<p>You have a pointer at index <code>0</code> in an array of size <code>arrLen</code>. 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).</p>

<p>Given two integers <code>steps</code> and <code>arrLen</code>, return the number of ways such that your pointer is still at index <code>0</code> after <strong>exactly</strong> <code>steps</code> steps. Since the answer may be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> steps = 3, arrLen = 2
<strong>Output:</strong> 4
<strong>Explanation: </strong>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
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> steps = 2, arrLen = 4
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are 2 differents ways to stay at index 0 after 2 steps
Right, Left
Stay, Stay
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> steps = 4, arrLen = 2
<strong>Output:</strong> 8
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= steps &lt;= 500</code></li>
	<li><code>1 &lt;= arrLen &lt;= 10<sup>6</sup></code></li>
</ul>

class Solution:
    def numWays(self, steps: int, arrLen: int) -> int:
        # Define a modulus value for taking the modulo operation to avoid overflow.
        mod = 1000000007
        
        # Calculate the maximum position the pointer can reach, which is the minimum of steps/2 and arrLen - 1.
        maxPosition = min(steps // 2, arrLen - 1)
        
        # Create a 2D array dp to store the number of ways to reach a specific position at each step.
        dp = [[0] * (maxPosition + 1) for _ in range(steps + 1)]
        
        # Initialize the number of ways to stay at position 0 after 0 steps to 1.
        dp[0][0] = 1
        
        # Loop through the number of steps.
        for i in range(1, steps + 1):
            for j in range(maxPosition + 1):
                # Initialize the number of ways to stay at the current position with the number of ways to stay at the same position in the previous step.
                dp[i][j] = dp[i - 1][j]
                
                # If the current position is greater than 0, add the number of ways to reach it by moving left.
                if j > 0:
                    dp[i][j] = (dp[i][j] + dp[i - 1][j - 1]) % mod
                
                # If the current position is less than the maximum position, add the number of ways to reach it by moving right.
                if j < maxPosition:
                    dp[i][j] = (dp[i][j] + dp[i - 1][j + 1]) % mod
        
        # The final result is stored in dp[steps][0], representing the number of ways to reach the initial position after taking 'steps' steps.
        return dp[steps][0]
18389