Python 3 || 7 lines, Counter || T/S: 98% / 29%
Here\'s the plan:\n\n\n1. We use Counter to determine the counts of the characters as the list vals.\n\n1. We construct a function minimumDeletions that, given 
Here\'s the plan:\n\n\n1. We use `Counter` to determine the counts of the characters as the list `vals`.\n\n1. We construct a function `minimumDeletions` that, given a range [*num... num+k*], determines the minimum count of deletions to ensure that  character counts remaining after these deletions are in that interval.\n\n1. We evaluate the function for each element in `vals` and return the the minimum result. \n\n\n```\nclass Solution:\n    def minimumDeletions(self, word: str, k: int) -> int:\n\n        def deletions(num: int, res = 0)-> int:\n\n            for v in vals:\n                if v < num: res+= v\n                if v > num+k: res+= v - num- k\n\n            return res     \n        \n\n        vals = Counter(word).values()\n        \n        return min(map (deletions,vals))\n```\n[https://leetcode.com/problems/minimum-deletions-to-make-string-k-special/submissions/1206582720/](https://leetcode.com/problems/find-trending-hashtags/submissions/1206775504/)\n\n\n\nI could be wrong, but I think that time complexity is *O*(*N*) and space complexity is *O*(1), in which *N* ~ `len(word)`. (Recall `len(vals.keys())` <= 26)
['Python3']