2019-06-13

Day 11: 2D Arrays

Objective

Today, we’re building on our knowledge of Arrays by adding another dimension. Check out the Tutorial tab for learning materials and an instructional video!

Context

Given a 6 X 6 2D Array, A:

1 1 1 0 0 0
0 1 0 0 0 0
1 1 1 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0

We define an hourglass in A to be a subset of values with indices falling in this pattern in A‘s graphical representation:

1
2
3

a b c
  d
e f g

There are 16 hourglasses in A, and an hourglass sum is the sum of an hourglass’ values.

Task

Calculate the hourglass sum for every hourglass in A, then print the maximum hourglass sum.

Input Format

There are 6 lines of input, where each line contains 6 space-separated integers describing 2D Array A; every value in A will be in the inclusive range of -9 to 9.

Constraints

-9 <= A[i][j] <= 9
0 <= i,j <= 5

Output Format

Print the largest (maximum) hourglass sum found in A.

Sample Input

1 1 1 0 0 0
0 1 0 0 0 0
1 1 1 0 0 0
0 0 2 4 4 0
0 0 0 2 0 0
0 0 1 2 4 0

Sample Output

Explanation

A contains the following hourglasses:

1 1 1   1 1 0   1 0 0   0 0 0
  1       0       0       0
1 1 1   1 1 0   1 0 0   0 0 0

0 1 0   1 0 0   0 0 0   0 0 0
  1       1       0       0
0 0 2   0 2 4   2 4 4   4 4 0

1 1 1   1 1 0   1 0 0   0 0 0
  0       2       4       4
0 0 0   0 0 2   0 2 0   2 0 0

0 0 2   0 2 4   2 4 4   4 4 0
  0       0       2       0
0 0 1   0 1 2   1 2 4   2 4 0

The hourglass with the maximum sum (19) is:

1
2
3

2 4 4
  2
1 2 4

Solutions

Solution 1

function main() {
    let arr = Array(6);

    for (let i = 0; i < 6; i++) {
        arr[i] = readLine().split(' ').map(arrTemp => parseInt(arrTemp, 10));
    }

    let result = arr.reduce((target, rows, index) => {
        rows.reduce((subTarget, item, subIndex) => {
            (
                index < arr.length - 2 && subIndex < arr.length - 2
            ) &&
                target.push(
                    arr[index][subIndex] + arr[index][subIndex + 1] + arr[index][subIndex + 2] + arr[index + 1][subIndex + 1] + arr[index + 2][subIndex] + arr[index + 2][subIndex + 1] + arr[index + 2][subIndex + 2]
                );
        }, []);

        return target;
    }, []);

    console.log(Math.max.apply(null, result));
}

Solution 2

function main() {
     let arr = Array(6);
     let maxSum;
 
     for (let i = 0; i < 6; i++) {
         arr[i] = readLine().split(' ').map(arrTemp => parseInt(arrTemp, 10));
     }
 
     const findMaxSum = (n, m) => {
         let sum = 0;
         for (let i = 0; i < 3; i++) {
             for (let j = 0; j < 3; j++) {
                 sum += !(i === 1 && (j === 0 || j === 2)) && arr[n + i][m + j];
             }
         }
         return sum;
     }
 
     const search = () => {
         let sum;
         for (let i = 0; i < 4; i++) {
             for (let j = 0; j < 4; j++) {
                 sum = findMaxSum(i, j);
                 (i === 0 && j === 0) && (maxSum = sum);
                 maxSum < sum && (maxSum = sum);
             }
         }
     }
 
     search();
     console.log(maxSum);
}

Solution 2

function main() {
    let arr = Array(6);

    for (let i = 0; i < 6; i++) {
        arr[i] = readLine().split(' ').map(arrTemp => parseInt(arrTemp, 10));
    }

    const temp = []
    for (let i = 0; i < 4; i++) {
        for (let j = 0; j < 4; j++) {
            temp.push(arr[i][j] + arr[i][j + 1] + arr[i][j + 2] + arr[i + 1][j + 1] + arr[i + 2][j] + arr[i + 2][j + 1] + arr[i + 2][j + 2])
        }
    }
    
    let max = temp.reduce(function (previous, current) {
        return previous > current ? previous : current;
    });
    
    console.log(max);
}

# 30Days of Code, Algorithm, ES6, HackerRank, JavaScript

Day 11: 2D Arrays

Objective

Context

Task

Input Format

Constraints

Output Format

Sample Input

Sample Output

Explanation

Solutions

Solution 1

Solution 2

Solution 2

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