<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:
        # Ideas:
        # recursion
        # f(s, k) = # ways to start at 0, and end up at k in s steps.
        # k = 0,....,arrLen - 1
        # f(s,k) = 0 if s<k.
        # Want f(steps, 0)
        # f(0,0) = 1
        # if k = 0
        # f(s,0) = f(s-1,0) + f(s-1,1)
        # if k = arrLen - 1
        # f(s,arrLen-1) = f(s-1, arrLen-1) + f(s-1,arrLen-2)
        # If 0<k<arrLen-1:
        # f(s, k) = f(s-1, k) + f(s-1, k+1), f(s-1, k-1)
        
        if arrLen == 1:
            return 1
        
        cache = {}
        def helper(s,k):
            if (s,k) in cache:
                return cache[(s,k)]
            elif s==0 and k==0:
                return 1
            elif s<k:
                return 0
            elif k==0:
                ans = helper(s-1,0) + helper(s-1,1)
            elif k==arrLen-1:
                ans = helper(s-1,arrLen-1) + helper(s-1,arrLen-2)
            else:
                ans = helper(s-1,k) + helper(s-1,k+1) + helper(s-1,k-1)
            cache[(s,k)] = ans
            return ans
        
        return helper(steps,0)%(10**9+7)
                
        
62155