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  2. Data Structures
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  4. 2D Array - DS
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2D Array - DS

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  • ismaelsousafcas1
     + 0 comments
    #!/bin/python3

import math
import os
import random
import re
import sys

#
# Complete the 'hourglassSum' function below.
#
# The function is expected to return an INTEGER.
# The function accepts 2D_INTEGER_ARRAY arr as parameter.
#

def hourglassSum(arr):
    arrsum = []
    def hourglass(arr,i,j):
        hg = []
        for n in range(3): 
            if j + n >= len(arr[0]):
                return []
            else:
                hg.append(arr[i][j+n])
                print(arr[i][j+n])
                
        if i + 1 >= len(arr):
            return []
        else:
            hg.append(arr[i+1][j+1])
            print(arr[i+1][j+1])
        
        if i + 2 < len(arr):
            for n in range(3): 
                hg.append(arr[i+2][j+n])
        else:
            return []
        return hg
        
    for i in range (len(arr)):
        for j in range (len(arr[0])):                    
            hg =  hourglass(arr,i,j)
            if hg != []:
                arrsum.append(sum(hg))
                
                
    return max(arrsum)
                
    
            


if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    arr = []

    for _ in range(6):
        arr.append(list(map(int, input().rstrip().split())))

    result = hourglassSum(arr)

    fptr.write(str(result) + '\n')

    fptr.close()

  • shrivastavabhis6
     + 0 comments

    Java code:

    public static int hourglassSum(List> arr) {

    int maxSum = Integer.MIN_VALUE;

for(int i=0;i<4;i++) {
        for(int j=0;j<4;j++) {
                int sum = 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);

                if(sum > maxSum) {
                        maxSum = sum;
                }
        }
    }
    return maxSum;
}

  • jasonreynaruiz
     + 0 comments

    C# SOLUTION

    Brief explanation it checks both the row and column count since there could be more than 6 in either, added guards just in case. For the maxHourglassRow and maxHourglassCol I remove the first and last position a long side with one more for indexing giving to a total of 3 hence the maxHourglassRow - 3 (The exercise has 6 numbers, so if I remove 3, I am left with 3 indexing starts with 0 and I want to start counting to sum them i.e. row[0][0], row[0][1], row[0][2] <- the 0, 1 , 2 represent a column in a 2 dimensional array). In the sum I reused a method to sum the 3 columns from the top of the hourglass and the bottom as well, any comment is welcomed. I believe there could be improvements but for now this is what I came up with.

    public static int hourglassSum(List<List<int>> arr)
    {
        int rowCount = arr.Count;
        
        if (rowCount == 0)
        {
            throw new ArgumentException(); 
        }
				
        int colCount = arr[0].Count;
				
        if (colCount == 0)
        {
            throw new ArgumentException(); 
        }
				
        List<int> finalSumArray = [];
        int maxHourglassRow = rowCount - 3;
        int maxHourglassCol = colCount - 3;
        for (var r = 0; r <= maxHourglassRow; r++) {
            for (var c = 0; c <= maxHourglassCol; c++) {
                List<int> firstRow = arr[r];
                List<int> secondRow = arr[r + 1];
                List<int> thirdRow = arr[r + 2];
                finalSumArray.Add(rowSum(firstRow, c) + secondRow[c + 1] + rowSum(thirdRow, c));
            } 
        }
        return finalSumArray.Max();
		}
    
    public static int rowSum(List<int> row, int column)
    {
        int secondColumn = column + 1;
        int thirdColumn = column + 2;
        return row[column] + row[secondColumn] + row[thirdColumn];
    }

  • i_am_amar
     + 0 comments

    My solution in JavaScript :-)

    function hourglassSum(arr) {
  let sum = -Infinity;

  for (let i = 0; i <= 3; i++) {
    for (let j = 0; j <= 3; j++) {
      let a = arr[i][j];
      let b = arr[i][j + 1];
      let c = arr[i][j + 2];

      let d = arr[i + 1][j + 1];

      let e = arr[i + 2][j];
      let f = arr[i + 2][j + 1];
      let g = arr[i + 2][j + 2];

      if (a + b + c + d + e + f + g > sum) {
        sum = a + b + c + d + e + f + g;
      }
    }
  }
  return sum;
}

  • charancuz008
     + 0 comments
    int hourglassSum(vector<vector<int>> arr) {
vector<int>sums(16);

    int t=0;
for(int i=0;i<16;i++){
   int j=i,c=i%4; 
   if(i%4==0&&i!=0)t++;
        for(int k=c;k<c+3;k++){
            sums[j]+=arr[t][k]+arr[t+2][k];
        }
        sums[i]+=arr[t+1][c+1];
}
int maxi=INT_MIN;
for(int i=0;i<16;i++)if(maxi<sums[i])maxi=sums[i];
return maxi;
}