<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 {
    vector<long long int> help(vector<int>& nums1, vector<int>& nums2, long long int l, long long int r){
        long long int p1 = 0;
        long long int p2 = 0;

        long long int mod = 1e9 + 7;

        long long int n = nums1.size();
        long long int m = nums2.size();
        long long int match1 = -1, match2 = -1;
        while(l<n && r<m){
            if(nums1[l]<nums2[r]){
                p1 += nums1[l];
                l++;
            } else if(nums1[l]>nums2[r]){
                p2+=nums2[r];
                r++;
            } else {
                if(match1==-1){
                    match1 = l;
                } 

                if(match2==-1){
                    match2 = r;
                }

                if(r+1<m && nums2[r+1]==nums2[r]){
                    p2+=nums2[r];
                    r++;
                    continue;
                }

                if(l+1<n && nums1[l+1]==nums1[l]){
                    p1+=nums2[l];
                    l++;
                    continue;
                }

                if(l-match1>r-match2){
                    match1 = l;
                    p2 = p2 - (r-match2)*nums2[r];
                } else {
                    match2 = r;
                    p1 = p1- (l-match1)*nums2[r];
                }

                return {p1+nums1[match1],p2+nums2[r],match1,match2};
            }
        }

        while(l<n){
            p1+=nums1[l];
            l++;
        }

        while(r<m){
            p2+=nums2[r];
            r++;
        }

        return {p1,p2,l,r};
    }
public:
    int maxSum(vector<int>& nums1, vector<int>& nums2) {
        long long int n = nums1.size();
        long long int m = nums2.size();

        long long int mod = 1e9 + 7;

        long long int l = 0;
        long long int r = 0;
        long long int profit = 0;
        while(l<=n || r<=m){
            vector<long long int> curr_ans = help(nums1,nums2,l,r);
            profit = (profit + max(curr_ans[0],curr_ans[1]))%(mod);
            l = curr_ans[2]+1;
            r = curr_ans[3] + 1;
        }

        return profit;
    }
};
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