Forming a Magic Square

We define a magic square to be an n X n matrix of distinct positive integers from 1 to n² where the sum of any row, column, or diagonal of length n is always equal to the same number: the magic constant.

You will be given a 3 X 3 matrix s of integers in the inclusive range [1,9]. We can convert any digit a to any other digit b in the range [1,9] at cost of |a - b|. Given s, convert it into a magic square at minimal cost. Print this cost on a new line.

Note: The resulting magic square must contain distinct integers in the inclusive range [1,9].

For example, we start with the following matrix s:

1
2
3

5 3 4
1 5 8
6 4 2

We can convert it to the following magic square:

1
2
3

8 3 4
1 5 9
6 7 2

This took three replacements at a cost of |5 - 8| + |8 - 9| + |4 - 7| = 7.

Function Description

Complete the formingMagicSquare function in the editor below. It should return an integer that represents the minimal total cost of converting the input square to a magic square.

formingMagicSquare has the following parameter(s):

• s: a 3 X 3 array of integers

Input Format

Each of the lines contains three space-separated integers of row s[i].

Constraints

Output Format

Print an integer denoting the minimum cost of turning matrix s into a magic square.

Sample Input 0

1
2
3

4 9 2
3 5 7
8 1 5

Sample Output 0

1

1

Explanation 0

If we change the bottom right value, s[2][2], from 5 to 6 at a cost of |6 - 5| = 1, s becomes a magic square at the minimum possible cost.

Sample Input 1

1
2
3

4 8 2
4 5 7
6 1 6

Sample Output 1

1

4

Explanation 1

Using 0-based indexing, if we make

• s[0][1]->9 at a cost of |9 - 8| = 1
• s[1][0]->3 at a cost of |3 - 4| = 1
• s[2][0]->8 at a cost of |8 - 6| = 2,

then the total cost will be 1 + 1 + 2 = 4.

Solution

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// Complete the formingMagicSquare function below.
function formingMagicSquare(s) {
    const squares = ['618753294', '816357492', '834159672', '438951276', '672159834', '276951438', '294753618', '492357816'];
    let min = 100;
    let cost = (s, squares) => {

        return [...s.map(value => value.join('')).join('')].reduce((target, item, index) => {
            target += Math.abs(+item - +squares[index])

            return target;
        }, 0)
    };

    squares.forEach((item, index) => {
        let value = cost(s, squares[index]);

        (value < min) && (min = value);
    });

    return min;
}