Best Solution and explanation Cpp
Intuition\n Describe your first thoughts on how to solve this problem. \n\n\n# Approach\nTo solve this problem, we can use dynamic programming. We\'ll define a 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n\n# Approach\nTo solve this problem, we can use dynamic programming. We\'ll define a 2D array dp of size n x 2, where dp[i][0] represents the length of the longest non-decreasing subarray ending at index i when we choose nums1[i], and dp[i][1] represents the length of the longest non-decreasing subarray ending at index i when we choose nums2[i].\n\nWe can initialize dp[0][0] and dp[0][1] to 1 since we can always choose a single element as a non-decreasing subarray. Then, for each index i from 1 to n - 1, we\'ll calculate dp[i][0] and dp[i][1] based on the values at previous indices.\n\nThe recurrence relation is as follows:\n\nIf nums1[i] >= nums1[i-1] or nums1[i] >= nums2[i-1], we can extend the non-decreasing subarray when choosing nums1[i], so \n\ndp[i][0] = max(dp[i][0], dp[i-1][0] + 1)if prev element is of nums1 .\ndp[i][0] = max(dp[i][0], dp[i-1][1] + 1);if prev element is of nums2.\n\nIf nums2[i] >= nums2[i-1] or nums2[i] >= nums2[i-1], we can extend the non-decreasing subarray when choosing nums2[i],\n so dp[i][1] = max(dp[i][1], dp[i-1][1] + 1)if prev element is of nums2.\nelse dp[i][1] = max(dp[i][1], dp[i-1][0] + 1)if prev element is of nums1\nFinally, the maximum length of the longest non-decreasing subarray will be max(dp[n-1][0], dp[n-1][1]).\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(N)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    int maxNonDecreasingLength(vector<int>& nums1, vector<int>& nums2) {\n        int n = nums1.size();\n        vector<vector<int>> dp(n,vector<int>(2, 1));\n        int maxlen = 1;\n        for (int i = 1; i < n; i++) {\n            if (nums1[i] >= nums1[i-1])\n                dp[i][0] = max(dp[i][0], dp[i-1][0] + 1);\n            if(nums1[i] >= nums2[i-1])\n                dp[i][0] = max(dp[i][0], dp[i-1][1] + 1);    \n            if (nums2[i] >= nums2[i-1])\n                dp[i][1] = max(dp[i][1], dp[i-1][1] + 1);\n            if(nums2[i] >= nums1[i-1])\n                dp[i][1] = max(dp[i][1], dp[i-1][0] + 1);\n            maxlen = max({dp[i][0],dp[i][1],maxlen});\n        }\n\n        return maxlen;\n    }\n};\n```
['C++']