<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>

// long long  dp[100001][2];


// long long  mod = 1e9 + 7;

// class Solution {
// public:


//     unordered_map<long long ,long long > hash1;
//     unordered_map<long long ,long long > hash2;

//     long long  helper(long long  index,long long  isFirst,vector<int > &arr1,vector<int > &arr2){

//         if(isFirst){
//             if(index>=arr1.size()){
//                 return 0;
//             }
//         }else{
//             if(index>=arr2.size()){
//                 return 0;
//             }
//         }
//         if(dp[index][isFirst]!=-1) return dp[index][isFirst];
//         long long  ans = 0;

//         if(isFirst){

//             long long  temp = (arr1[index] + helper(index+1,isFirst,arr1,arr2))%mod;
//             ans = max(ans,temp);
//             if(hash1.find(arr1[index])!=hash1.end() && hash2.find(arr1[index])!=hash2.end()){

//                 long long  temp = (arr1[index] + helper(hash2[arr1[index]]+1,0,arr1,arr2))%mod;
//                 ans = max(ans,temp);
//             }

//         }else{

//             long long  temp = (arr2[index] + helper(index+1,isFirst,arr1,arr2))%mod;
//             ans = max(ans,temp);

//             if(hash1.find(arr2[index])!=hash1.end() && hash2.find(arr2[index])!=hash2.end()){

//                 long long  temp = (arr2[index] + helper(hash1[arr2[index]]+1,1,arr1,arr2))%mod;
//                 ans = max(ans,temp);

//             }

//         }

//         return dp[index][isFirst] = ans%mod;



//     }
    


//     int  maxSum(vector<int >& nums1, vector<int >& nums2) {

        
//         for(long long  i =0 ; i<nums1.size() ; i++){
//             hash1[nums1[i]] = i;
//         }
//         for(long long  i =0 ; i<nums2.size() ; i++){
//             hash2[nums2[i]] = i;
//         }
        

//         memset(dp,-1,sizeof(dp));
   

//         long long  ans1 = helper(0,1,nums1,nums2)%mod;
//         long long  ans2 = helper(0,0,nums1,nums2)%mod;


//         return max(ans1,ans2);

        

//     }
// };


#include <vector>
#include <unordered_map>
#include <cstring>
#include <algorithm>

using namespace std;

class Solution {
public:
    unordered_map<long long, long long> hash1;
    unordered_map<long long, long long> hash2;
    long long dp[100001][2];
    long long mod = 1e9 + 7;

    long long helper(long long index, long long isFirst, vector<int> &arr1, vector<int> &arr2) {
        if (isFirst) {
            if (index >= arr1.size()) {
                return 0;
            }
        } else {
            if (index >= arr2.size()) {
                return 0;
            }
        }
        if (dp[index][isFirst] != -1) return dp[index][isFirst];
        long long ans = 0;

        if (isFirst) {
            long long temp = (arr1[index] + helper(index + 1, isFirst, arr1, arr2)) ;
            ans = max(ans, temp);
            if (hash1.find(arr1[index]) != hash1.end() && hash2.find(arr1[index]) != hash2.end()) {
                temp = (arr1[index] + helper(hash2[arr1[index]] + 1, 0, arr1, arr2)) ;
                ans = max(ans, temp);
            }
        } else {
            long long temp = (arr2[index] + helper(index + 1, isFirst, arr1, arr2)) ;
            ans = max(ans, temp);
            if (hash1.find(arr2[index]) != hash1.end() && hash2.find(arr2[index]) != hash2.end()) {
                temp = (arr2[index] + helper(hash1[arr2[index]] + 1, 1, arr1, arr2));
                ans = max(ans, temp);
            }
        }

        return dp[index][isFirst] = ans ;
    }

    int maxSum(vector<int>& nums1, vector<int>& nums2) {
        for (long long i = 0; i < nums1.size(); i++) {
            hash1[nums1[i]] = i;
        }
        for (long long i = 0; i < nums2.size(); i++) {
            hash2[nums2[i]] = i;
        }

        memset(dp, -1, sizeof(dp));

        long long ans1 = helper(0, 1, nums1, nums2) % mod;
        long long ans2 = helper(0, 0, nums1, nums2) % mod;

        return max(ans1, ans2);
    }
};

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