Python3    ||  3 lines, Counter  ||   T/S: 96% / 70%
(We use nums = [ 1,3,2,4,1,4] as an example.).\n\nHere\'s the plan: \n1. We compute all the sums:  \n\n        ((1,3),(3,2),(4,1),(1,4)) -->(1,5,5,5)\n\n2. We d
(We use *nums* = [ 1,3,2,4,1,4] as an example.).\n\nHere\'s the plan: \n1. We compute all the sums:  \n```\n        ((1,3),(3,2),(4,1),(1,4)) -->(1,5,5,5)\n```\n2. We determine whether any sum occurs twice:\n ```\n        (1,5,5,5) --> Counter({5: 3, 4: 1})\n``` \n3. We return the result.\n___\nHere\'s the code:\n```\nclass Solution:\n    def findSubarrays(self, nums: List[int]) -> bool:\n\n        sums = (x+y for x,y in zip(nums[1:], nums))  # <-- 1)\n\n        counts = Counter(sums).values()              # <-- 2)\n\n        return max(counts) > 1                       # <-- 3)\n```\n_____\n\nOr (if you must):\n```\n    def findSubarrays(self, nums: List[int]) -> bool:\n        return max(Counter([x+y for x,y in zip(nums[1:], nums)]).values()) > 1\n```\n\n[https://leetcode.com/problems/find-subarrays-with-equal-sum/submissions/1287506500/](https://leetcode.com/problems/find-subarrays-with-equal-sum/submissions/1287506500/)\n\nI could be wrong, but I think that time complexity is *O*(*N*) and space complexity is *O*(*N*), in which *N* ~ `len(nums)`.
['Python', 'Python3']