Binary Search | Python | Easy to understand
Idea:\nThe pretty straightforward solution, sort potions and for each spell find an index of minimal potions that will make success and we know that everything 
Idea:\nThe pretty straightforward solution, sort potions and for each spell find an index of minimal potions that will make success and we know that everything greater will make a success too. \n\n\n```\nclass Solution:\n    def successfulPairs(self, spells: List[int], potions: List[int], success: int) -> List[int]:\n        potions.sort()\n        ans = []\n        for i in spells:\n            cur = success/i\n            ans.append(len(potions)-bisect.bisect_left(potions, cur))\n        return ans\n```
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