We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
Loading...
  1. Practice
  2. Algorithms
  3. Warmup
  4. Diagonal Difference
  5. Discussions

Diagonal Difference

  • ishwor + 3 comments

    I still have one extra variable, which is unnecessary

    static void Main(String[] args) {
        int n = Convert.ToInt32(Console.ReadLine());
        int[][] a = new int[n][];
        for(int a_i = 0; a_i < n; a_i++){
           string[] a_temp = Console.ReadLine().Split(' ');
           a[a_i] = Array.ConvertAll(a_temp,Int32.Parse);
        }
        
        int totalSum=0;
        int j=n-1;
        for(int i=0;i<n;i++){
            totalSum=a[i][i]-a[j][i]+totalSum;
            j--;
        }
        
        Console.WriteLine(Math.Abs(totalSum));
    }
    • [DELETED] + 1 comment
      [deleted]
      • rautnilesh8 + 1 comment

        What happen if input values are changed. I mean couple of more negative values in matrix

        • max_ragmo + 0 comments

          Does not affect the outcome, since it is the difference we are after, it is always a positive value

    • garg_gourav + 4 comments

      can be done in ((n-1)/2)+1 iteration

      for (int i = 0; i < ((n - 1) / 2) + 1; i++)
{
   total = ar[i][i] - ar[i][n - 1 - i] + ((i != n - 1 - i) ? ar[n - 1 - i][n - 1 - i] - ar[n - 1 - i][i] : 0) + total;
}
      • kps_forums + 0 comments

        For a more readable format:

        int squareDiffHalf(vector<vector<int>> matrix, int n)
{
    int diff = 0;
    for (int i = 0; i < n/2; i++)
    {
        diff += (matrix[i][i] + matrix[n-i-1][n-i-1] - matrix[i][n-i-1] - matrix[n-i-1][i]);
    }
    return abs(diff);
}
      • neostarteon + 0 comments

        Cannot read this. always go for a solutions others can read

      • mohan_krish_dk_1 + 0 comments

        could you explain?

      • A_Band + 0 comments

        Very good love you

    • gragest + 7 comments

      just in case you didnt know the for loop couldve been written like this for a cleaner look 

      for(int i = 0, j = n-1; i < n; i++, j--){
    sum1 += a[i][i];
    sum2 += a[j][i];
}

      or as 

      for(int i = 0; i < n; i++){
    sum1 += a[i][i];
    sum2 += a[(n-1)-i][i];
}
      • nikhilraynade + 0 comments

        Great

      • niyaziozturkg + 2 comments

        Yes, it is a good thinking .. If anyone has read Herbert Schildt or Deitel's Java books would have recognized it.. my solution is the same.. some one else should be able to read it quickly..

        static int diagonalDifference(int[][] a) {
    int primeDi = 0, secondDi = 0;
    // could be improved to 
            // int sum = 0;
    for (int i = 0 , j = a[0].length-1; i < a[0].length ; i++, j--) {
        primeDi += a[i][i] ;
        secondDi += a[i][j] ;
                    // could be improved to 
                    // sum = a[i][i] - a[i][j] ;
    }

    return Math.abs(primeDi - secondDi);
            // could be improved to
            // return Math.abs(sum);
}
        • mjperry94 + 0 comments

          This comment has helped me understand not only the solution, but how to get to the solution. From a rookie programmer, thank you for the insight!

        • muhammed_saafan + 0 comments

          this is awesome thanks

      • josesalvador_pe1 + 0 comments

        This is more efficient:

        n--; for(int i = 0; i <= n; i++){ sum1 += a[i][i]; sum2 += a[n-i][i]; }

      • josesalvador_pe1 + 1 comment

        Like this is more efficient:

        n--;

        for(int i = 0; i <= n; i++){

        sum1 += a[i][i];
sum2 += a[n-i][i];

        }

      • imran_nazir + 0 comments

        I think this is the most elegant a clear solution. Well done

      • micahbales + 0 comments

        This is a fantastic solution.

      • JuandiM + 0 comments

        Beautiful code