<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 {
public:
    int maxSum(vector<int>& nums1, vector<int>& nums2) {

        int n = nums1.size();
        int m = nums2.size();
        unordered_map<int, pair<int, int>> mp;
        for (int i = 0; i < n; i++) {
            mp[nums1[i]] = {i, -1};
        }

        for (int i = 0; i < m; i++) {

            if (mp.contains(nums2[i])) {
                mp[nums2[i]].second = i;
            } else {
                mp[nums2[i]] = {-1, i};
            }
        }

        const long long THRESHOLD = 1LL << 50;
        const long long INF = 1LL << 60;

        vector<int> nums;
        for (int v : nums1) {
            nums.push_back(v);
        }

        for (int v : nums2) {
            nums.push_back(v);
        }

        sort(nums.begin(), nums.end());
        nums.erase(unique(nums.begin(), nums.end()), nums.end());

        int size = nums.size();

        vector<vector<long long>> dp(2, vector(size + 1, -1 * INF));
        dp[0][0] = 0;
        dp[1][0] = 0;

        for (int i = 0; i < size; i++) {

            pair<int, int> p = mp[nums[i]];
            dp[0][i + 1] = max(dp[0][i + 1], dp[0][i]);
            dp[1][i + 1] = max(dp[1][i + 1], dp[1][i]);

            if (p.first != -1 && p.second != -1) {

                for (int j = 0; j < 2; j++) {
                    dp[j][i + 1] = max(dp[j][i + 1], dp[0][i] + nums[i]);
                    dp[j][i + 1] = max(dp[j][i + 1], dp[1][i] + nums[i]);
                }
            } else if (p.first != -1) {
                dp[0][i + 1] = max(dp[0][i + 1], dp[0][i] + nums[i]);
            } else {
                dp[1][i + 1] = max(dp[1][i + 1], dp[1][i] + nums[i]);
            }
        }

        int MOD = 1e9 + 7;
        return max(dp[0][size], dp[1][size]) % MOD;
    }
};
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