Watch 10 video solutions for Number of Ways to Stay in the Same Place After Some Steps, a hard level problem involving Dynamic Programming. This walkthrough by NeetCodeIO has 6,864 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
You have a pointer at index 0 in an array of size arrLen. At each step, you can move 1 position to the left, 1 position to the right in the array, or stay in the same place (The pointer should not be placed outside the array at any time).
Given two integers steps and arrLen, return the number of ways such that your pointer is still at index 0 after exactly steps steps. Since the answer may be too large, return it modulo 109 + 7.
Example 1:
Input: steps = 3, arrLen = 2 Output: 4 Explanation: There are 4 differents ways to stay at index 0 after 3 steps. Right, Left, Stay Stay, Right, Left Right, Stay, Left Stay, Stay, Stay
Example 2:
Input: steps = 2, arrLen = 4 Output: 2 Explanation: There are 2 differents ways to stay at index 0 after 2 steps Right, Left Stay, Stay
Example 3:
Input: steps = 4, arrLen = 2 Output: 8
Constraints:
1 <= steps <= 5001 <= arrLen <= 106Problem Overview: You start at index 0 of a 1D array. For each step you can move left, move right, or stay in place. After exactly steps moves, count how many sequences keep you at index 0. The array length limits movement, and results must be returned modulo 1e9 + 7.
Approach 1: Recursive with Memoization (Top-Down DP) (Time: O(steps * min(arrLen, steps)), Space: O(steps * min(arrLen, steps)))
Model the process as a recursive state dfs(step, pos) representing the number of ways to reach the final state from a given step and position. From each state you have three transitions: stay (pos), move left (pos - 1), or move right (pos + 1). Memoize results in a cache to avoid recomputing overlapping subproblems. A key optimization: you never need positions beyond min(arrLen - 1, steps) because you cannot travel farther than the remaining steps. This drastically reduces the state space and keeps the recursion efficient.
Approach 2: Dynamic Programming (Bottom-Up) (Time: O(steps * min(arrLen, steps)), Space: O(steps * min(arrLen, steps)) or O(min(arrLen, steps)))
Define dp[s][i] as the number of ways to reach position i after s steps. The transition mirrors the allowed moves: dp[s][i] = dp[s-1][i] + dp[s-1][i-1] + dp[s-1][i+1]. Bound the position range to maxPos = min(arrLen - 1, steps) since larger indices are unreachable. Initialize dp[0][0] = 1 and iterate step by step while applying modulo arithmetic. A rolling array optimization reduces memory to one dimension because each row depends only on the previous step. This bottom-up approach avoids recursion overhead and is typically faster in production implementations of dynamic programming.
Recommended for interviews: The bottom-up dynamic programming approach is what interviewers typically expect. It clearly demonstrates understanding of state design, transition rules, and boundary constraints. Mentioning the position bound optimization (min(arrLen, steps)) shows deeper insight into reducing the DP state space. The recursive solution with memoization still helps demonstrate how the problem naturally decomposes into overlapping subproblems before converting it into iterative DP.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Recursive with Memoization | O(steps * min(arrLen, steps)) | O(steps * min(arrLen, steps)) | When reasoning about the problem recursively or converting recursion to DP |
| Bottom-Up Dynamic Programming | O(steps * min(arrLen, steps)) | O(min(arrLen, steps)) with rolling array | Preferred for interviews and production due to predictable iteration and memory optimization |