Description

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

Example 1:

Input: nums = [3,2,3]
Output: [3]

Example 2:

Input: nums = [1]
Output: [1]

Example 3:

Input: nums = [1,2]
Output: [1,2]

Constraints:

1 <= nums.length <= 5 * 10^4
-10^9 <= nums[i] <= 10^9

Solution

Boyer-Moore Majority Vote, find the first 2 majorities, then check if they appear more than n/3 times.

For boyer-moore majority vote, if we want to find multiple majorities, use new element to decrease all previous counters.

Time complexity: $o(n)$
Space complexity: $o(1)$

Code

class Solution:def majorityElement(self, nums: List[int]) -> List[int]:n1, c1, n2, c2 = 0, 0, 1, 0for each_num in nums:if each_num == n1:c1 += 1elif each_num == n2:c2 += 1elif c1 == 0:n1, c1 = each_num, 1elif c2 == 0:n2, c2 = each_num, 1else:c1 -= 1c2 -= 1c1, c2 = 0, 0for each_num in nums:if each_num == n1:c1 += 1elif each_num == n2:c2 += 1res = []if c1 > len(nums) / 3:res.append(n1)if c2 > len(nums) / 3:res.append(n2)return res