This is my solution (python):

!/bin/python3

import math import os import random import re import sys import itertools

#

Complete the 'formingMagicSquare' function below.

#

The function is expected to return an INTEGER.

The function accepts 2D_INTEGER_ARRAY s as parameter.

#

def formingMagicSquare(arr): all_val = [] for row in arr: for col in row: all_val.append(col)

const = list(range(1, 10))
all_comb_r = list(itertools.permutations(const, 9))

all_valid_comb = []

for comb in all_comb_r:
    c = list(comb)
    temp = []
    curr = []
    curr_col = []
    curr_d = []
    for val in c:
        temp.append(val)
        if len(temp) == 3:
            curr.append(temp)
            temp = []

    temp = []
    for i in range(len(curr)):
        for j in range(len(curr[i])):
            temp.append(curr[j][i])
            if len(temp) == 3:
                curr_col.append(temp)
                temp = []
    temp = []
    for i in range(len(curr)):
        temp.append(curr[i][i])
        if len(temp) == 3:
            curr_d.append(temp)
            temp = []

    s = 0
    e = 2
    for i in range(len(curr)):
        temp.append(curr[s][e])
        if len(temp) == 3:
            curr_d.append(temp)
            temp = []
        s += 1
        e -= 1

    sum_curr = [sum(i) for i in curr]
    sum_curr_col = [sum(i) for i in curr_col]
    sum_curr_d = [sum(i) for i in curr_d]

    all_val = []
    for num in range(3):
        all_val.append(sum_curr[num])
        all_val.append(sum_curr_col[num])
        if num != 2:
            all_val.append(sum_curr_d[num])

    if all(i == all_val[0] for i in all_val):
        all_valid_comb.append(curr)

all_ans = []
for i in range(len(all_valid_comb)):
    curr_ans = 0
    for j in range(len(all_valid_comb[i])):
        for a in range(len(all_valid_comb[i][j])):
            if all_valid_comb[i][j][a] != arr[j][a]:
                curr_ans += abs(all_valid_comb[i][j][a] - arr[j][a])
    all_ans.append(curr_ans)

return min(all_ans)

if name == 'main': fptr = open(os.environ['OUTPUT_PATH'], 'w')

s = []

for _ in range(3):
    s.append(list(map(int, input().rstrip().split())))

result = formingMagicSquare(s)

fptr.write(str(result) + '\n')

fptr.close()

I do like this. 1st. Unroll the input matrix to long list. (imagine the list 9 items was rolled in clockwise direction to create a 3x3 matrix),

2nd. Create the unrolled List of the Magic Square , since magic square can rotate or flip, so the List has some repeat pattern, and I have to create another reverse List.

``List<int> rightRotateList 
        = new List<int>{8,3,4,9,2,7,6,1,8,3,4,9,2,7,6};
    List<int> leftRotateList
        = new List<int>(rightRotateList);
    leftRotateList.Reverse();

Notice I have exclude the center number of magic square, number 5, since it's unchanged in every version of magic square 3x3.

3rd. Calculate the cost and compare. Here is the full code in C#

   ````
 public static int formingMagicSquare(List<List<int>> s)

{
    List<int> costList = new List<int>();

    List<int> sToList = s[0];
    sToList.Add(s[1][2]);
    sToList.Add(s[2][2]);
    sToList.Add(s[2][1]);
    sToList.Add(s[2][0]);
    sToList.Add(s[1][0]);

    List<int> rightRotateList 
        = new List<int>{8,3,4,9,2,7,6,1,8,3,4,9,2,7,6};
    List<int> leftRotateList
        = new List<int>(rightRotateList);
    leftRotateList.Reverse();

    for (int i = 0 ; i < 4 ; i++)
    {
        var costRight = 0 ;
        var costLeft = 0 ;
        for (int j = 0 ; j < sToList.Count ; j++)
        {
            costRight += Math.Abs(sToList[j] - rightRotateList[i*2 + j]);
            costLeft += Math.Abs(sToList[j] - leftRotateList[i*2 + j]);
        }
        var centerMatrixDiff = Math.Abs(s[1][1] - 5);
        costList.Add(costRight + centerMatrixDiff);
        costList.Add(costLeft + centerMatrixDiff);

    }

    return costList.Min();
    // return costList;

}

` 4th. It's easier to explain in image, but I don't know how to upload here. So I uploaded to this image hosting service https://ibb.co/sjwBpwZ

JS/Javascript solution:-

function formingMagicSquare(s) {
    // Flatten the input 2D array
    const flatInput = s.flat();

    // Define all possible 3x3 magic squares
    const magicSquares = [
        [8, 1, 6, 3, 5, 7, 4, 9, 2],
        [6, 1, 8, 7, 5, 3, 2, 9, 4],
        [4, 9, 2, 3, 5, 7, 8, 1, 6],
        [2, 9, 4, 7, 5, 3, 6, 1, 8],
        [8, 3, 4, 1, 5, 9, 6, 7, 2],
        [4, 3, 8, 9, 5, 1, 2, 7, 6],
        [6, 7, 2, 1, 5, 9, 8, 3, 4],
        [2, 7, 6, 9, 5, 1, 4, 3, 8]
    ];

    // Calculate the minimum cost to convert to any magic square
    let minCost = Infinity;

    for (const magic of magicSquares) {
        let cost = 0;
        for (let i = 0; i < 9; i++) {
            cost += Math.abs(flatInput[i] - magic[i]);
        }
        minCost = Math.min(minCost, cost);
    }

    return minCost;
}