Leetcode 229. Majority Element II

Problem

Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.

Algorithm

When an element's frequency is not among the top two, its count is reset to zero. Otherwise, the algorithm continues tracking that element. This process retains two candidate elements whose potential frequencies may exceed ⌊n/3⌋, followed by a final validation step to confirm whether both have reached the threshold.

Code

class Solution:
    def majorityElement(self, nums: List[int]) -> List[int]:
        num1, num2 = None, None
        cnt1, cnt2 = 0, 0
        for num in nums:
            if num1 == num:
                cnt1 += 1
            elif num2 == num:
                cnt2 += 1
            elif cnt1 == 0:
                num1 = num
                cnt1 =  1
            elif cnt2 == 0:
                num2 = num
                cnt2 =  1
            else:
                cnt1 -= 1
                cnt2 -= 1

        cnt1, cnt2 = 0, 0
        for num in nums:
            if num1 == num:
                cnt1 += 1
            if num2 == num:
                cnt2 += 1
        
        ans = []
        if cnt1 > len(nums) // 3:
            ans.append(num1)
        if cnt2 > len(nums) // 3:
            ans.append(num2)
        
        return ans