<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 recur(self, i, step_i):
        """
        step_i index of step
        i      index in array
        """
        
        # avail steps
        avail_steps = self.S - step_i
        if avail_steps == 0:
            return i == 0
        
        if self.memo[i][step_i] != -1:
            return self.memo[i][step_i]
        
        ways = 0
        
        # if have to turn back now
        if avail_steps == i:
            ways = 1
        # elif avail_steps < i: # no chance of making it back
        #     ways = 0
        else: # have avail steps
            # do nothing
            ways += self.recur(i, step_i + 1)
            # right
            if i < self.N - 1: # can go right
                ways += self.recur(i + 1, step_i + 1)
            if i > 0: # can go left
                ways += self.recur(i - 1, step_i + 1)
                
        self.memo[i][step_i] = ways
        return ways
        
        
        
    def numWays(self, steps: int, arrLen: int) -> int:
        
        self.S = steps
        self.N = arrLen
        """
        steps number of slots
        place left and right or just spacing
        every right must be accompanied by a left
        balanced parentheses
        """
        mod = 1000000007
        self.memo = defaultdict(lambda: defaultdict(lambda: -1))
        return self.recur(0, 0) % mod
        
28999