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  1. Practice
  2. Data Structures
  3. Arrays
  4. 2D Array - DS
  5. Discussions

2D Array - DS

  • revathymanulal + 4 comments

    I used a similar approach but dint keep on storing all of the values of hour glass instead just kept on swaping the values incase the value of new hour glass was greater than the previous one.

    import sys

a = []
for arr_i in xrange(6):
    arr_temp = map(int,raw_input().strip().split(' '))
    a.append(arr_temp)
    
max_sum =-63
    
for i in range(4):
    for j in range(4):
        check_sum = a[i][j]+a[i][j+1]+a[i][j+2]+a[i+1][j+1]+a[i+2][j]+a[i+2][j+1]+a[i+2][j+2]
                    
        if check_sum > max_sum:
            max_sum = check_sum
            
print max_sum
    • tppiotrowski + 0 comments
      [deleted]
    • nr_selcuk + 4 comments

      why the range(4)? I think it should be the range(len(a)-2) in order to satisfy any size...

      • DarkC_oder + 0 comments
        [deleted]
      • sukhraj5 + 0 comments

        The problem specifies the matrix is always 6x6

      • sandeepini + 0 comments

        Because our array is of 6x6. when we run arr[i][j+2] In j+2=5(When loop run second last time (3+2=5)) which valid . This is the reason.

      • sandeepini + 0 comments

        Because our array is of 6x6. when we run arr[i][j+2] In j+2=5(When loop run second last time (3+2=5)) which valid . This is the reason.

    • pathansuhana969 + 0 comments

      why it is not working in python3? any one pls explain.

    • ankitiitism + 0 comments

      only 4/9 test cases passed , can any one identify bug.

      include

      using namespace std; int main() { int arr[6][6] ; int t=6; while(t--) { for(int i=0; i<6; i++) cin>>arr[6-t][i]; } vectorv; int n=6; for(int i= 0; i<4;i++) for(int j= 0; j<4; j++) { int sum=0; 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]); v.push_back(sum); } int max = *max_element(v.begin(), v.end()); cout<