<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 mod = 1000000007;

int fn(int idx, int cnt,  int len, int steps, int **dp) {

    if(idx<0 or idx>=len) {
        return 0;
    }
    if(cnt>=steps) {
        if(idx==0) {
            return 1;
        }
         return 0;
    }

    if(dp[idx][cnt]!=-1) return dp[idx][cnt];

    int val1 = fn(idx, cnt+1, len, steps, dp)%mod; // for 0


    int val2 = fn(idx+1, cnt+1, len, steps, dp)% mod; // for right

    int val3 = fn(idx-1, cnt+1, len, steps, dp)%mod; // for left


    int ans = val1%mod;
    ans = ((ans%mod) + (val2%mod))%mod;
    ans = ((ans%mod) + (val3%mod))%mod;

    cout<<idx<< " "<<cnt<<" "<<ans<<endl;

    return dp[idx][cnt]=ans;
}

    int numWays(int steps, int arrLen) {

            vector<int>vv;
            int *dp[510];
            for(int i=0;i<510;++i) dp[i] = new int[510];

            for(int i=0;i<510;++i)
            for(int j=0;j<510;++j)
            dp[i][j] = -1;

        return fn(0, 0, arrLen, steps, dp);
        
    }
};
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