Dynamic Programming Approach
Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind this problem lies in recognizing that at each position i in the ar
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this problem lies in recognizing that at each position i in the arrays, we have a choice: we can either take the value from nums1[i] or nums2[i] for our nums3 array. The goal is to make these choices in such a way that maximizes the length of the longest non-decreasing subarray in nums3. Dynamic programming is suitable here because the decision at each position depends on the decisions made up to that point, and there are overlapping subproblems that can be solved once and reused.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n**Dynamic Programming State Definition:** Define a 2D DP array dp where dp[i][0] represents the length of the longest non-decreasing subarray ending at i using the element from nums1, and dp[i][1] represents the same but using the element from nums2.\n\n**Initialization:** Initialize dp[i][0] and dp[i][1] to 1 for all i, because a single element by itself forms a non-decreasing subarray of length 1.\n\n**Transition:** For each position i from 1 to n - 1, update dp[i][0] and dp[i][1] based on the following conditions:\n\nIf nums1[i] is greater than or equal to both nums1[i-1] and nums2[i-1], then dp[i][0] can be incremented by considering the longest subarray lengths ending at i-1 with elements from both arrays.\n\nSimilarly, update dp[i][1] if nums2[i] is greater than or equal to both nums1[i-1] and nums2[i-1].\n\nThe key is to choose the maximum length possible by considering both previous elements from nums1 and nums2.\n\n**Result:** After filling the DP table, the maximum length of the longest non-decreasing subarray in nums3 is the maximum value in the dp table.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(n) - The solution iterates through each element of the arrays nums1 and nums2 exactly once. The decisions at each step are made in constant time, leading to a linear time complexity.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(n) - The solution utilizes a 2D DP array of size 2n (since there are n positions and for each position, two choices are tracked). Therefore, the space complexity is linear in terms of the input size n.\n\n# Code\n```\nclass Solution(object):\n    def maxNonDecreasingLength(self, nums1, nums2):\n        """\n        :type nums1: List[int]\n        :type nums2: List[int]\n        :rtype: int\n        """\n        n = len(nums1)\n        # dp[i][0] represents the length of the longest non-decreasing subarray ending with nums1[i]\n        # dp[i][1] represents the length of the longest non-decreasing subarray ending with nums2[i]\n        dp = [[1 for _ in range(2)] for _ in range(n)]\n        \n        for i in range(1, n):\n            # Update dp for nums1[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            \n            # Update dp for nums2[i]\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        \n        # The maximum length of the longest non-decreasing subarray is the max value in dp\n        max_length = max(max(row) for row in dp)\n        return max_length\n\n\n\n```
['Python']