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

  • 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.