<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 sol(vector<int>& nums1, vector<int>& nums2, vector<int>& p1, vector<int>& p2,vector<int>&v, int n1,int n2,int n, int x, int y, int z, vector<vector<vector<int>>>&dp){
    //     int a, b ;
    //     if(x==n-1){
    //         if(z==0){
    //             if(y==0){
    //                 a = p1[n1-1] -p1[v[x].first] + 2*num1[v[x].first] ;
    //                 b=  p2[n2-1] -p2[v[x].second] + num2[v[x].second] ;
    //             }
    //             else{
    //                 a = p1[n1-1] -p1[v[x].first] + num1[v[x].first] ;
    //                 b=  p2[n2-1] -p2[v[x].second] + 2*num2[v[x].second] ;
    //             }

    //         }
    //         else{
    //             if(y==0){
    //                 a = p1[n1-1] -p1[v[x].first]  ;
    //                 b=  p2[n2-1] -p2[v[x].second] + num2[v[x].second] ;
    //             }
    //             else{
    //                 a = p1[n1-1] -p1[v[x].first] + num1[v[x].first] ;
    //                 b=  p2[n2-1] -p2[v[x].second] 
    //             }
    //         }
    //         return max(a,b) ;
    //     }

    //     if(dp[x][y][z]!=-1)return dp[x][y][z] ;

    //     int ans=0 ;
    //     if(z==0){
    //         if(y==0){
                
    //             ans = max(ans,sol(nums1, nums2,p1,p2,v,n1,n2,n,x+1,y,z)+p1[v1[x+1].first]-p1[v1[x+1].first]) ;
    //             ans = max(ans,sol(nums1, nums2,p1,p2,v,n1,n2,n,x+1,y,z)+p1[v1[x+1].first]-p1[v1[x+1].first]) ;
    //         }
    //     }

    // }
    int maxSum(vector<int>& nums1, vector<int>& nums2) {
        int n1 = nums1.size() ;
        int n2 = nums2.size() ;
        vector<long long>p1(n1), p2(n2) ;
        long long ans = (long long)0 ;
        p1[0] = (long long)nums1[0] ;
        p2[0] = (long long)nums2[0] ;
        unordered_map<int,int>m1,m2 ;
        m1[nums1[0]] = 1;
        m2[nums2[0]] = 1 ;
        for(int i=1 ;i<n1 ;i++){
            p1[i] = (long long)nums1[i] + p1[i-1] ;
            m1[nums1[i]] = i+1 ;
        }
        for(int i=1 ;i<n2 ;i++){
            p2[i] = (long long)nums2[i] + p2[i-1] ;
            m2[nums2[i]] = i+1 ;
        }

        vector<pair<int,int>> v ;
        for(int i=0 ;i<n1 ;i++){
            if(m2[nums1[i]]>0){
                v.push_back({i,m2[nums1[i]]-1}) ;
            }
        }

        int n = v.size() ;
       
        int x, y, z=0 ;
        v.push_back({n1-1,n2-1}) ;
        ans+= max((long long)p1[v[0].first],(long long)p2[v[0].second]) ;
        for(int i=0 ;i<n ;i++){
            cout<<v[i].first<<" "<<v[i].second<<endl ;
            ans += max((long long)(p1[v[i+1].first]-p1[v[i].first]),(long long)(p2[v[i+1].second]-p2[v[i].second])) ;
            ans = ans%(1000000007) ;
        }
        cout<<ans<<endl ;

        return ans ;
    }
};
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