1269. number of options left in place

There is a length of arrLen Array of , Start with a pointer in the index 0 It's about .

In every step of the operation , You can move the pointer to the left or right 1 Step , Or stay where you are （ The pointer cannot be moved out of the range of the array ）.

Here are two integers steps and arrLen , Please calculate and return ： Just in time to execute steps After the first operation , The pointer still points to the index 0 The number of schemes at .

Because the answer could be big , Please return the number of schemes model 10^9 + 7 After the results of the .

Example 1：

 Input ：steps = 3, arrLen = 2
 Output ：4
 explain ：3  Step after , All in all  4  There are two different ways to stop at the index  0  It's about .
 towards the right , towards the left , Immobility 
 Immobility , towards the right , towards the left 
 towards the right , Immobility , towards the left 
 Immobility , Immobility , Immobility 

Example 2：

 Input ：steps = 2, arrLen = 4
 Output ：2
 explain ：2  Step after , All in all  2  There are two different ways to stop at the index  0  It's about .
 towards the right , towards the left 
 Immobility , Immobility 

Example 3：

 Input ：steps = 4, arrLen = 2
 Output ：8

Tips ：

1 <= steps <= 500
1 <= arrLen <= 10^6

class Solution {
    
public:
    int numWays(int steps, int arrLen) {
    
        int mod=1000000007;
        int len=min(arrLen,steps+1);
        //vector<vector<int>>dp(steps+1,vector<int>(len));
// int dp[steps+1][arrLen+2];
        vector<int>dp(len);
        vector<int>dp2(len);
        dp[0]=1;
        for(int i=1;i<=steps;i++){
    
            for(int j=0;j<len;j++){
    
                dp2[j]=dp[j];
                if(j>0)dp2[j]=(dp2[j]+dp[j-1])%mod;
                if(j<len-1)dp2[j]=(dp2[j]+dp[j+1])%mod;
            }
            dp=dp2;
            //cout<<endl;
        }
        return dp[0];

    }
};