Minimum Index of a Valid Split Python Solution

lcpy gen -n 2780   # by problem number
lcpy gen -s minimum_index_of_a_valid_split   # by problem name

Input: nums = [1,2,2,2]
Output: 2
Explanation: We can split the array at index 2 to obtain arrays [1,2,2] and [2].
In array [1,2,2], element 2 is dominant since it occurs twice in the array and 2 * 2 > 3.
In array [2], element 2 is dominant since it occurs once in the array and 1 * 2 > 1.
Both [1,2,2] and [2] have the same dominant element as nums, so this is a valid split.
It can be shown that index 2 is the minimum index of a valid split.

Input: nums = [2,1,3,1,1,1,7,1,2,1]
Output: 4
Explanation: We can split the array at index 4 to obtain arrays [2,1,3,1,1] and [1,7,1,2,1].
In array [2,1,3,1,1], element 1 is dominant since it occurs thrice in the array and 3 * 2 > 5.
In array [1,7,1,2,1], element 1 is dominant since it occurs thrice in the array and 3 * 2 > 5.
Both [2,1,3,1,1] and [1,7,1,2,1] have the same dominant element as nums, so this is a valid split.
It can be shown that index 4 is the minimum index of a valid split.

Input: nums = [3,3,3,3,7,2,2]
Output: -1
Explanation: It can be shown that there is no valid split.

class Solution:
    # Time: O(n)
    # Space: O(1)
    def minimum_index(self, nums: list[int]) -> int:
        dominant = 0
        count = 0
        for num in nums:
            if count == 0:
                dominant = num
                count = 1
            elif num == dominant:
                count += 1
            else:
                count -= 1

        total = 0
        for num in nums:
            if num == dominant:
                total += 1

        left = 0
        for i, num in enumerate(nums):
            if num == dominant:
                left += 1
            right = total - left
            if left * 2 > i + 1 and right * 2 > len(nums) - i - 1:
                return i
        return -1