<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 {
public:
    int numWays(int steps, int arrLen) {
        const int mod = 1e9 + 7;
        arrLen = min(arrLen, steps);

        // dp[i] - no. of ways to reach at i after steps

        // dp[i][j] -> dp[i + 1][j - 1]
        // dp[i][j] -> dp[i][j]
        // dp[i][j] -> dp[i + 1][j + 1]

        vector<int> dp(arrLen);
        dp[0] = 1;

        for (int i = 0; i < steps; i++) {
            vector<int> ndp(arrLen, 0);

            for (int j = 0; j < arrLen; j++) {
                if (j - 1 >= 0)
                ndp[j - 1] = (ndp[j - 1] + dp[j]) % mod;

                ndp[j] = (ndp[j] + dp[j]) % mod;

                if (j + 1 < arrLen)
                ndp[j + 1] = (ndp[j + 1] + dp[j]) % mod;
            }
            
            dp = ndp;
        }

        return dp[0];
    }
};
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