efficient python solution

  • 0
        def maximumProduct(self, nums):
            :type nums: List[int]
            :rtype: int
            from operator import mul
            def product(arr):
                return reduce(mul, arr)
            max_nums, min_nums = heapq.nlargest(3, nums), heapq.nsmallest(2, nums)
            return max(product(max_nums), product(max_nums[:1] + min_nums))

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