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  4. 2D Array - DS
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2D Array - DS

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  • evanbechtol
     + 0 comments

    Here is my soln in JavaScript. I'm sure that there is a more efficient way to do this, but this was quick and easy.

    function hourglassSum(arr) {
    let maxSum = -9999;
    let currSum = 0;
    
    for (let row = 0; row < arr.length - 2; row++) {
        for (let col = 0; col < arr.length - 2; col++) {
            currSum = getHourglassSum(arr, row, col);
            
            if (currSum > maxSum) {
                maxSum = currSum;
            }
        }
    }
    
    return maxSum;
}

function getHourglassSum(arr, row, col) {
    return arr[row][col] 
            + arr[row][col + 1] 
            + arr[row][col + 2] 
            + arr[row + 1][col + 1] 
            + arr[row + 2][col] 
            + arr[row + 2][col + 1] 
            + arr[row + 2][col + 2];
}

  • jmcparland
     + 0 comments

    Haskell

    module Main where

import qualified Data.Array  as A

type Hourglass = A.Array (Int, Int) Int

inputToHourglass :: String -> Hourglass
inputToHourglass s = A.listArray ((0, 0), (5, 5)) $ map read $ words s

hsum :: Hourglass -> Int -> Int -> Int
hsum h x y = sum [h A.! (x', y') | x' <- [x .. x + 2], y' <- [y .. y + 2], (x', y') /= (x + 1, y), (x', y') /= (x + 1, y + 2)]

maxHourglassSum :: Hourglass -> Int
maxHourglassSum h = maximum [hsum h x y | x <- [0 .. 3], y <- [0 .. 3]]

main :: IO ()
main = interact $ show . maxHourglassSum . inputToHourglass

  • ivan_romero_dev1
     + 0 comments
    public static int hourglassSum(List<List<Integer>> arr) {
        int maxSum = Integer.MIN_VALUE;
        for (int i = 0; i < 4; i++) {
            for (int j = 0; j < 4; j++) {
                int result = sumArray(arr, i, j);
                maxSum = Math.max(maxSum, result);
            }
        }
        return maxSum;
    }

    private static int sumArray(List<List<Integer>> arr, int i, int j) {
        return arr.get(i).get(j) + arr.get(i).get(j + 1) + arr.get(i).get(j + 2)
                + arr.get(i + 1).get(j + 1) +
                arr.get(i + 2).get(j) + arr.get(i + 2).get(j + 1) + arr.get(i + 2).get(j + 2);
    }

  • lonewolf63729900
     + 0 comments

    python: def hourglassSum(arr): l=[] for i in range(1,5): for j in range(1,5): l.append(a[i-1][j-1]+a[i-1][j]+a[i-1][j+1]+ a[i][j]+ a[i+1][j-1]+a[i+1][j]+a[i+1][j+1]) return(max(l))

  • alban_tyrex
     + 0 comments

    Here is my c++ solution you can watch video explanation here : https://www.youtube.com/watch?v=aJv6KLHL3Qk

    int hourglassSum(vector<vector<int>> arr) {
    int result = -63;
    for(int r = 1; r <= 4; r++){
        for(int c = 1; c <= 4; c++){
          int s = arr[r][c] + arr[r-1][c] + arr[r-1][c-1] + arr[r-1][c+1];
          s += arr[r+1][c] + arr[r+1][c-1] + arr[r+1][c+1];
          result = max(result,s);
        }
    }
    
    return result;
}