<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>

#define ll long long
#define vl vector<ll>

class Solution {
public:
    int maxSum(vector<int>& n1, vector<int>& n2) {
        vl nums1; for (auto nn : n1) nums1.push_back(nn);
        vl nums2; for (auto nn : n2) nums2.push_back(nn);
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        ll n = nums1.size(), m = nums2.size();
        ll M = 1000000007;
        vl dp1(1, 0), dp2(1, 0);
        ll i1 = 0, i2 = 0;
        while (i1<n && i2<m) {
            if (nums1[i1] < nums2[i2]) {
                ll temp = (dp1.back() + nums1[i1]);
                dp1.push_back(temp);
                i1++;
            } else if (nums1[i1] > nums2[i2]) {
                ll temp = (dp2.back() + nums2[i2]);
                dp2.push_back(temp);
                i2++;
            } else {
                ll temp = max(dp1.back(), dp2.back()) + nums1[i1];
                dp1.push_back(temp);
                dp2.push_back(temp);
                i1++, i2++;
            }
        }
        while (i1<n) {
            ll temp = (dp1.back() + nums1[i1]);
            dp1.push_back(temp);
            i1++;
        }
        while (i2<m) {
            ll temp = (dp2.back() + nums2[i2]);
            dp2.push_back(temp);
            i2++;
        }
        ll ans = max(dp1.back(), dp2.back())%M;
        return ans;
    }
};
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