<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:
        mod = 10 ** 9 + 7
        
        # dp[i]: 
        # nways of staying at index 0 at the end if current index is i and have used j steps
        # where j depends on which iteration we are at (space compression)
        dp = [0] * arrLen

        # basecase: when used up all steps -> dp[0] = 0
        dp[0] = 1
        arrLen = min(arrLen, steps)
        # tabulate
        for used_steps in range(steps - 1, -1, -1):
            prev_dp = dp[:]
            for idx in range(arrLen):
                # # early stop: if remain steps < distance back to index 0 -> no way to back to 0
                # if steps - used_steps < idx:
                #     break
                # 3 choices: 
                # since dp[idx] means stay we we just do the other 2 choices and add to dp
                if idx < arrLen - 1:
                    dp[idx] += prev_dp[idx + 1]
                if idx > 0:
                    dp[idx] += prev_dp[idx - 1]
                
                # mod
                dp[idx] %= mod
        
        return dp[0]
                     




        
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