<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:
    def maxSum(self, nums1: List[int], nums2: List[int]) -> int:
        #------------------------------
        d = defaultdict(int)
        nums1.insert(0,0)
        nums2.insert(0,0)
        mod = 10**9+7
        #------------------------------
        for i,v in enumerate(nums1):
            d[v]=i
        #------------------------------
        arr = []
        for j,w in enumerate(nums2):
            if w in d:
                idx = d[w]
                arr.append([idx,j])
        #------------------------------
        arr.append([ len(nums1)-1 , len(nums2)-1 ])
        #------------------------------
        p1 = list(accumulate(nums1))
        p2 = list(accumulate(nums2))
        # print(p1)
        # print(p2)
        #-------------------------------
        def rangeq(arrr,i,j):
            if i==0 : 
                return arrr[j]
            return arrr[j]-arrr[i-1]
        #-------------------------------
        sm  = sum(nums1)
        res = 0
        # print(res)
        for i in range(1,len(arr)):
            i1,i2 = arr[i-1]
            j1,j2 = arr[i]
            # print(arr[i-1],arr[i])
            a = rangeq(p1,i1,j1)
            b = rangeq(p2,i2,j2)
            # sm-=min(a,b)
            # sm+=max(a,b)
            # print(min(a,b),max(a,b),"sub","add")
            # res = max(res,sm)
            res+=max(a,b)
            res%=mod
            # print("max=",max(a,b))
        for j,w in enumerate(nums2):
            if w in d:
                res-=nums1[d[w]]
                res%=mod
        # for i in range(0,len(arr)):
        #     #--------
        #     # one common case
        #     idx1,idx2 = arr[i]
        #     val = nums1[idx1]
        #     res = max(
        #         res,
        #         rangeq(p1,0,idx1) + rangeq(p2,idx2,len(p2)-1)-val,
        #         rangeq(p2,0,idx2) + rangeq(p2,idx2,len(p2)-1)-val
        #     )
        
        return res
        
        
        
        
        
38800