Simple Solution WITH HELPFUL COMMENTS
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:\n    int maxNonDecreasingLength(vector<int>& nums1, vector<int>& nums2) {\n        // If there\'s only one element, the longest non-decreasing subarray is just that element.\n        if(nums1.size() == 1)\n            return 1;\n\n        // DP table to store the length of the longest non-decreasing subarray ending at index i\n        // dp[i][0] -> if we pick nums1[i]\n        // dp[i][1] -> if we pick nums2[i]\n        vector<vector<int>> dp(nums1.size(), vector<int>(2, 1));\n\n        int maxLength = 1; // Variable to track the maximum length of non-decreasing subarray\n\n        // Traverse through the arrays starting from the second element\n        for(int i = 1; i < nums1.size(); ++i){\n\n            // Case 1: If we choose nums1[i] for the current position in nums3\n            if(nums1[i] >= nums1[i-1]) {\n                dp[i][0] = dp[i-1][0] + 1; // Extend the subarray from nums1[i-1]\n            }\n\n            if(nums1[i] >= nums2[i-1]) {\n                dp[i][0] = max(dp[i][0], dp[i-1][1] + 1); // Extend the subarray from nums2[i-1]\n            }\n\n            // Update the maximum length so far\n            maxLength = max(maxLength, dp[i][0]);\n\n            // Case 2: If we choose nums2[i] for the current position in nums3\n            if(nums2[i] >= nums2[i-1]) {\n                dp[i][1] = dp[i-1][1] + 1; // Extend the subarray from nums2[i-1]\n            }\n\n            if(nums2[i] >= nums1[i-1]) {\n                dp[i][1] = max(dp[i][1], dp[i-1][0] + 1); // Extend the subarray from nums1[i-1]\n            }\n\n            // Update the maximum length so far\n            maxLength = max(maxLength, dp[i][1]);\n        }\n\n        return maxLength; // Return the maximum length of any non-decreasing subarray\n    }\n};\n\n```
['C++']