Get the Maximum Score

You are given two sorted arrays of distinct integers nums1 and nums2.

A valid path is defined as follows:

Choose array nums1 or nums2 to traverse (from index-0).
Traverse the current array from left to right.
If you are reading any value that is present in nums1 and nums2 you are allowed to change your path to the other array. (Only one repeated value is considered in the valid path).

Score is defined as the sum of uniques values in a valid path.

Return the maximum score you can obtain of all possible valid paths.

Since the answer may be too large, return it modulo 10^9 + 7.

Example 1:

Input: nums1 = [2,4,5,8,10], nums2 = [4,6,8,9]
Output: 30
Explanation: 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 [2,4,6,8,10].

Example 2:

Input: nums1 = [1,3,5,7,9], nums2 = [3,5,100]
Output: 109
Explanation: Maximum sum is obtained with the path [1,3,5,100].

Example 3:

Input: nums1 = [1,2,3,4,5], nums2 = [6,7,8,9,10]
Output: 40
Explanation: There are no common elements between nums1 and nums2.
Maximum sum is obtained with the path [6,7,8,9,10].

Example 4:

Input: nums1 = [1,4,5,8,9,11,19], nums2 = [2,3,4,11,12]
Output: 61

Constraints:

1 <= nums1.length <= 10^5
1 <= nums2.length <= 10^5
1 <= nums1[i], nums2[i] <= 10^7
nums1 and nums2 are strictly increasing.

Solution:

Intuition

We must take the common elements and won't miss them;
And there will be two path between the common elements,
and we will take and only take one path.

We calculate the sum of both path, and take the bigger one.

Explanation

So we apply two points solutions,
and always take the step in the smaller element.

If two elements are the same,
we compare the accumulated sum in the both paths,
and we pick the bigger one.

Complexity

Time O(N)
Space O(1)

class Solution {
    public int maxSum(int[] nums1, int[] nums2) {
        int i = 0, j = 0, m = nums1.length, n = nums2.length;
        long a = 0, b = 0, res = 0;
        int mod = (int) (1e9 + 7);
        while (i < m || j < n) {
            if (i < m && (j == n || nums1[i] < nums2[j])) {
                a += nums1[i ++];
            } else if (j < n && (i == m || nums2[j] < nums1[i])) {
                b += nums2[j ++];
            } else {
                res += Math.max(a, b) + nums1[i];
                i ++;
                j ++;
                a = 0;
                b = 0;
            }
        }
        return (int) ((res + Math.max(a, b)) % mod);
    }
}