<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 dp(self, remain, idx, memo, arrLen):
        if idx < 0 or idx >= arrLen: return 0
        if idx > remain: return 0
        if remain == 0:
            if idx == 0:
                return 1
            return 0
        if (remain, idx) in memo: return memo[(remain, idx)]
        res = 0
        res += self.dp(remain-1, idx-1, memo, arrLen)
        res += self.dp(remain-1, idx, memo, arrLen)
        res += self.dp(remain-1, idx+1, memo, arrLen)
        memo[(remain, idx)] = res
        return res
        
    def numWays(self, steps: int, arrLen: int) -> int:
        memo = {}
        mod = 10 ** 9 + 7
        return self.dp(steps,0,memo, arrLen) % mod
    

    # def numWays(self, steps: int, arrLen: int) -> int:
    #     # complexity: time O(steps^2), mem O(steps)
    #     mod = 10 ** 9 + 7
    #     dp = [0] * min(steps // 2 + 1, arrLen)
    #     dp[0] = 1
    #     for i in range(steps):
    #         newdp = [0] * min(steps // 2 + 1, arrLen)
    #         for j, ways in enumerate(dp):
    #             if ways > 0:
    #                 for dx in (-1,0,1):
    #                     idx = j + dx
    #                     if idx < 0 or idx >= len(dp): continue
    #                     newdp[idx] += ways
    #                     newdp[idx] %= mod
    #         dp = newdp
    #     return dp[0]
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