Simple 4-line solution, faster than 80%
Key Insight\n\nWanting the smallest possible starting value is the same as wanting the smallest possible ending value.\n\n# Algorithm\nTherefore sort the tasks 
# Key Insight\n\nWanting the smallest possible _starting_ value is the same as wanting the smallest possible _ending_ value.\n\n# Algorithm\nTherefore sort the tasks in order of what is the smallest possible energy value that can exist after the task runs. This is _minimum MINUS actual_.\n\nThen loop through the sorted tasks, starting from the smallest possible ending value. [This is equivalent to playing back a movie of the optimal solution, starting from the end of the movie.]\n\nStart the loop (i.e. the end of the movie) with an optimistic hope that you might need zero energy at that time point. \n\nAt every task (as you loop backwards through the movie of optimally arranged tasks):\n* if the task needs to end with more energy than you currently have, increase the energy to match that requirement\n* add the actual energy used by the task\nThis gives the minimum amount of energy that was needed _before_ this task.\n\nOnce you have looped through all, that energy value you have been cumulating, is now the minimum needed for the entire group of tasks.\n\n\n```\nclass Solution:\n    def minimumEffort(self, tasks: List[List[int]]) -> int:\n                \n        tasks.sort(key = lambda task:task[1]-task[0])\n        \n        energy = 0\n        for task in tasks:\n            energy = max(energy, task[1]-task[0]) + task[0]\n \n        return energy\n```\n\n\n# Please LIKE if it helped you!    8-)
['Python']