[Python 3] O(n) Time, O(1) Space, Detailed Explanation
Intuition\n Describe your first thoughts on how to solve this problem. \nThere are 3 different cases to consider when looking at the border between two pairs of
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThere are 3 different cases to consider when looking at the border between two pairs of numbers (a pair being numbers in nums1 and nums2 with the same index):\n1. The numbers are increasing, and if either pair was flipped they would still be increasing (ex. nums1 = [1, 3] and nums2 = [2, 4])\n2. The numbers are increasing, but if either pair flipped they would not be increasing (ex. nums1 = [1, 2] and nums2 = [3, 4])\n3. The numbers are not increasing (since there is a guaranteed solution this means that if either pair is flipped they will be increasing, ex. nums1 = [6, 5] and nums2 = [4, 7])\n\nTwo observations that lead to this approach:\n1. In the first case above, these borders can be used to divide the problem into independent subproblems, as the chosen flips of each will not affect the other\n2. Within each subsection, Every boundary between numbers now falls into the second or third case above. As a result, there are only two sets of flips that are valid, and they are mirrors of each other.\n\nIf the second observation is confusing, try drawing out some examples to build intuition. Think of every flippable pair as being "aligned" a certain direction and all pairs needing to be "aligned" the same way for both arrays to be increasing\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThese two observations give us a relatively straightforward solution:\n1. Divide the problem into sections on borders where the arrays are increasing no matter whether the numbers are flipped or not\n2. In each subsection, take a greedy approach and flip any pair of numbers when necessary, counting the number of flips needed to make the subsection strictly increasing.\n3. After making all necessary flips in a subsection, add the count to the total. Note that the exact opposite set of flips would also results in an increasing subsection, so the real count is min(count, length of subsection - count)\n\nThe code below keeps track of each subsection as it goes, allowing the algorithm to be completed in one pass and without any extra memory.\n\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n    def minSwap(self, nums1: List[int], nums2: List[int]) -> int:\n        sectionSwaps = 0\n        sectionLength = 1\n        minSwaps = 0\n        for x in range(0, len(nums1) - 1):\n            # If the end of the subsection has been reached,\n            # add the minimum number of steps to the total\n            if max(nums1[x], nums2[x]) < min(nums1[x + 1], nums2[x + 1]):\n                minSwaps += min(sectionSwaps, sectionLength - sectionSwaps)\n                sectionLength = 1\n                sectionSwaps = 0\n                continue\n            sectionLength += 1\n            # Flip the next pair if it does not align with the current one\n            if nums1[x] >= nums1[x + 1] or nums2[x] >= nums2[x + 1]:\n                sectionSwaps += 1\n                temp = nums1[x + 1]\n                nums1[x + 1] = nums2[x + 1]\n                nums2[x + 1] = temp\n        # The last subsection is not handled in the for loop\n        minSwaps += min(sectionSwaps, sectionLength - sectionSwaps)\n        return minSwaps\n```
['Python3']