<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 n1,n2;
long long int dp[100002][2];
long long int mod=1e9+7;
map<int,int>mp0,mp1;
long long int f(int ind,int which,vector<int>& nums1, vector<int>& nums2){
    if(((ind==n1) && (which==0)) || ((ind==n2) && (which==1))) return 0;
    if(dp[ind][which]!=-1) return dp[ind][which];
    if(which==0){
        long long int same=(nums1[ind]+f(ind+1,which,nums1,nums2));
        long long int jump=0;
        if(mp1.find(nums1[ind])!=mp1.end()){
            jump=(nums1[ind]+f(mp1[nums1[ind]]+1,1,nums1,nums2));
        }
        return dp[ind][which]=max(same,jump);
    }else{
        long long int same=(nums2[ind]+f(ind+1,which,nums1,nums2));
        long long int jump=0;
        if(mp0.find(nums2[ind])!=mp0.end()){
            jump=(nums2[ind]+f(mp0[nums2[ind]]+1,0,nums1,nums2));
        }
        return dp[ind][which]=max(same,jump);
    }
}
    int maxSum(vector<int>& nums1, vector<int>& nums2) {
        n1=nums1.size();
        n2=nums2.size();
        for(int i=0;i<n1;i++) mp0[nums1[i]]=i;
        for(int i=0;i<n2;i++) mp1[nums2[i]]=i;

        memset(dp,-1,sizeof(dp));
        long long int x1=f(0,0,nums1,nums2);
        
        memset(dp,-1,sizeof(dp));
        long long int x2=f(0,1,nums1,nums2);
        return max(x1,x2)%mod;
    }
};
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