<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 {
    const int MOD = 1e9 + 7;
public:
    int numWays(int steps, int n) {
        if (n == 1) return 1;
        vector<vector<int>> dp(2, vector<int>(n, 0));
        dp[0][0] = dp[0][1] = 1;

        for (int s=1; s<steps; ++s) {
            for (int i=0; i<min(n, steps); ++i) {
                dp[1][i] = 0;
                if (i > 0) dp[1][i] = (dp[1][i] + dp[0][i-1]) % MOD;;
                dp[1][i] = (dp[1][i] + dp[0][i]) % MOD;
                if (i < n-1) dp[1][i] = (dp[1][i] +dp[0][i+1]) % MOD;
            }

            swap(dp[0], dp[1]);
        }

        return dp[0][0];
    }
};
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