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

  • nancy345 + 4 comments

    how to do this task in C???

    • mdumlupinar + 0 comments

      sorry, i am not familiar with c language.

    • azheruddin617mr1 + 8 comments
      int main()
{
    int matrix[6][6];
    for(int i = 0;i < 6;i++) {
        for(int j = 0;j < 6;j++) {
             scanf("%d",&matrix[i][j]);
        }
    }
    int maxsum = -1000,jj = 0;
    for(int i = 0;i < 4;i++) {
        int sum = 0;
        for(int j = jj;j < jj+3;j++) {
            sum += matrix[i][j];
            if(j==jj) sum += matrix[i+1][jj+1];
            sum += matrix[i+2][j];
        }
        jj = (jj < 3) ? jj+1 : 0;
        if(sum > maxsum) maxsum = sum;
        if(jj != 0) i--;
    }
    printf("%d",maxsum);
    return 0;
}
      • dattagates + 2 comments

        why have u assigned maxsum=-1000 ? can u elobarate

        • azheruddin617mr1 + 0 comments

          the inputs also given in negitive....

        • flash793 + 0 comments

          The constraints given say that any element in the array will be at least -9 and be at most 9. The smallest hourglass sum will be -9*7 = -63. So in this case, you just need to initialize max sum to be any value <= -63.

      • meerakrish + 0 comments

        why did you use j=jj

      • Sarath_95 + 2 comments

        what the below statement does in the code?

        if(jj != 0) i--;

        • franklinvidal + 0 comments
          jj = (jj < 3) ? jj+1 : 0; // set 0 if jj(collumn pointer) set in one position which allows calculate an hourglass
if(sum > maxsum) maxsum = sum;
if(jj != 0) i--; // if jj is in a position where a hourglass can be calculated, the line pointer(i) continue in the same line
        • ragulravi1999 + 0 comments

          To run loop for same value of i with different values for j untill j maximum limit is reached

      • bhavgothadiya + 0 comments

        why not take -1

      • pa1v_kar + 0 comments
        [deleted]
      • pa1v_kar + 2 comments

        could anyone identify the mistake

        int main(){
    int arr[6][6];
    for(int arr_i = 0; arr_i < 6; arr_i++){
       for(int arr_j = 0; arr_j < 6; arr_j++){
          
          scanf("%d",&arr[arr_i][arr_j]);
       }
    }
    int sum[30];
    int x=0;
    int arr_i,arr_j;
    int largest;
    
    for(arr_i=0;arr_i<3;arr_i++)
    {
        sum[x]=0;
        int count=0;
        for(arr_j=arr_i;count<3;arr_j++)
        {
            if(count==0)
            {
                  sum[x]+=arr[arr_i+1][arr_j+1];
            }
            sum[x]+=arr[arr_i][arr_j];
            sum[x]+=arr[arr_i+2][arr_j];
            
            ++count;
        }
          
        ++x;
     }
     largest=sum[0];
    for(int i=1;i<=x;i++)
    {   
        if(largest<sum[i])
            largest=sum[i];
    }
    printf("%d",largest);
    return 0;
}
        • Nidheesh_P + 1 comment

          in for loop try i<4 instead of i<3

          • manny_capacidade + 1 comment

            Whats the reason behind i<4..Not able to get the reason.Can you explain please

            • fhasson91 + 0 comments

              there are 4 hourglasses vertically and 4 hourglasses horizontally

        • uniquechhetrii + 2 comments
          for(arr_i=0;arr_i<3;arr_i++)
    {
        sum[x]=0;
        int count=0;
        for(arr_j=arr_i;count<3;arr_j++)
        {
            if(count==0)
            {
                  sum[x]+=arr[arr_i+1][arr_j+1];
            }
            sum[x]+=arr[arr_i][arr_j];
            sum[x]+=arr[arr_i+2][arr_j];
            
            ++count;
        }
          
        ++x;

          I didnt understand this part. Can somebody help me out in this?

          • chadams + 0 comments

            His algorithm is getting the middle element of the hourglass on the first pass through and then incrementing the top and bottom rows of the hourglass. He increments count to track the amount of loops needed to read a hourglass and increments x to correspond to the position in the hourglass; 0,1,2 respectively.

