<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>

#include <vector>
#include <unordered_map>
#include <utility> // for std::pair
#include <algorithm> // for std::sort
using namespace std;

class Solution {
public:
    int maxSum(vector<int>& nums1, vector<int>& nums2) {
        const int MOD = 1e9 + 7;

        // Calculate suffix sums for nums1
        vector<long long> suffixSum1(nums1.size());
        suffixSum1[nums1.size() - 1] = nums1[nums1.size() - 1];
        for (int i = nums1.size() - 2; i >= 0; --i) {
            suffixSum1[i] = suffixSum1[i + 1] + nums1[i];
        }

        // Calculate suffix sums for nums2
        vector<long long> suffixSum2(nums2.size());
        suffixSum2[nums2.size() - 1] = nums2[nums2.size() - 1];
        for (int i = nums2.size() - 2; i >= 0; --i) {
            suffixSum2[i] = suffixSum2[i + 1] + nums2[i];
        }

        // Map to store indices of nums1 elements
        unordered_map<int, int> indexMap1;
        for (int i = 0; i < nums1.size(); ++i) {
            indexMap1[nums1[i]] = i;
        }

        // Vector to store matching index pairs
        vector<pair<int, int>> matchingIndices;
        for (int i = 0; i < nums2.size(); ++i) {
            if (indexMap1.find(nums2[i]) != indexMap1.end()) {
                // If nums2[i] exists in nums1, store the indices
                matchingIndices.push_back({indexMap1[nums2[i]], i});
            }
        }

        // Sort the vector of pairs
        sort(matchingIndices.begin(), matchingIndices.end());

        long long score1 = 0, score2 = 0;
        int p1 = 0, p2 = 0;

        // Traverse using the matching indices
        for (const auto& [i, j] : matchingIndices) {
            // Add values from p1 to i in nums1
            while (p1 <= i) {
                score1 += nums1[p1++];
            }
            // Add values from p2 to j in nums2
            while (p2 <= j) {
                score2 += nums2[p2++];
            }
            // Take the maximum score so far
            long long maxScore = max(score1, score2);
            score1 = score2 = maxScore;
        }

        // Complete the path for remaining elements
        while (p1 < nums1.size()) {
            score1 += nums1[p1++];
        }
        while (p2 < nums2.size()) {
            score2 += nums2[p2++];
        }

        // Return the maximum score modulo 10^9 + 7
        return max(score1, score2) % MOD;
    }
};

80549