169. Majority Element

Given an array nums of size n, return the majority element.

The majority element is the element that appears more than ⌊n / 2⌋ times. You may assume that the majority element always exists in the array.

Example 1:

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

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

Constraints:

n == nums.length
1 <= n <= 5 * 104
-109 <= nums[i] <= 109

approach 1

class Solution(object):
    def majorityElement(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        mid = len(nums)/2
        Dictionary = {}
        for num in nums:
            if num in Dictionary.keys():
                Dictionary[num] = Dictionary[num] + 1
            else:
                Dictionary[num] = 1
            print(Dictionary[num])
            if Dictionary[num] > mid:
                return num

we use a data structure called a HashMap to keep track of the count of each element in the array. As we iterate through the array, we update the count for each element in the HashMap. If the count of any element becomes greater than half of the array length, we return that element as the majority element.

Here’s a step-by-step explanation:

1,Initialize an empty HashMap called Dictionary.
2,Initialize a variable called mid to save half of the array length (len(nums) // 2).
3,Iterate through each element num in the array.
4,If num is already present in Dictionary, count equals count plus one
increment its count; otherwise, add num to Dictionary with a count of 1.
5,Check if the count of num is greater than mid.
If the condition is met, return num as the majority element.
This approach ensures that we keep track of the count of each element efficiently, and the final majority element is the one that appears more than half of the time in the array.

approach 2

https://gregable.com/2013/10/majority-vote-algorithm-find-majority.html

tally

count equals count plus one