<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 <iostream>
#include <vector>
#include <unordered_set>
#include <ranges>
#include <span>
#include <numeric>
#include <concepts>

template <typename T>
bool myempty(const T& to_check)
{
    for (const auto& _ : to_check) { return false; }
    return true;
}

template <typename T>
long long sum(const T& to_sum)
{
    long long result = 0;
    for (int i : to_sum) { result += i; }
    return result;
}

template <typename T>
int size(const T& t)
{
    int res = 0;
    for (const auto& _ : t) { res++; }
    return res;
}

// template <typename T>
// void print(const T& to_print)
// {
//     std::cout << "PRINTING: ";
//     for (int n : to_print) { std::cout << n << " "; } std::cout << std::endl;
// }

std::span<const int> nonempty_span(std::span<const int> s, int offset)
{
    if (offset) s = s.subspan(offset);
    if (!myempty(s)) { s = s.subspan(1); }
    return s;
}

long long getPathSum(std::unordered_set<int>& shared, std::span<const int> s1, std::span<const int> s2, long long acc)
{
    if (myempty(s1)) { return acc + sum(s2); }
    if (myempty(s2)) { return acc + sum(s1); }
    
    auto range_before_shared = [&shared](std::span<const int> s)
    {
        return s | std::views::take_while([&shared](int k){ return !shared.contains(k); });
    };
    
    auto range1 = range_before_shared(s1);
    auto range2 = range_before_shared(s2);
    // print(range1);
    // print(range2);
    long long sum1 = sum(range1);
    long long sum2 = sum(range2);
    return getPathSum(
        shared,
        nonempty_span(s1, size(range1)),
        nonempty_span(s2, size(range2)),
        acc + std::max(sum1, sum2));
}


class Solution {
public:
    int maxSum(const std::vector<int>& nums1, const std::vector<int>& nums2)
    {
        std::unordered_set<int> set1(nums1.begin(), nums1.end());
        auto set2 = nums2 | std::views::filter([&set1](int k){ return set1.contains(k); });
        std::unordered_set<int> shared(set2.begin(), set2.end());
        //print(shared);
        return (sum(shared) + getPathSum(shared, nums1, nums2, 0ll)) % 1'000'000'007;
    }
};
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