<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>

'''
Brute Force - You try all combination of paths possible - O(2^n)

Pseudocode
    - keep track of the max_sum from path1 and path2
    - 2 pointers, iterate through each nums1 and nums2, iterating on the pointer that has the smaller value
    - if the pointer reaches the same value 
        - set path2_sum to be path1_sum if it is bigger (you moved to path2), but still keep path1_sum the same
        - same thing with path1
        
        
A few implementation notes
- THE MOD MUST BE TAKEN AT THE END, otherwise you might mod and the result will be a smaller number resulting in something weird when you compare the a/b path sums
- You can embed the loops inside the while loop by changing it to an or and just changing the condition instead of having 2 outside again
- I didn't change this part but realize that if path1_sum > path2_sum, you set path2_sum to be path1_sum which basically means the same as setting both to be the max of each other...sort of like a meeting point location...this can simplify a lot of the code
        
Time Complexity: O(n)
Space Complexity: O(1)
'''

class Solution:
    def maxSum(self, nums1: List[int], nums2: List[int]) -> int:
        path1_sum = 0
        path2_sum = 0
        ptr1 = 0
        ptr2 = 0
        mod = pow(10, 9) + 7
        
        while ptr1 < len(nums1) or ptr2 < len(nums2):
            if ptr1 < len(nums1) and (ptr2 == len(nums2) or nums1[ptr1] < nums2[ptr2]):
                path1_sum += nums1[ptr1]
                ptr1 += 1
            elif ptr2 < len(nums2) and (ptr1 == len(nums1) or nums1[ptr1] > nums2[ptr2]):
                path2_sum += nums2[ptr2]
                ptr2 += 1
            else:
                path1_sum += nums1[ptr1]
                path2_sum += nums2[ptr2]
                
                # If path1 was bigger, set path2 to be from path1, BUT KEEP PATH1 the same
                if path1_sum > path2_sum:
                    path2_sum = path1_sum
                else:
                    path1_sum = path2_sum                    
                ptr1 += 1
                ptr2 += 1
                
                '''
                Easier solution ehre
                path1_sum = path2_sum = max(path1_sum, path2_sum) + nums1[ptr1]
                ptr1 += 1
                ptr2 += 1
                '''
        
        return max(path1_sum, path2_sum) % mod
28500