Python Super Short, Simple & Clean Solution 99% faster
the main idea is that when iterating over the intervals there are three cases:\n\n1. the new interval is in the range of the other interval\n2. the new interval
the main idea is that when iterating over the intervals there are three cases:\n\n1. the new interval is in the range of the other interval\n2. the new interval\'s range is before the other\n3. the new interval is after the range of other interval\n\n```\nclass Solution:\n    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:\n        result = []\n        \n        for interval in intervals:\n\t\t\t# the new interval is after the range of other interval, so we can leave the current interval baecause the new one does not overlap with it\n            if interval[1] < newInterval[0]:\n                result.append(interval)\n            # the new interval\'s range is before the other, so we can add the new interval and update it to the current one\n            elif interval[0] > newInterval[1]:\n                result.append(newInterval)\n                newInterval = interval\n            # the new interval is in the range of the other interval, we have an overlap, so we must choose the min for start and max for end of interval \n            elif interval[1] >= newInterval[0] or interval[0] <= newInterval[1]:\n                newInterval[0] = min(interval[0], newInterval[0])\n                newInterval[1] = max(newInterval[1], interval[1])\n\n        \n        result.append(newInterval); \n        return result\n```
['Python3']