<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 mod = 1e9 + 7;
    long long int dp[100001][2];

    long long int find(int id, vector<int>& nums1, vector<int>& nums2, map<int, int>& mp1, map<int, int>& mp2, int on) {
        if ((on == 0 && id == nums1.size()) || (on == 1 && id == nums2.size())) {
            return 0;
        }

        if (dp[id][on] != -1) {
            return dp[id][on];
        }

        int num = (on == 0) ? nums1[id] : nums2[id];
        long long int result1 = 0, result2 = 0;

        // Check for the number in the first map
        if (on == 0 && mp2.find(num) != mp2.end()) {
            int nextIndex = mp2[num] + 1;
            result1 = num + find(nextIndex, nums1, nums2, mp1, mp2, 1);
        }
        if (on == 1 && mp1.find(num) != mp1.end()) {
            int nextIndex = mp1[num] + 1;
            result1 = num + find(nextIndex, nums1, nums2, mp1, mp2, 0);
        }

        // If the number doesn't exist in the other map or we choose not to switch
        result2 = num + find(id + 1, nums1, nums2, mp1, mp2, on);

        dp[id][on] = max(result1, result2);
        return dp[id][on];
    }

    int maxSum(vector<int>& nums1, vector<int>& nums2) {
        int n = nums1.size();
        int m = nums2.size();
        map<int, int> mp1, mp2;

        memset(dp, -1, sizeof(dp));

        // Populate maps with indices of nums1 and nums2
        for (int i = 0; i < n; i++) {
            mp1[nums1[i]] = i;
        }
        for (int i = 0; i < m; i++) {
            mp2[nums2[i]] = i;
        }

        long long int ans1 = find(0, nums1, nums2, mp1, mp2, 0) % mod;
        long long int ans2 = find(0, nums1, nums2, mp1, mp2, 1) % mod;

        return max(ans1, ans2) % mod;
    }
};

121994