<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 maxSum(vector<int>& nums1, vector<int>& nums2) {
        long long dp1[100001]={0},dp2[100001]={0};
        // int mod=1;
        dp1[0]=nums1[0];
        dp2[0]=nums2[0];
        unordered_map<int,int>m2,m1;
        int pos=0;
        for(auto &num:nums2){
            m2[num]=pos;
            pos++;
        }
        pos=0;
        for(auto &num:nums1){
            m1[num]=pos;
            pos++;
        }
        int i=1,j=1;
        int n1=nums1.size(),n2=nums2.size();
        while(i<n1 && j<n2){
            dp2[j]=(nums2[j]+dp2[j-1]);
            //dp1
            int id2=-1;
            
            dp1[i]=(dp1[i-1]+nums1[i]);
            if(m2.count(nums1[i])){
                id2=m2[nums1[i]];
            }
            if(id2!=-1 && id2<=j ){
                if(dp2[id2]<dp1[i]){
                    j=id2;
                    dp2[j]=dp1[i];
                }
                else{
                    dp1[i]=dp2[id2];
                }
            }
            int id1=-1;
            if(m1.count(nums2[j])){
                id1=m1[nums2[j]];
            }
            if(id1!=-1 && id1<=i ){
                if(dp1[id1]<dp2[j]){
                    i=id1;
                    dp1[i]=dp2[j];
                }
                else{
                    dp2[j]=dp1[id1];
                }
            }
            // cout<<i<<' '<<dp1[1]<<"  "<<j<<' '<<dp2[j]<<endl;
            if(i<n1-1)
            i++;
            if(j<n2-1)
            j++;
            if(i==n1-1 && j==n2-1)break;

        }
        // c
        // cout<<n1-1<<' '<<n2-1<<endl;;
        // cout<<i<<' '<<j<<endl;;
        while(i<n1){
            dp1[i]=(nums1[i]+dp1[i-1]);
            i++;
        }
        while(j<n2){
            dp2[j]=(nums2[j]+dp2[j-1]);
            j++;
        }
        // for(int i=0;i<n1;i++){
        //     cout<<dp1[i]<<' ';
        // }
        // cout<<endl;
        // for(int j=0;j<n2;j++){
        //     cout<<dp2[j]<<' ';
        // }
        int m=(1e9+7);
        return max(dp1[n1-1],dp2[n2-1])%m;

    }
};
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