The most inefficient but easy to read solution
IntuitionEasier to read, easier to write, don't care about efficiency if it worksApproach
Let's find 3 or less candidates
Let's calculate the solution for all c
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Easier to read, easier to write, don't care about efficiency if it works

# Approach
<!-- Describe your approach to solving the problem. -->
1. Let's find 3 or less candidates
2. Let's calculate the solution for all candidates and find the smallest one.

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

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

# Code
```python3 []
import numpy as np
from typing import List

class Solution:
    def minimumCost(self, nums: List[int]) -> int:
        
        median = int(np.median(nums))

        candidates = set() # unique candidates palindromic numbers for values of all indices. Maximum 3 values
        
        if str(median) == str(median)[-1::-1]:
            candidates.add(median)
        
        m = median # right side candidate from median
        while True:
            if str(m+1) == str(m+1)[-1::-1]:
                candidates.add(m+1)
                break
            else:
                m += 1

        m = median # left side candidate from median
        while True:
            if str(m-1) == str(m-1)[-1::-1]:
                candidates.add(m-1)
                break
            else:
                m -= 1

            
        d = dict()
        for i in candidates: # any unique candidate of palindromic numbers might be the best
            d[i] = 0
            for n in nums:
                d[i] += abs(n-i)

        return min([v for k,v in d.items()])
```
['Python3']