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)).

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.

Description Hints 2 Solution Video 20:57

1269. Number of Ways to Stay in the Same Place After Some Steps

Q: 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.

Hard50.2% acceptance1 company

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

Asked at 1 company

Google

Topics

Dynamic Programming

Hints open one at a time

Hint 1

Try with Dynamic programming, dp(pos,steps): number of ways to back pos = 0 using exactly "steps" moves.

Hint 2

Notice that the computational complexity does not depend of "arrlen".

20:57

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