Challenge link: https://www.hackerrank.com/challenges/magic-square-forming/problem

This is my day 2 of the challenge. What I'm trying to solve is the forming a magic square problem from Hackerrank. So here's the break down on how I solve the problem:

**Predefined Magic Squares:** First, I define all possible configurations of a 3x3 magic square. Each configuration is represented as a one-dimensional array consisting of nine integers. These integers range from 1 to 9, with each number appearing exactly once. This step establishes the set of target magic squares we aim to compare against.

```
const magicSquares = [
[8, 1, 6, 3, 5, 7, 4, 9, 2],
[4, 3, 8, 9, 5, 1, 2, 7, 6],
[2, 9, 4, 7, 5, 3, 6, 1, 8],
[6, 7, 2, 1, 5, 9, 8, 3, 4],
[6, 1, 8, 7, 5, 3, 2, 9, 4],
[8, 3, 4, 1, 5, 9, 6, 7, 2],
[4, 9, 2, 3, 5, 7, 8, 1, 6],
[2, 7, 6, 9, 5, 1, 4, 3, 8]
];
```

**Merging the Input Matrix:** The input function takes a 3x3 square (denoted as s). To simplify operations on this matrix, I merge it into a one-dimensional array. This flattening process facilitates easier access and comparison of the input square's elements with those of the predefined magic squares.

```
const allInOne = [...s[0], ...s[1], ...s[2]];
```

**Calculating the Transformation Cost:** For each predefined magic square, I calculate the cost of transforming the input square into this magic square. The reduce method accumulates the absolute differences between corresponding elements of the input square and a magic square. After iterating through all possible magic squares, the minimum transformation cost is determined.

```
let lesserCost = Number.MAX_SAFE_INTEGER;
magicSquares.forEach(x => {
let totalCost = x.reduce((acc, magicValue, index) => {
return acc + Math.abs(magicValue - allInOne[index]);
}, 0);
lesserCost = Math.min(totalCost, lesserCost);
});
```

Here's the final output

```
function formingMagicSquare(s) {
// will convert this one programatically
const magicSquares = [
[8, 1, 6, 3, 5, 7, 4, 9, 2],
[4, 3, 8, 9, 5, 1, 2, 7, 6],
[2, 9, 4, 7, 5, 3, 6, 1, 8],
[6, 7, 2, 1, 5, 9, 8, 3, 4],
[6, 1, 8, 7, 5, 3, 2, 9, 4],
[8, 3, 4, 1, 5, 9, 6, 7, 2],
[4, 9, 2, 3, 5, 7, 8, 1, 6],
[2, 7, 6, 9, 5, 1, 4, 3, 8]
];
const allInOne = [...s[0], ...s[1], ...s[2]]
let lesserCost = Number.MAX_SAFE_INTEGER;
magicSquares.forEach(x => {
let totalCost = x.reduce((acc, magicValue, index) => {
return acc + Math.abs(magicValue - allInOne[index]);
}, 0);
lesserCost = Math.min(totalCost, lesserCost);
})
return lesserCost;
}
```

VoilÃ ! That's it. I hope you like my solution :)

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