C++ Two Pointer approach explained.
Intuition\n Describe your first thoughts on how to solve this problem. \nThis problem can be solved using a two pointer approach\n# Approach\n Describe your app
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThis problem can be solved using a two pointer approach\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe approach for this problem is to use two pointers, one for each array. We will compare the elements from both the arrays at each iteration, and move the pointer ahead accordingly. We will also keep track of the maximum sum of elements from both the arrays at each iteration and take the maximum of the two sums. Finally, we will return the maximum of the two sums.\n# Complexity\n- Time complexity: $$O(n + m)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nWe iterate through both the arrays once, so the time complexity is equal to the sum of the lengths of the arrays.\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\nWe only use constant space, so the space complexity is O(1).\n# Code\n```\nclass Solution {\npublic:\n    int maxSum(vector<int>& nums1, vector<int>& nums2) {\n        int i = 0, j = 0, n = nums1.size(), m = nums2.size(), mod = 1e9 + 7;\n        long a = 0, b = 0;\n        while (i < n || j < m) {\n            if (i < n && (j == m || nums1[i] < nums2[j])) {\n                a += nums1[i++];\n            } else if (j < m && (i == n || nums1[i] > nums2[j])) {\n                b += nums2[j++];\n            } else {\n                a = b = max(a, b) + nums1[i];\n                i++, j++;\n            }\n        }\n        return max(a, b) % mod;\n    }\n};\n```
['Array', 'Two Pointers', 'Dynamic Programming', 'Greedy', 'C++']