<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>

/*
at each step -> i can choose to number from my array , or swap the array
dp[n][2]

use binary search to find the number
*/

class Solution {
public:
    int maxSum(vector<int>& nums1, vector<int>& nums2) {
        
        int n = max(nums1.size(), nums2.size());
        vector<vector<long long>> dp(n+1, vector<long long>(2, -1));


        function<long long(long long,long long)> solve = [&](long long i, long long opt){
            
            if(dp[i][opt] != -1) return dp[i][opt];

            if(opt == 0){
                if(i == nums1.size()) return 0LL;

                long long curr = nums1[i];
                long long stay = curr + solve(i+1, opt);

                long long j = lower_bound(nums2.begin(), nums2.end(), curr) - nums2.begin();
                long long move = LLONG_MIN;

                if(j < nums2.size() && nums2[j] == curr){
                    move = curr + solve(j+1, 1);
                }

                return  dp[i][opt]  = max(stay, move);
            }


            else{
                if(i == nums2.size()) return 0LL;

                long long curr = nums2[i];
                long long stay = curr + solve(i+1, opt);

                long long j = lower_bound(nums1.begin(), nums1.end(), curr) - nums1.begin();
                long long move = LLONG_MIN;

                if(j < nums1.size() && nums1[j] == curr){
                    move = curr + solve(j+1, 0);
                }

                return  dp[i][opt] = max(stay, move);
            }

        };


        return max(solve(0, 0), solve(0,1)) % (int)(1e9+7);

    }
};
119691