<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 mod = 1e9 + 7;
    int binarySearch(vector<int>&nums1 , int key ){
        int n = nums1.size();
        int left = 0 ; 
        int right = n - 1; 
        while(left <= right ) {
            int mid = (left + right )/2; 
            if(nums1[mid] == key ) return mid;
            else if(nums1[mid] < key ) left = mid + 1;
            else right = mid - 1;
        }
        return -1;
    }

    long long solve(vector<int>&nums1 , vector<int>&nums2 , int n , int m , int i , int arr , vector<vector<long long>>&dp){
        
        if(arr == 0 ){
            if(i == n ) return 0;
            if(dp[i][arr] != -1) return dp[i][arr];
            long long ans = (long long)nums1[i] + solve(nums1 , nums2 , n , m , i + 1 , arr , dp ) ; 
            int index = binarySearch(nums2 , nums1[i]);
            if(index != -1 ) ans = max(ans , (long long)nums1[i] + solve(nums1 , nums2 , n , m , index + 1 , 1 - arr, dp ));
            return dp[i][arr] = ans;
        }
        if(i == m ) return 0;
        if(dp[i][arr] != -1) return dp[i][arr];
        long long ans = (long long)nums2[i] + solve(nums1 , nums2 , n , m , i+1 , arr , dp) ; 
        int index = binarySearch(nums1 , nums2[i]);
        if(index != -1 ) ans = max(ans, (long long)nums2[i] + solve(nums1 , nums2 , n , m , index + 1 , 1- arr , dp));
        return dp[i][arr] = ans;
    }

    int maxSum(vector<int>& nums1, vector<int>& nums2) {
        int n = nums1.size();
        int m = nums2.size();
        vector<vector<long long>> dp1( max(n , m) , vector<long long>(2 , -1 ));
        vector<vector<long long>> dp2( max(n , m) , vector<long long>(2 , -1 ));
        long long ans1 = solve(nums1 , nums2 , n , m , 0 , 0 ,dp1) ; 
        long long ans2 = solve(nums1, nums2 , n , m , 0 , 1 , dp2) ; 
        long long ans = max(ans1 , ans2);
        ans = ans % mod;
        return (int)ans;     
    }
};
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