Find Number of Ways to Reach the K-th Stair - Problem

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: It is possible that Alice reaches stair k, and performs some operations to reach stair k again.

Input & Output

Example 1 — Basic Case

$ Input: k = 1

› Output: 2

💡 Note: Alice starts at stair 1 (already at target). She can also go up to 2 then down to 1. Two ways total.

Example 2 — Multiple Paths

$ Input: k = 2

› Output: 2

💡 Note: Path 1: Start at 1, go up by 2^0=1 to reach 2. Path 2: Start at 1, go up by 2^0=1 to 2, up by 2^1=2 to 4, down to 3, down to 2.

Example 3 — Edge Case

$ Input: k = 0

› Output: 1

💡 Note: Alice starts at 1, can only go down to 0 (one way since she cannot go down consecutively).

Constraints

0 ≤ k ≤ 10⁹

Visualization

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Asked in

G Google 15 f Meta 12 Apple 8

The key insight is that Alice can place at most one down move between consecutive up moves, creating a combinatorial pattern. The mathematical approach using C(n,k) combinations provides the optimal O(log k) solution by counting valid sequences directly. Time: O(log k), Space: O(1)

Common Approaches

✓ Optimized

⏱️ Time: N/A Space: N/A

Dp 2d

⏱️ Time: N/A Space: N/A

Mathematical Combinatorics

⏱️ Time: O(log k) Space: O(1)

Observe that Alice can only go down at most once between consecutive up moves. This creates a pattern where we can use combinatorial mathematics to count the valid sequences directly without simulation.

Memoized Recursion

⏱️ Time: O(k × log k) Space: O(k × log k)

Use the same recursive approach but store results for each (position, jump, lastDown) state to avoid recalculating the same subproblems multiple times.

Recursive Exploration

⏱️ Time: O(2^n) Space: O(n)

Try all possible moves from each position - going down (if allowed) or jumping up by 2^jump power. Count all ways that eventually reach target k.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

const int MOD = 1000000007;

int min(int a, int b) {
    return a < b ? a : b;
}

int solution(int* nums, int n) {
    if (n == 0) return 0;
    
    // dp[arr1_val] = count of ways with arr1[pos] = arr1_val
    int* prevDp = calloc(nums[0] + 1, sizeof(int));
    
    // Initialize first position
    for (int arr1Val = 0; arr1Val <= nums[0]; arr1Val++) {
        prevDp[arr1Val] = 1;
    }
    
    // Process each subsequent position
    for (int pos = 1; pos < n; pos++) {
        int* currDp = calloc(nums[pos] + 1, sizeof(int));
        int prevSize = nums[pos-1] + 1;
        
        for (int currArr1 = 0; currArr1 <= nums[pos]; currArr1++) {
            int currArr2 = nums[pos] - currArr1;
            
            // Check all valid previous arr1 values
            for (int prevArr1 = 0; prevArr1 <= min(currArr1, prevSize - 1); prevArr1++) {
                int prevArr2 = nums[pos-1] - prevArr1;
                
                // Check if arr2 maintains non-increasing order
                if (prevArr2 >= currArr2) {
                    currDp[currArr1] = ((long long)currDp[currArr1] + prevDp[prevArr1]) % MOD;
                }
            }
        }
        
        free(prevDp);
        prevDp = currDp;
    }
    
    long long result = 0;
    for (int i = 0; i <= nums[n-1]; i++) {
        result = (result + prevDp[i]) % MOD;
    }
    
    free(prevDp);
    return result;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int n = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] >= '0' && token[0] <= '9') {
            nums[n++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    printf("%d\n", solution(nums, n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

⚡ Linearithmic

Space Complexity

✓ Linear Space

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Medium Frequency

~35 min Avg. Time

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Ln 1, Col 1

Smart Actions

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