Detailed Python Solution
IntuitionThe goal is to replace each number in nums with the sum of its digits.ApproachCreate a helper function that calculates the digit sums then use it to re
# Intuition
The goal is to replace each number in nums with the sum of its digits.

# Approach
Create a helper function that calculates the digit sums then use it to return the minimum sum. For a more flowy solution, see the brute force

# Complexity
- Time complexity:
For the Optimized solution, $$O(n.log(k))$$
For the Brute Force, $$O(n.log(k)) + O(n)$$

- Space complexity:
For the Optimized solution, $$O(1)$$
For the Brute Force, $$O(n)$$
# Code
```python3 []
class Solution:
    def minElement(self, nums: List[int]) -> int:
        # Optimized solution
        # Step 01: We'll create a helper function that will calculate the digit sum of each 
        # number in the nums list
        def digitSum(num):
            return sum(int(digit) for digit in str(num))
        
        # Step 02: Return the minimum sum from the array
        return min(digitSum(num) for num in nums)

        # Time complexity for Optimized Solution
        # Step 01: It takes O(log(num)) to loop through the digits in the str and O(1)
                #  to calculate the sum of all digits. Its O(log(num)) because the time
                # it takes depends on the number of digits rather than the size of the number.
        # Step 02: It takes O(n) to loop through the nums array and find the minimum. n is the
        # length of nusm array
        # Combined: It takes O(n.log(k)) to find the digit sums of all numbers and return the min. n = len(nums) and k is len(num)
        # Space Complexity: O(1) for the summation of digits

        # # Brute Force:
        # # Step 01: Use mapping to convert nums to a string array
        # nums = list(map(str, nums))
        # numsSum = [] # where we'll store the sums of the digits
        # # Step 02: Loop the the new nums and add the digits in each position
        # curr_sum = 0
        # for val in nums:
        #     print(val)
        #     for x in val:
        #         print(x)
        #         curr_sum += int(x)
        #         print(curr_sum)
        #     numsSum.append(curr_sum)
        #     curr_sum = 0 #reset current sum to 0
        # # Step 03: Return the minimum element in numsSum
        # return min(numsSum)

        # Time complexity for Brute Force:
        # Step 01: O(n) to map all numbers in num to strings. n is the length of nums
        # Step 02: O(n.log(k)) to loop through each digit of elements in nums. n is the 
        # length of of nums and k is the number of digits in num
        # Step 03: O(n) to get the minimum sum n = len(nums) = len(numsSum)
        # Space complexity: O(n) to store the digit sums in numsSum array


```
['Python3']