Maximize Longest Non-Decreasing Subarray Length
Intuition\nThe problem involves constructing the longest non-decreasing subarray by choosing values from either of two arrays, nums1 or nums2, at each index. Th
# Intuition\nThe problem involves constructing the longest non-decreasing subarray by choosing values from either of two arrays, nums1 or nums2, at each index. The challenge is to keep the subarray non-decreasing by picking values optimally while maintaining the longest possible sequence. Observing each possible choice at each step and tracking the maximum achievable length forms the basis of our approach.\n\n# Approach\nUse dynamic programming to keep track of the longest non-decreasing subarray ending at each index. For each position i, calculate the longest sequence using nums1[i] or nums2[i] by checking if each value can continue a non-decreasing subarray from previous elements. Update the maximum length accordingly.\n\n# Complexity\n- Time complexity: $$O(n)$$, as we iterate through the arrays once.\n\n- Space complexity: $$O(1)$$, as only a constant amount of extra space is used for tracking lengths.\n\n# Code\n```python3 []\nclass Solution:\n\n    # chatgpt solution - tested method\n    def maxNonDecreasingLength(self, nums1: List[int], nums2: List[int]) -> int:\n        n = len(nums1)\n        dp1 = dp2 = 1  # Track longest non-decreasing subarray length ending at i using nums1 or nums2\n        max_len = 1    # Initialize maximum length\n\n        for i in range(1, n):\n            new_dp1 = new_dp2 = 1  # Reset for the current index i\n\n            # Check if nums1[i] continues non-decreasing subarray from nums1[i - 1]\n            if nums1[i] >= nums1[i - 1]:  \n                new_dp1 = dp1 + 1\n\n            if nums1[i] >= nums2[i - 1]:  # Check if nums1[i] can continue from nums2[i - 1]\n                new_dp1 = max(new_dp1, dp2 + 1)\n\n            if nums2[i] >= nums2[i - 1]:  # Check if nums2[i] continues from nums2[i - 1]\n                new_dp2 = dp2 + 1\n\n            if nums2[i] >= nums1[i - 1]:  # Check if nums2[i] can continue from nums1[i - 1]\n                new_dp2 = max(new_dp2, dp1 + 1)\n\n            dp1, dp2 = new_dp1, new_dp2  # Update dp values\n            max_len = max(max_len, dp1, dp2)  # Update maximum length\n\n        return max_len\n    \n    # my solution - not tested method as not working\n    def MyVersion_maxNonDecreasingLength(self, nums1: List[int], nums2: List[int]) -> int:\n        \n        n = len(nums1) # same length as nums2\n\n        # safety net as we have different treatment for first and last iteration \n        # so we expect them to be different\n        if n==1:\n            return n\n\n        nums3 = []\n        length_non_decreasing_sub_array_list = []\n        length_non_decreasing_sub_array = 0\n        \n        # For the first element we pick the smallest of both lists\n        nums3.append(min(nums1[0], nums2[0]))\n        length_non_decreasing_sub_array += 1\n        length_non_decreasing_sub_array_list.append(length_non_decreasing_sub_array)\n\n        # for the following n-1 elements but not the last one, we will only add from nums1[i] and nums2[i], \n        # the one closer to nums3[i-1] if they both bigger other the only one bigger \n        # then increment the length_non_decreasing_sub_array\n        # if non bigger we append to nums3 but no increment of length_non_decreasing_sub_array\n\n        for i in range(1, n-1):\n            # nums1[i] is bigger than nums3[i-1] but not nums2[i]\n            # we can directly add nums1[i] to nums3 and increase length_non_decreasing_sub_array by 1\n            if nums1[i] >= nums3[i-1] and nums3[i-1] > nums2[i]:\n                nums3.append(nums1[i])\n                length_non_decreasing_sub_array += 1\n                length_non_decreasing_sub_array_list.append(length_non_decreasing_sub_array)\n\n\n            # nums2[i] is bigger than nums3[i-1] but not nums1[i]\n            # we can directly add nums1[i] to nums3 and increase length_non_decreasing_sub_array by 1\n            elif nums2[i] >= nums3[i-1] and nums3[i-1] > nums1[i]:\n                nums3.append(nums2[i])\n                length_non_decreasing_sub_array += 1\n                length_non_decreasing_sub_array_list.append(length_non_decreasing_sub_array)\n\n            # both nums1[i] and nums2[i] are bigger than nums3[i-1]\n            # we now have to compare which one is the closest to nums3[i-1] and append to nums3\n            # increase length_non_decreasing_sub_array by 1\n            elif nums1[i] >= nums3[i-1] and nums2[i] >= nums3[i-1]: \n                nums3.append(min(nums2[i], nums1[i]))\n                length_non_decreasing_sub_array += 1\n                length_non_decreasing_sub_array_list.append(length_non_decreasing_sub_array)\n\n            # None of nums1[i] and nums2[i] is bigger than nums3[i-1]\n            # we now have to compare which one is the closest to nums3[i-1] and append to nums3\n            # but no increase of length_non_decreasing_sub_array\n            # since we are not ask to return nums3 we could just here return length_non_decreasing_sub_array\n            # previous sentence is wrong as we can have another a break before going again to built the longest\n            # we will reset length_non_decreasing_sub_array to 1\n            else: \n                nums3.append(min(nums2[i], nums1[i]))\n                length_non_decreasing_sub_array = 1\n                length_non_decreasing_sub_array_list.append(length_non_decreasing_sub_array)\n\n        # for last iteration we pick the biggest of both lists\n        nums3.append(max(nums1[n-1], nums2[n-1]))\n        if nums3[n-1] >= nums3[n-2]:\n            length_non_decreasing_sub_array += 1\n            length_non_decreasing_sub_array_list.append(length_non_decreasing_sub_array)\n        else:\n            length_non_decreasing_sub_array = 1\n            length_non_decreasing_sub_array_list.append(length_non_decreasing_sub_array)           \n    \n        # now we will return the max of length_non_decreasing_sub_array_list who captured the longest streak\n        return max(length_non_decreasing_sub_array_list)\n        \n\n```
['Array', 'Two Pointers', 'Dynamic Programming', 'Greedy', 'Python3']