          • vigneshiyer66661 + 0 comments
            [deleted]
      • chernandez2 + 5 comments

        This code passes all the test cases

        #include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
using namespace std;

int main(){
    vector< vector<int> > arr(6,vector<int>(6));
    for(int arr_i = 0;arr_i < 6;arr_i++){
       for(int arr_j = 0;arr_j < 6;arr_j++){
          cin >> arr[arr_i][arr_j];
       }
    }

    vector<int> res;
    res.reserve(18);

    for(unsigned int j=0; j<4;++j){
        for(unsigned int i=0; i<4;++i){
            res.push_back(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]);
        }
    }
    cout<<*max_element(res.begin(),res.end())<<endl;

    return 0;
}
        • 160302032CSE + 2 comments

          can anyone explain this part

          res.reserve(18); for(unsigned int j=0; j<4;++j){ for(unsigned int i=0; i<4;++i){ res.push_back(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]); } } cout<<*max_element(res.begin(),res.end())<

          • revathymanulal + 0 comments
            • res.reserve(18) is like declaring an array of size 18 though am not sure why size 18 was choosen ,I think 16 would have done the job.

            -the two for loops with contraints <4 is for the purpose of forming the hour glass (had it been <6 it would have resulted in indexing issues)

            -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] these are the indexes which form an hour glass, the sum of every hour glass is calculated and push to the res

            -cout<<*max_element(res.begin(),res.end()) the maximum value in res is choosen as the output

          • revathymanulal + 1 comment

            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
        • typicallucas + 1 comment

          Hi, you're including a bunch of extraneous libraries. You are just using vector, iostream, and algorithm.

          The rest could be deleted :) Cheers

          • banerjeeunmesha + 1 comment

            I didnot include any extraneous libraries, I did my code in c++,kindly help me...thank you for your time and help my code:

            include

            using namespace std; void sum(int[][6]); int main() { int a[6][6]; int i,j; cout<<"enter val\n"; for(i=0;i<6;i++) { for(j=0;j<6;j++) { cin>>a[i][j]; } } sum(a); return 0; } void sum(int a[][6]) { int l=0,m=0; int i,j,max; int sum[4][4]; for(i=0,j=0;i<=3,j<=3;j++) { if(j==3) { sum[l][m]=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]; i++; j=-1; } else { sum[l][m]=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]; } m++; if(m==4) { l++; m=0; } } max=sum[0][0]; for(i=0;i<4;i++) { for(j=0;j<4;j++) { if(sum[i][j]>=max) { max=sum[i][j]; } } } cout<<"max"<

            • typicallucas + 1 comment

              hi, i think you're more likely to get help if your code is more readable. Wrap your code with three back-ticks and that should help improve the

              • banerjeeunmesha + 1 comment

                include

                using namespace std; void sum(int[][6]); int main() { int a[6][6]; int i,j; cout<<"enter val\n"; for(i=0;i<6;i++) { for(j=0;j<6;j++) { cin>>a[i][j]; } } sum(a); return 0; } void sum(int a[][6]) { int l=0,m=0; int i,j,max; int sum[4][4]; for(i=0,j=0;i<=3,j<=3;j++) { if(j==3) { sum[l][m]=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]; i++; j=-1; } else { sum[l][m]=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]; } m++; if(m==4) { l++; m=0; } } max=sum[0][0]; for(i=0;i<4;i++) { for(j=0;j<4;j++) { if(sum[i][j]>=max) { max=sum[i][j]; } } } cout<<"max"<

                • banerjeeunmesha + 1 comment

                  how to add back ticks? i am unable to add the pdf file as well plzz help.

        • ramin_shahab + 0 comments

          :O Are you using all those includes?

        • arka_cool1996 + 0 comments

          Simple and efficient and also easy to understand . Thank you sir.

        • arka_cool1996 + 0 comments

          Is there any specific reason why you ran the j loop instead of the i loop first while finding the sum of the hourglass or is it just a matter of choice.

      • joshknows09 + 0 comments

        Can you go over this a little bit?What are we trying to solve and how?^

    • 7jdeep + 1 comment

      include

      int main() { int a[6][6],hg[3][3]; int i,j,x,y,sum=0,max=-63; for(i=0;i<6;i++) { for(j=0;j<6;j++) { scanf("%d",&a[i][j]); } }

      for(i=0;i<4;i++)
{
    for(j=0;j<4;j++)
    {
        for(x=i;x<i+3;x++)
        {
            for(y=j;y<j+3;y++)
            {
                if(x==i+1&&y==j)
                continue;
                else if(x==i+1&&y==j+2)
                continue;
                else hg[x][y]=a[x][y];
                sum+=hg[x][y];
            }
        }
        if(sum>max)
        {
            max=sum;
        }

        sum=0;
    }
}
printf("%d",max);
return 0;

      }

      • fooboogoobear + 0 comments

        Holy for-loops, Batman

    • 7jdeep + 0 comments
      [deleted]