate d = 2^j - k. If 0 <= d <= j+1, calculate C(j+1, d) and add it to the total count.
• The loop can be optimized to stop when 2^j - j - 1 > k.

This approach leverages a key insight: the jump value for the i-th "go up" move is always i-1, regardless of any "go down" moves made in between.

If we make exactly j "go up" moves, the total positive displacement is sum_{i=0}^{j-1} 2^i = 2^j - 1. Starting from stair 1, these moves alone would land us on stair 2^j. To end up on stair k, we need to make d = 2^j - k "go down" moves.

The number of down moves d must be non-negative, so 2^j >= k.

The rule against consecutive "go down" moves means each down-move must be separated by at least one up-move. This gives us j+1 potential slots to place the d down-moves: one before each of the j up-moves, and one after the last one. _ U_0 _ U_1 _ ... _ U_{j-1} _

We need to choose d of these j+1 slots to place our down-moves. This is a combination problem, and the number of ways is C(j+1, d). This also implies that d cannot be greater than j+1.

So, the algorithm is to sum C(j+1, 2^j - k) for all j that satisfy 0 <= 2^j - k <= j+1. We can iterate j from 0 up to a small limit (around 31) and add the results.

1class Solution {2    public int waysToReachStair(int k) {3        int ways = 0;4        // j is the number of 'up' moves. 2^30 > 10^9, so j won't exceed ~31.5        for (int j = 0; j < 32; j++) {6            long powerOf2 = 1L << j;7            long downs = powerOf2 - k;8 9            if (downs < 0) {10                continue;11            }12            13            // The number of available slots for down moves is j + 1.14            // One slot before all jumps, and one slot after each of the j jumps.15            if (downs > j + 1) {16                // For j > 1, 2^j - k > j + 1 is equivalent to 2^j - j - 1 > k.17                // Since 2^j - j - 1 is an increasing function for j>=1,18                // we can break the loop if this condition is met.19                if (j > 1 && (powerOf2 - j - 1 > k)) {20                    break;21                }22                continue;23            }24 25            ways += combinations(j + 1, (int) downs);26        }27        return ways;28    }29 30    // Helper function to calculate combinations C(n, k)31    private int combinations(int n, int k) {32        if (k < 0 || k > n) {33            return 0;34        }35        if (k == 0 || k == n) {36            return 1;37        }38        if (k > n / 2) {39            k = n - k;40        }41        long res = 1;42        for (int i = 1; i <= k; i++) {43            res = res * (n - i + 1) / i;44        }45        return (int) res;46    }47}