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  1. Practice
  2. Algorithms
  3. Warmup
  4. Diagonal Difference
  5. Discussions

Diagonal Difference

  • drew_adorno + 8 comments

    You do not need to have any conditions to complete this problem.

    int main()
{
    int n;
    int firstDiagonal=0;
    int secondDiagonal=0;

    cin >> n;
    int count=n-1;
    vector< vector<int> > a(n,vector<int>(n));
    for(int a_i = 0;a_i < n;a_i++){
       for(int a_j = 0;a_j < n;a_j++){
          cin >> a[a_i][a_j];
       }
    }

    for(int i=0; i<n; i++)
    { 
        firstDiagonal+=a.at(i).at(i);
        secondDiagonal+=a.at(count).at(i);
        count--;
    }

    cout << abs(firstDiagnol-secondDiagnol);

    return 0;
}

    Because the diagonal are always going to be positions we know when we have the size of the matrix.

    • thriggle + 2 comments

      You've actually got 3xN conditions, counting all your loops, but I like your thinking.

      Here's my solution from JavaScript:

      var n = parseInt(readLine());
var suma = 0, sumb = 0;
for(var i = 0; i < n; i++){
   var line = readLine().split(' ');
   suma += Number(line[i]);
   sumb += Number(line[n-i-1]);
}
console.log(Math.abs(sumb-suma));
      • ProFurkan + 0 comments

        Brilliant

      • darkem + 0 comments

        Excellent!! My code have problem, the problem is because not use Math.abs!

        Excellent Work. :D

    • wei972008 + 0 comments
      [deleted]
    • JocoMagic + 0 comments

      Hey there i'm pretty newbie and don't quite understand what the vectors are for and what they are doing, and if you could also explainwhat is going on with your third FOR statement i would appreciate it. Thanks in advance.

    • Puneet3821 + 3 comments

      no need to use extra for loop.

      for(int a_i = 0;a_i < n;a_i++){
       for(int a_j = 0;a_j < n;a_j++){
          cin >> a[a_i][a_j];
       }
        d1 = d1 + a[a_i][a_i];
        d2 = d2 + a[a_i][n-1-a_i];
    }
      • honeyd725 + 1 comment

        is this in java

        • sarinarw + 0 comments

          Yes it is in Java, many other languages have some version of split though. Scala and Python have split() that works in the same way, C has str_tok <- apparently works like split

      • sumit_dutta + 1 comment

        there is a segmentaion fault in your code.

        • ben_nxumalo + 0 comments
          $first = $second=0;
$j = $n-1;
for($i=0;$i<$n; $i++){
    $first +=$a[$i][$i];
    $second +=$a[$i][$j];
    $j--;
} 


$result=  abs($first-$second);
print abs($result);
      • Turings_Ghost + 1 comment

        Slightly cleaner and more readable:

                int diag1 = 0, diag2 = 0;
        for(int p = 0; p<n; p++)
        {
            diag1 += a[p][p];
            diag2 += a[p][n - (p + 1)];
        }
        • prachigoel1598 + 1 comment

          can u please explain why n-j-1...

          • post4visakhvspm + 0 comments

            because n will be length and always +1 than index

    • bhargavisri + 0 comments

      Can you please explain me
      firstDiagonal+=a.at(i).at(i); secondDiagonal+=a.at(count).at(i);

    • gaganbaheti_it20 + 1 comment

      you will get an error. i tested your code and it's showing error

      • jesus_rpalos + 0 comments

        I had no problem running my code when I submitted.

    • Turings_Ghost + 0 comments

      Every "for" has a condition. If you mean no "if" or other explicit condition statements, sure.

    • ckastam + 0 comments

      that's in int main....not in the function....