Python Solution beats 99%


  • 0
    B
    class Solution(object):
        def majorityElement(self, nums):
    
            ans = []
            l = len(nums)
            nums_set = set(nums)
            
            for n in nums_set:
                if nums.count(n) > l / 3:
                    ans.append(n)
                    
            return ans
    

  • 0
    J

    Doesn't that have a worst case Space complexity of O(n) if no number is repeated ?
    (num_set would basically be the same size as nums if there's no repeated number, so it will not be constant space)


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