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  1. Prepare
  2. Data Structures
  3. Arrays
  4. 2D Array - DS
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2D Array - DS

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  • dastagirwajahat1
     + 0 comments
    def hourglassSum(arr):
    # Write your code here
    sol=float("-inf")
    for i in range(len(arr)-2):
        for j in range(len(arr)-2):
            top=arr[i][j]+arr[i][j+1]+arr[i][j+2]
            middle=arr[i+1][j+1]
            bottom=arr[i+2][j]+arr[i+2][j+1]+arr[i+2][j+2]
            current_sol=top+middle+bottom
            if current_sol>sol:
                sol=current_sol
    return sol

  • aliabbas55506
     + 0 comments

    This is my CPP Solution what i did was traverse the 2D array in an hourglass form using 2 loops. As the last 2 indices of both i and j can't make an hourglass shape, so we run the loop till the first 4. int hourglassSum(vector> arr) { int maxSum=INT_MIN; int sum; for(int i=0;i<4;i++){ for(int j=0 ; j<4; j++){ sum = 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]); if(sum>maxSum) maxSum=sum; } } return maxSum; }

  • phanvantienpvt01
     + 0 comments

    My python sollution`

    def hourglassSum(a):
    ma = -float('inf')
    for i in range(1,5):
        for j in range(1,5):
            top = bottom = 0
            for dy in [-1,0,1] :
                top += a[i-1][j+dy]
                bottom += a[i+1][j+dy]
            ma = max(ma,top + a[i][j] + bottom)
    return ma

  • victor_escoto
     + 0 comments

    A solution in Python that respects the constraints proposed by the challenge (and focuses on readability):

    """
Constraints:
    -9 <= arr[i][j] <= 9
    0 <= i,j <= 5
"""

def get_hour_glass_sum(arr: List[int], x: int, y: int) -> int:
    first_line = sum([arr[x][y + i] for i in range(3)])
    second_line = arr[x + 1][y + 1]
    third_line = sum([arr[x + 2][y + i] for i in range(3)])
    
    return first_line + second_line + third_line

def hourglassSum(arr):
    greater_sum = None
    
    for x in range(4):
        for y in range(4):
            current_sum = get_hour_glass_sum(arr, x, y)
            
            if greater_sum is None:
                greater_sum = current_sum
                continue
            
            greater_sum = max(current_sum, greater_sum)
            
    return greater_sum

  • nugrahanggara016
     + 0 comments

    this is my solution : Image

    **My Code In Java: **

    int temporary = Integer.MIN_VALUE;
        for( int i = 0; i < arr.size() - 2; i++){
            for( int j = 0 ; j < arr.size() - 2; j++){
                int a = arr.get(i).get(j);
                int b = arr.get(i).get(j+1);
                int c = arr.get(i).get(j+2);
                int d = arr.get(i+1).get(j+1);
                int e = arr.get(i+2).get(j);
                int f = arr.get(i+2).get(j+1);
                int g = arr.get(i+2).get(j+2);
                
                int result = a + b + c + d + e + f + g;
                if (result > temporary){
                    temporary = result;
                }
            }
        }
        
        return temporary;