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Number of Ways to Stay in the Same Place After Some Steps - Solution & Explanation

HardDynamic Programming22 min readAsked at: Google
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Problem Statement

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 <= 500
  • 1 <= arrLen <= 106

Approach Overview

Problem 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 1: Dynamic Programming Approach

This approach leverages dynamic programming to find the number of ways to stay at index 0 after a given number of steps. We define a 2D table dp[i][j] where i represents the number of steps remaining, and j represents the current position of the pointer.

To optimize computation, we can limit the table size to the minimum of steps and arrLen since going beyond these positions is unnecessary.

This code implements a dynamic programming solution in C. It initializes a DP table where dp[i][j] represents the number of ways to be at position j after i steps. The transition involves considering staying at the same position, moving left, or moving right.

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Complexity

Time Complexity: O(steps * min(steps, arrLen))

Space Complexity: O(steps * min(steps, arrLen))

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Approach 2: Recursive with Memoization Approach

This approach utilizes recursion combined with memoization to optimize the recursive calls. Here, recursion is used to explore all possible paths dynamically adjusting by staying at, moving left, or moving right from each position in every step.

The results of the recursive calls are stored in a memoization table to avoid redundant calculations.

This C implementation uses recursion with memoization. The recursive function explores all directions (stay, left, right) and stores results in a cache to prevent redundant computation, ensuring efficiency.

Code

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Complexity

Time Complexity: O(steps * min(steps, arrLen))

Space Complexity: O(steps * min(steps, arrLen))

Try this approach in the editor →

Approach 3: Memoization Search

We observe the data range of the problem and find that steps does not exceed 500, which means that we can only go to the right for up to 500 steps.

We can design a function dfs(i, j), which represents the number of schemes when we are currently at position i and the remaining steps are j. So the answer is dfs(0, steps).

The execution process of the function dfs(i, j) is as follows:

  1. If i \gt j or i geq arrLen or i \lt 0 or j \lt 0, then return 0.
  2. If i = 0 and j = 0, then the pointer has stopped in place and there are no remaining steps, so return 1.
  3. Otherwise, we can choose to move one step to the left, one step to the right, or stay still, so return dfs(i - 1, j - 1) + dfs(i + 1, j - 1) + dfs(i, j - 1). Note the modulo operation of the answer.

During the process, we can use memoization search to avoid repeated calculations.

The time complexity is O(steps times steps), and the space complexity is O(steps times steps). Where steps is the number of steps given in the problem.

Code

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Complexity Comparison

ApproachComplexity
Dynamic Programming Approach

Time Complexity: O(steps * min(steps, arrLen))

Space Complexity: O(steps * min(steps, arrLen))

Recursive with Memoization Approach

Time Complexity: O(steps * min(steps, arrLen))

Space Complexity: O(steps * min(steps, arrLen))

Memoization Search—

Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
Recursive with MemoizationO(steps * min(arrLen, steps))O(steps * min(arrLen, steps))When reasoning about the problem recursively or converting recursion to DP
Bottom-Up Dynamic ProgrammingO(steps * min(arrLen, steps))O(min(arrLen, steps)) with rolling arrayPreferred for interviews and production due to predictable iteration and memory optimization

Video Solution

Number of Ways to Stay in the Same Place After Some Steps - Leetcode 1269 - Python • NeetCodeIO • 6,969 views views

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Frequently Asked Questions

Is Number of Ways to Stay in the Same Place After Some Steps easy or hard?
The problem is classified as Hard because it requires careful dynamic programming state design and an optimization that limits the reachable positions. Without the position bound, the DP space becomes unnecessarily large.
Number of Ways to Stay in the Same Place After Some Steps Python/Java solution
Both Python and Java implementations typically use dynamic programming with either a 2D DP table or a rolling 1D array. The transition considers three moves per step and applies modulo 1e9+7. Time complexity remains O(steps * min(arrLen, steps)).
How to solve Number of Ways to Stay in the Same Place After Some Steps in O(n)?
A strict O(n) solution is not typical because the state depends on both steps and positions. The efficient solution reduces the state space using dp[step][pos] where pos is limited to min(arrLen, steps). This produces O(steps * min(arrLen, steps)) time, which is optimal for the constraints.
What is the best approach for Number of Ways to Stay in the Same Place After Some Steps?
The optimal approach uses dynamic programming with state dp[step][position]. Each state aggregates three transitions: stay, move left, and move right. By limiting positions to min(arrLen - 1, steps), the DP state space becomes manageable. This results in O(steps * min(arrLen, steps)) time and can be optimized to O(min(arrLen, steps)) space.
Is Number of Ways to Stay in the Same Place After Some Steps asked at Google/Amazon/Meta?
Dynamic programming problems with state transitions and movement constraints frequently appear in interviews at companies like Google, Amazon, and Meta. Variants of this problem test DP state modeling, boundary pruning, and modulo arithmetic.
What data structure is used in Number of Ways to Stay in the Same Place After Some Steps?
The core data structure is a dynamic programming array that stores the number of ways to reach each index after a certain number of steps. Some implementations also use memoization tables or hash maps for caching recursive states.
What is the time complexity of Number of Ways to Stay in the Same Place After Some Steps?
The optimal dynamic programming solution runs in O(steps * min(arrLen, steps)) time. The algorithm only processes positions that are reachable within the given number of steps. Space complexity is O(min(arrLen, steps)) with a rolling DP array.

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