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

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 still at index `0`

after **exactly **`steps`

steps.

Since the answer may be too large, return it **modulo** `10^9 + 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 <= 10^6`

class Solution { public int numWays(int steps, int arrLen) { int end = Math.min(steps, arrLen); int mod = 1000000000 + 7; long[][] dp = new long[steps+1][end+5]; dp[0][0] = 1; for (int i = 0; i < steps; i++) { dp[i+1][0] = (dp[i][0] + dp[i][1]) % mod; for (int j = 1; j < end; j++) { dp[i+1][j] = (dp[i][j-1] + dp[i][j] + dp[i][j+1]) % mod; } } return (int) (dp[steps][0]); } }

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