Example 2:

Input: arrays = [[1],[1]]
Output: 0

Constraints:

m == arrays.length
2 <= m <= 10⁵
1 <= arrays[i].length <= 500
-10⁴ <= arrays[i][j] <= 10⁴
arrays[i] is sorted in ascending order.
There will be at most 10⁵ integers in all the arrays.

22:T412,Perfromance: runtime 100.0%, memory 88.89%\nComplexity: runtime O(n), memory O(1)\n\nThe min/max for each array are at the first/last index, so finding the global min/max over all arrays is O(n). If the global min & max are from the same array, calculate the 2nd largest/smallest pick the largest distance of the two options.\n\n```\n\nclass Solution:\n def maxDistance(self, arrays: List[List[int]]) -> int:\n global_min, index_min = min((array[0], index) for index, array in enumerate(arrays))\n global_max, index_max = max((array[-1], index) for index, array in enumerate(arrays))\n \n if index_min != index_max:\n return global_max - global_min\n \n second_min, index_second_min = min((array[0], index) for index, array in enumerate(arrays) if index != index_min)\n second_max, index_second_max = max((array[-1], index) for index, array in enumerate(arrays) if index != index_max)\n \n return max(abs(global_max - second_min), abs(second_max - global_min))\n```23:T709,# Intuition and Approach\n\nFirst you should be clear to get the maximum difference we need a minimum and then a maximum element.\nI first tried to solve the problem by storing the maximum for each array in a sorted container like map. \n\nThen for each array, I removed it\'s maximum from the sorted container(because we need a maxmimum from different array). Then, I considered th

Maximum Distance in Arrays

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Problem Description

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