Majority Element Solutions in Python

# Source : https://leetcode.com/problems/majority-element/
# Author : penpenps
# Time   : 2019-07-22

from typing import List

# Sort nums and pick the (n/2)-th number
# Time Complexity: O(nlogn)
# Space Complexity: O(n)

class Solution:
    def majorityElement(self, nums: List[int]) -> int:
        return sorted(nums)[len(nums)//2]

# Source : https://leetcode.com/problems/majority-element/
# Author : penpenps
# Time   : 2019-07-22

from typing import List

# Count occurance for each number and find out the one whose occurance more than n/2 times
# Time Complexity: O(n)
# Space Complexity: O(n)

import collections
class Solution:
    def majorityElement(self, nums: List[int]) -> int:
        m = collections.Counter(nums)
        n = len(nums)
        for k,v in m.items():
            if v >= n/2:
                return k

# Source : https://leetcode.com/problems/majority-element/
# Author : penpenps
# Time   : 2019-07-22

from typing import List

# Randomly pick one to check whether the answer
# Time Complexity: Worst: O(infinity) Average: O(n)
# Space Complexity: O(1)

import random
class Solution:
    def majorityElement(self, nums: List[int]) -> int:
        def occurance(nums, target):
            return sum(1 if x == target else 0 for x in nums)
        
        n = len(nums)
        while True:
            index = random.randint(0, n-1)
            if occurance(nums, nums[index]) >= n / 2:
                return nums[index]

# Source : https://leetcode.com/problems/majority-element/
# Author : penpenps
# Time   : 2019-07-22

from typing import List

# Recursivly divide two half parts and search most occurance number
# Time Complexity: O(nlogn)
# Space Complexity: O(logn)

class Solution:
    def majorityElement(self, nums: List[int]) -> int:
        def occurance(nums, left, right, target):
            return sum(1 if x == target else 0 for x in nums[left:right+1])
        def helper(nums, left, right):
            length = right - left + 1
            if length <= 2:
                return nums[left]
            mid = (left+right) // 2
            leftMaj = helper(nums, left, mid)
            rightMaj = helper(nums, mid+1, right)
            if leftMaj == rightMaj:
                return leftMaj
            leftMajCnt = occurance(nums, left, mid, leftMaj)
            rightMajCnt = occurance(nums, mid+1, right, rightMaj)
            
            return leftMaj if leftMajCnt > rightMajCnt else rightMaj
        return helper(nums, 0, len(nums)-1)

# Source : https://leetcode.com/problems/majority-element/
# Author : penpenps
# Time   : 2019-07-22

from typing import List

# Boyer-Moore Voting Algorithm, count for one number, if not equal to current number, count minus 1 until ot 0 then try another number
# Time Complexity: O(n)
# Space Complexity: O(1)

class Solution:
    def majorityElement(self, nums: List[int]) -> int:
        ans = nums[0]
        count = 0
        for n in nums:
            if count == 0:
                ans = n
                count += 1
            elif n == ans:
                count += 1
            else:
                count -= 1
        return ans