<p>You are given two <strong>sorted</strong> arrays of distinct integers <code>nums1</code> and <code>nums2</code>.</p>

<p>A <strong>valid<strong><em> </em></strong>path</strong> is defined as follows:</p>

<ul>
	<li>Choose array <code>nums1</code> or <code>nums2</code> to traverse (from index-0).</li>
	<li>Traverse the current array from left to right.</li>
	<li>If you are reading any value that is present in <code>nums1</code> and <code>nums2</code> you are allowed to change your path to the other array. (Only one repeated value is considered in the valid path).</li>
</ul>

<p>The <strong>score</strong> is defined as the sum of unique values in a valid path.</p>

<p>Return <em>the maximum score you can obtain of all possible <strong>valid paths</strong></em>. Since the answer may be too large, return it modulo <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/07/16/sample_1_1893.png" style="width: 500px; height: 151px;" />
<pre>
<strong>Input:</strong> nums1 = [2,4,5,8,10], nums2 = [4,6,8,9]
<strong>Output:</strong> 30
<strong>Explanation:</strong> Valid paths:
[2,4,5,8,10], [2,4,5,8,9], [2,4,6,8,9], [2,4,6,8,10],  (starting from nums1)
[4,6,8,9], [4,5,8,10], [4,5,8,9], [4,6,8,10]    (starting from nums2)
The maximum is obtained with the path in green <strong>[2,4,6,8,10]</strong>.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [1,3,5,7,9], nums2 = [3,5,100]
<strong>Output:</strong> 109
<strong>Explanation:</strong> Maximum sum is obtained with the path <strong>[1,3,5,100]</strong>.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [1,2,3,4,5], nums2 = [6,7,8,9,10]
<strong>Output:</strong> 40
<strong>Explanation:</strong> There are no common elements between nums1 and nums2.
Maximum sum is obtained with the path [6,7,8,9,10].
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums1.length, nums2.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums1[i], nums2[i] &lt;= 10<sup>7</sup></code></li>
	<li><code>nums1</code> and <code>nums2</code> are strictly increasing.</li>
</ul>

class Solution {
public:
    int M=1000000007;
    long long dp[100000][2];
    long long f(int ind,int an,int& n1,int& n2,vector<int>& v1, vector<int>& v2){
        if(an==0 && ind==n1)
            return 0;
        if(an==1 && ind==n2)
            return 0;
        if(dp[ind][an]!=-1)
            return dp[ind][an];

        long long ni,s1,s2;
        s1=s2=INT_MIN;

        if(an==0){
            //stay in same array
            s1=(1LL*v1[ind]+f(ind+1,0,n1,n2,v1,v2));

            //shift to other array
            ni=lower_bound(v2.begin(),v2.end(),v1[ind])-v2.begin();
            if(ni<n2){
                if(v2[ni]==v1[ind])
                    s2=(1LL*v1[ind]+f(ni+1,1,n1,n2,v1,v2));
            }
            else
                s2=v1[ind];
        }
        else{
            //stay in same array
            s1=(1LL*v2[ind]+f(ind+1,1,n1,n2,v1,v2));

            //shift to other array
            ni=lower_bound(v1.begin(),v1.end(),v2[ind])-v1.begin();
            
            if(ni<n1){
                if(v1[ni]==v2[ind])
                    s2=(1LL*v2[ind]+f(ni+1,0,n1,n2,v1,v2));
            }
            else
                s2=v2[ind];
        }

        // if(ind==0)
        //     cout<<an<<" "<<s1<<" "<<s2<<"\n";
        long long mxs=max(s1,s2);

        return dp[ind][an]=mxs;
    }
    int maxSum(vector<int>& v1, vector<int>& v2) {
        memset(dp,-1,sizeof(dp));
        int n1=v1.size(),n2=v2.size();
        long long s1,s2;
        
        s1=f(0,0,n1,n2,v1,v2)%M;
        s2=f(0,1,n1,n2,v1,v2)%M;

        return max(s1,s2);
    }
};
65583