Python 3 || 11 lines, 2ptrs  || T/S: 90% / 26%
Here\'s the plan:\n\n1. We unzip events, rezip vals with each beg and end, and then sort each of these lists.\n\n1. We iterate through the lists similtaneously 
Here\'s the plan:\n\n1. We unzip `events`, rezip `vals` with each `beg` and `end`, and then sort each of these lists.\n\n1. We iterate through the lists similtaneously keeping track of the max value of all the events to the left and determining the value immediately to the right; if the sum of the left max value and the immediate right value is greater than those previous, we store it in `ans`.\n\n1. We return ans upon the finish of the iteration.\n```\nclass Solution:\n    def maxTwoEvents(self, events: List[List[int]]) -> int:\n        \n        mx, ans = 0, 0\n\n        beg, end, val = zip(*events)\n        beg2ndMtg = sorted(zip(beg, val))\n        end1stMtg = iter(sorted(zip(end, val)))\n\n        end1st, val1st = next(end1stMtg)\n\n        for beg2nd, val2nd in beg2ndMtg:\n\n            while end1st < beg2nd:\n                mx = max(mx, val1st)\n                end1st, val1st = next(end1stMtg)\n\n            ans = max(ans, mx + val2nd)\n\n        return ans  \n```\n[https://leetcode.com/problems/two-best-non-overlapping-events/submissions/1225229981/](https://leetcode.com/problems/two-best-non-overlapping-events/submissions/1225229981/)\n\n\nI could be wrong, but I think that time complexity is *O*(*N* log *N*) and space complexity is *O*(*N*), in which *N* ~ `len(events)`.
['Python3']