<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 helper(vector<int>& nums1, vector<int>& nums2, vector<int>& A, vector<int>& B) {
        long sum1 = 0, sum2 = 0;
        long maxSum = 0;
        int i = 0, j = 0;

        while (i < nums1.size() && j < nums2.size()) {
            if (A[i] != -1 && B[j] != -1) {
                if (A[i] == j) {
                    if (sum1 > sum2) {
                        maxSum += sum1 + nums1[i];
                    } else {
                        maxSum += sum2 + nums2[j];
                    }
                    sum1 = 0;
                    sum2 = 0;
                    i++;
                    j++;
                } else if (A[i] < j) {
                    sum1 += nums1[i++];
                } else {
                    sum2 += nums2[j++];
                }
            } else {
                if (A[i] == -1) {
                    sum1 += nums1[i++];
                }
                if (B[j] == -1) {
                    sum2 += nums2[j++];
                }
            }
        }

        while (i < nums1.size()) sum1 += nums1[i++];
        while (j < nums2.size()) sum2 += nums2[j++];

        maxSum += max(sum1, sum2);
        return maxSum % 1000000007; // Optional: Return the result modulo 10^9+7 if needed
    }

    int maxSum(vector<int>& nums1, vector<int>& nums2) {
        vector<int> A(nums1.size(), -1);
        vector<int> B(nums2.size(), -1);
        unordered_map<int, int> mapA;
        unordered_map<int, int> mapB;

        for (int i = 0; i < nums1.size(); i++) {
            mapA[nums1[i]] = i;
        }
        for (int i = 0; i < nums2.size(); i++) {
            mapB[nums2[i]] = i;
        }
        for (int i = 0; i < nums1.size(); i++) {
            if (mapB.find(nums1[i]) != mapB.end()) {
                A[i] = mapB[nums1[i]];
            }
        }
        for (int i = 0; i < nums2.size(); i++) {
            if (mapA.find(nums2[i]) != mapA.end()) {
                B[i] = mapA[nums2[i]];
            }
        }
        return helper(nums1, nums2, A, B);
    }
};

94364