<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 M  = pow(10, 9) + 7;
    int ways(int steps, int pos, int len, vector<vector<int>> &dp){
        if(steps == 0 && pos == 0)
            return 1;
        if(steps == 0)
            return 0;

        if(dp[pos][steps] != -1)
            return dp[pos][steps];

        int res = 0;

        //stay at the same place
        res = ((res%M)  + ways(steps -1, pos, len, dp) % M) %M;

        //move left if possible
        if(pos > 0)
            res = ((res % M) +  ways(steps - 1, pos -1, len, dp) % M) % M;
        
        //move right if possible
        if(pos + 1 < len)
            res = ((res % M) + ways(steps - 1, pos + 1, len, dp) % M) % M;

        return dp[pos][steps] =  res;
    }
    int numWays(int s, int a) {
        int pos = 0;
        int maxLen = pow(10, 6), maxSteps = 500;
        vector<vector<int>> dp(maxSteps + 1, vector<int> (maxSteps +1, -1));
       return  ways(s, pos, a, dp);
    }
};
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