# def majorityElement(nums):
#     """
#     :type nums: List[int]
#     :rtype: int
#     """
#     majority = {}
#     # Start initial position
#     for i in range(len(nums)):
#         k = 1
#         if nums[i] in majority:
#             # print(f'found {nums[i]} already')
#             continue
#         # Count the duplicates
#         for j in range(i+1, len(nums)):
#             if nums[i] == nums[j]:
#                 k += 1
#         majority[nums[i]] = k
#         print(majority)
#     # print(max(majority.values()))
#     # print(max(majority.items(), key=lambda x: x[1])[0])
#     return max(majority.items(), key=lambda x: x[1])[0]
# majorityElement([2,2,1,1,1,2,2])

# Faster runtime and less memory
def majorityElement(nums):
    majority = {}  # Create an empty dictionary to store the counts of each number
    for num in nums:  # Iterate over each element in the input list
        if num in majority:  # Check if the number is already in the dictionary
            majority[num] += 1  # If it is, increment its count by 1
        else:
            majority[num] = 1  # If it is not, add it to the dictionary with a count of 1
        if majority[num] > len(nums) / 2:  # Check if the count of the number exceeds half the length of the list
            return num  # If it does, return the number as the majority element
    return None  # If no majority element is found, return None

result = majorityElement([2, 2, 1, 1, 1, 2, 2])  # Call the function with the input list [2, 2, 1, 1, 1, 2, 2]
print(result)  # Print the result