class Solution:
    '''
    Time Complexity: O(n^2)
    Space Complexity: O(n)
    where, n is the length of nums
    NOTE: You did not understand what minimum sum was in the question for which, 
    you can't be blamed cause there was nobody to ask when solo leet-coding, 
    and you were trying to achieve better time complexity without fully understanding the problem,
    which killed a lot of time!
    '''
    def minimumPairRemoval(self, nums: List[int]) -> int:
        def is_non_decreasing(arr):
            for i in range(1, len(arr)):
                if arr[i] < arr[i - 1]:
                    return False
            return True
        
        operations = 0
        while not is_non_decreasing(nums):
            min_sum = float('inf')
            idx = 0
            for i in range(len(nums) - 1):
                s = nums[i] + nums[i + 1]
                if s < min_sum:
                    min_sum = s
                    idx = i
            nums[idx] = nums[idx] + nums[idx + 1]
            nums.pop(idx+1)
            operations += 1
        
        return operations