<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>

const long long MN = 1e5 + 10;
long long memo[MN][2];

const long long MOD = 1e9 + 7;

class Solution {
private:
    vector<int> nums1, nums2;
    unordered_map<long long, long long> um1, um2;
    
public:
    long long dp(long long i, long long state, long long n1, long long n2) {
        if (state == 0 && i >= n1 || state == 1 && i >= n2) {
            return 0;
        }
        
        if (memo[i][state] != -1) {
            return memo[i][state];
        }
        
        long long ans = 0;
        
        if (state == 0) {
            ans = max(ans, (nums1[i] + dp(i + 1, state, n1, n2)));
            if (um2.count(nums1[i])) {
                long long idx_nums2 = um2[nums1[i]];
                ans = max(ans, (nums1[i] + dp(idx_nums2 + 1, 1 - state, n1, n2)));
            }
        } 
        else {
            ans = max(ans, (nums2[i] + dp(i + 1, state, n1, n2)));
            if (um1.count(nums2[i])) {
                long long idx_nums1 = um1[nums2[i]];
                ans = max(ans, (nums2[i] + dp(idx_nums1 + 1, 1 - state, n1, n2)));
            }
        }
        
        return memo[i][state] = ans;
    }
    
    long long maxSum(vector<int>& nums1, vector<int>& nums2) {
        this->nums1 = nums1;
        this->nums2 = nums2;
        long long n1 = nums1.size();
        long long n2 = nums2.size();
        
        um1.clear();
        um2.clear();
        
        for (long long i = 0; i < n1; ++i) {
            um1[nums1[i]] = i;
        }
        
        for (long long i = 0; i < n2; ++i) {
            um2[nums2[i]] = i;
        }
        
        memset(memo, -1, sizeof(memo));
        long long ans1 = dp(0, 0, n1, n2);
        
        memset(memo, -1, sizeof(memo));
        long long ans2 = dp(0, 1, n1, n2);
        
        return max(ans1, ans2) % MOD;
    }
};
134658