Forming a Magic Square (Day 2 of 365)

xdmarttt — Sat, 09 Mar 2024 12:18:16 +0000

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;
}

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

1 of 365 Coding challenge

xdmarttt — Fri, 08 Mar 2024 07:09:22 +0000

Hey! So, here's my first article, diving into this whole 365-day coding challenge thing. I'm just getting started, so go easy on me. I'm not a pro at English, so thanks for putting up with that too.

I started early in the morning, all gung-ho, but I kinda dove into the coding without really getting what the problem was asking. Ended up with a wonky solution and took me a good two hours to realize where I messed up. Not the greatest start, but I learned something.

Now, my first challenge was 'Climbing the Leaderboard,' and you can find it here: https://www.hackerrank.com/challenges/climbing-the-leaderboard/problem.

So, let's delve into my solution:

The first part involves obtaining the unique ranks:

const uniqueRanked = [...new Set(ranked)];

Additionally, I created another array that serves as a placeholder for the indices:

const rankedSet = Array.from({ length: uniqueRanked.length });

Moving on, I iterated through the player scores:

player.forEach(score => {
    if (score < uniqueRanked[uniqueRanked.length - 1]) {
        finalRank.push(rankedSet[rankedSet.length - 1] + 1);
    } else if (score > uniqueRanked[0]) {
        finalRank.push(1);
    } else {
        const midPoint = Math.floor((rankedSet.length + 1) / 2) - 1;
        const start = score <= uniqueRanked[midPoint] ? midPoint : 0;

        for (let i = start; i <= rankedSet[rankedSet.length - 1]; i++) {
            if (score >= uniqueRanked[i]) {
                finalRank.push(rankedSet[i]);
                break;
            }
        }
    }
});

As indicated in the code above, I excluded the first and last indices to streamline the iteration process. The nested iteration involves selecting a middle point that divides the array into two. Although not strictly necessary, I implemented this approach to optimize performance in specific cases on Hackerrank where speed is crucial. The goal is to swiftly determine the correct placement of the rank.

I know this may not be the best solution, but I just want to share my journey here :) . I'm open to any new suggestions and constructive criticism to improve my solution. Hope you have a good day! Thanks, everyone!

DEV Community: xdmarttt

Forming a Magic Square (Day 2 of 365)

1 of 365 Coding challenge