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

  • mdumlupinar + 1 comment

    you can set -63 for the initial value of the max variable, because the possible minimum value of a hourglass can up to be -63. According to the given restrictions, the minimum hourglass can have maximum seven -9. So, this is the initial number for us: 7x-9=-63.

    • nancy345 + 4 comments

      how to do this task in C???

      • mdumlupinar + 0 comments

        sorry, i am not familiar with c language.

      • azheruddin617mr1 + 7 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 I 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.

      • 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]