Beats 100% | 1ms | Two  pointers
IntuitionThe problem involves merging two arrays by picking the maximum sum path when encountering common elements. This strategy minimizes missing out on poten

### Intuition
The problem involves merging two arrays by picking the maximum sum path when encountering common elements. This strategy minimizes missing out on potential larger sums.

---

### Approach
1. **Two Pointers:** Use two pointers (`i` and `j`) to traverse through both arrays (`nums1` and `nums2`).
2. **Tracking Sums:** Maintain two cumulative sums (`top` and `bottom`) for elements of `nums1` and `nums2`, respectively.
3. **Handling Common Elements:**
   - When a common element is encountered (`nums1[i] == nums2[j]`), add the maximum of `top` and `bottom` to the final answer `ans` and reset both sums to 0.
   - Then add the common element to `ans`.
4. **Processing Remaining Elements:**
   - Once the end of one array is reached, continue adding the remaining elements of the other array to its respective cumulative sum.
5. **Final Update:** Add the larger of `top` and `bottom` to the final answer `ans`.
6. **Modulo Operation:** Return the result modulo \(10^9 + 7\) as required.

---

### Complexity
- **Time Complexity:** \(O(n + m)\)
   - Each array is traversed once, where \(n\) and \(m\) are the lengths of `nums1` and `nums2`.
- **Space Complexity:** \(O(1)\)
   - Only a few variables are used to keep track of the cumulative sums and pointers.

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### Suggested Improvements
1. **Edge Cases:** Ensure edge cases like one array being empty or having no common elements are handled explicitly.
2. **Modulo Inside Loop:** Instead of applying the modulo operation at the end, apply it during updates to `ans` to avoid overflow when `nums1` and `nums2` have large elements.

---

### Updated Code
```cpp
class Solution {
public:
    int maxSum(vector<int>& nums1, vector<int>& nums2) {
        const int MOD = 1e9 + 7;
        long long ans = 0, top = 0, bottom = 0;
        int i = 0, j = 0;

        while (i < nums1.size() && j < nums2.size()) {
            if (nums1[i] < nums2[j]) {
                top += nums1[i++];
            } else if (nums1[i] > nums2[j]) {
                bottom += nums2[j++];
            } else {
                ans = (ans + max(top, bottom) + nums1[i]) % MOD;
                top = bottom = 0;
                i++; j++;
            }
        }

        while (i < nums1.size()) {
            top += nums1[i++];
        }

        while (j < nums2.size()) {
            bottom += nums2[j++];
        }

        ans = (ans + max(top, bottom)) % MOD;

        return ans;
    }
};
```

['C++']