<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 {
    int MOD = 1e9 + 7;
    vector<vector<long long>> dp;

    long long f(int ind1, int pres, vector<int>& nums1, vector<int>& nums2,
          vector<long long>& pre1, vector<long long>& pre2) {
        
        if (pres == 0 && ind1 == nums1.size())
            return 0;
        if (pres == 1 && ind1 == nums2.size())
            return 0;
        
        if(dp[ind1][pres]!=-1) return dp[ind1][pres];
        
        long long sw = 0, notsw = 0;
        
        if (pres == 0) {
            int ind2 = lower_bound(nums2.begin(), nums2.end(), nums1[ind1]) - nums2.begin();
            if (ind2 != nums2.size() && nums1[ind1] == nums2[ind2]) {
                sw = (nums1[ind1] + f(ind2 + 1, 1 , nums1, nums2, pre1, pre2));
            }
            notsw = (nums1[ind1] + f(ind1 + 1, 0, nums1, nums2, pre1, pre2));
        } else {
            int ind2 = lower_bound(nums1.begin(), nums1.end(), nums2[ind1]) - nums1.begin();
            if (ind2 != nums1.size() && nums2[ind1] == nums1[ind2]) {
                sw = (nums2[ind1] + f(ind2 + 1, 0, nums1, nums2, pre1, pre2));
            }
            notsw = (nums2[ind1] + f(ind1 + 1, 1, nums1, nums2, pre1, pre2));
        }

        return dp[ind1][pres] = max(sw, notsw);
    }

public:
    int maxSum(vector<int>& nums1, vector<int>& nums2) {
        int n1 = nums1.size(), n2 = nums2.size();
        vector<long long> pre1(n1, 0), pre2(n2, 0);
        pre1[0] = nums1[0];
        pre2[0] = nums2[0];
        for (int i = 1; i < n1; i++) {
            pre1[i] = pre1[i - 1] + nums1[i];
        }
        for (int i = 1; i < n2; i++) {
            pre2[i] = pre2[i - 1] + nums2[i];
        }
        dp = vector<vector<long long>>(max(n1, n2), vector<long long>(2,-1));
        return max(f(0, 0, nums1, nums2, pre1, pre2), f(0, 1, nums1, nums2, pre1, pre2)) % MOD;
    }
};

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