<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 , n;
    int mod = 1e9+7;
    map<int,int>mp1;
    map<int,int>mp2;
    long long dp[100001][2];
    long long f(int i, int j , vector<int>&v1, vector<int>&v2 , int start)
    {
        if(i>=n && j>=m) return 0;
        if(i>=n && start) return 0;
        if(j>=m && !start) return 0;
        if(start)
        {
            if(dp[i][start]!=-1) return dp[i][start];
        }
        if(!start)
        {
            if(dp[j][start]!=-1) return dp[j][start];
        }
        if(start)
        {
            int next=-1;
            if(mp2.count(v1[i]) && mp2[v1[i]]>=j) next = mp2[v1[i]];
            long long ans = (v1[i] + f(i+1, j , v1,v2,start));
            if(next!=-1)
            {
                ans = max(ans , (v1[i] + f(i+1,next+1,v1,v2,!start)));
            }
            return dp[i][start] = ans;
        }
        else
        {
            int next=-1;

            if(mp1.count(v2[j]) && mp1[v2[j]]>=i) next = mp1[v2[j]];

            long long ans = (v2[j] + f(i, j+1 , v1,v2,start));
            if(next!=-1)
            {
                ans = max(ans , (v2[j] + f(next+1,j+1,v1,v2,!start)));
            }
            return dp[j][start] = ans;
        }
        return 0;
    }
    int maxSum(vector<int>& v1, vector<int>& v2) {
         n = v1.size();
         m = v2.size();

        for(int i=0;i<n;i++)
        {
            mp1[v1[i]]=i;
        }
        for(int i=0;i<m;i++)
        {
            mp2[v2[i]]=i;
        }
        memset(dp , -1 ,sizeof(dp));
        return max(f(0,0,v1,v2 , 1) , f(0,0,v1,v2,0))%mod;
    }
};
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