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

Diagonal Difference

  • 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