Simple & Easy to understand code
Approach
Use a loop to check if the list and the sorted list are the same
find the minimum sum (minsum) and the index (mini)
remove the mini+1
set nums[mini] =
# Approach
<!-- Describe your approach to solving the problem. -->
1. Use a loop to check if the list and the sorted list are the same
2. find the minimum sum (minsum) and the index (mini)
3. remove the mini+1
4. set nums[mini] = minsum
5. increment count
6. return count


# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
O(n^2)

- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
O(1)

# Code
```python3 []
class Solution:
    def findminsum(self, nums: List[int]):
        minsum,mini = float("inf"),0
        for i in range(len(nums)-1):
            if minsum > nums[i] + nums[i+1] :
                minsum = nums[i] + nums[i+1]
                mini = i
        return minsum,mini

    def minimumPairRemoval(self, nums: List[int]) -> int:
        count = 0
        while nums != sorted(nums):
            minsum,mini = self.findminsum(nums)
            nums.pop(mini+1)
            nums[mini] = minsum
            count +=1
        return count

```
['Python3']