I have got such 2d jagged array (row is 1st dimension, column is 2nd dimension, bottom to up, left to right)

0   0   b   b   b   c
0   0   h   g   g   c
0   0   h   a   a   c
0   0   f   f   d   d
0   0   i   j   e   e
0   0   i   j   0   0

I would like to get a 2d jagged array summarize the positions of consecutive elements for each row, and ignore all 0 or single elements (assume that things like x x x y x x would not occur, so there won't be 2 blocks of same chars in a row)

the ideal outcome would be

{(b,3,5,2)}  // char, length, rowID, columnID
{(f,2,2,2), (d,2,2,4)}

this is my code, i feel it is not that readable, so I appreciate any suggestion in C# or F# (Linq extension method is welcomed as well)

   Tuple<char, int, int, int>[][] GetHorizontalBricks()
        List<Tuple<char, int, int, int>[]> ret = new List<Tuple<char, int, int, int>[]>();

        for (int rowID = 0; rowID < rowNum; rowID++)
            List<Tuple<char, int, int, int>> rowBricks = new List<Tuple<char, int, int, int>>();
            var row = myBoard[rowID];

            int recentCount = 1;
            int columnID = 0;
            while (columnID < columnNum)
                char recent = row[columnID];
                int next = columnID + 1;
                for (; next < columnNum; next++)
                    if (row[next] == recent)

                if (recentCount > 1 && recent != '0')
                    rowBricks.Add(Tuple.Create(recent, recentCount, rowID, columnID));

                columnID = next;
                recentCount = 1;

        return ret.ToArray();

1 Answer 1


First of all get rid of these tuples. After one week you will forget what each int represents. Define simple struct/class for your brick.

Second thing, small performance suggestion for big row count - explicit initial list size:

... = new List<Tuple<char, int, int, int>[]>(myBoard.Length);//assuming myBoard is an array

without it, list would be recreated and rewritten every 4 new elements.

Below is mine implementation with linq extensions. Probably it is not faster. I don't know if it is more readable, but at least line count is smaller :)

public struct Brick
    public char Symbol { get; set; }
    public int Length { get; set; }
    public int RowID { get; set; }
    public int ColumnID { get; set; }

    public override string ToString()
        return string.Format("{0} {1} {2} {3}", Symbol, Length, RowID, ColumnID);

public Brick[][] GetHorizontalBricks(char[][] board)
            .Reverse() //bottom up row order
            .Select((row,rowIndex) =>
                    .Select((c,i) => new { c = c, i = i })
                    .GroupBy(symbol => symbol.c) //assumption that we will never see 2 groups of same nonzero character in single row
                    .Where(symbol => symbol.Key != 0 && symbol.Count() > 1)
                    .Select(brick =>
                        new Brick
                            Symbol = brick.Key,
                            RowID = rowIndex,
                            ColumnID = brick.First().i,
                            Length = brick.Count()
            .Reverse() //reverse again to show result in form like you provided

and Console app test:

static void Main(string[] args)
    char[][] board = new char[][]
        new char[] { (char)0, (char)0, 'b' ,'b','b','c'},
        new char[] { (char)0, (char)0, 'h' ,'g','g','c'},
        new char[] { (char)0, (char)0, 'h' ,'a','a','c'},
        new char[] { (char)0, (char)0, 'f' ,'f','d','d'},
        new char[] { (char)0, (char)0, 'i' ,'j','e','e'},
        new char[] { (char)0, (char)0, 'i' ,'j',(char)0,(char)0}

    foreach (var row in new Program().GetHorizontalBricks(board))
        foreach (var brick in row)

  • \$\begingroup\$ thank u, I didn't know that Linq select supports index as well \$\endgroup\$
    – colinfang
    Oct 3, 2012 at 20:37
  • \$\begingroup\$ Small warning. Watch out for monster linq queries. It looks nice to do all in one chain but it can be pretty nasty to debug. \$\endgroup\$
    – Kuba
    Oct 3, 2012 at 20:40
  • 1
    \$\begingroup\$ If you don't specify size of a List, it's not going to be “recreated every 4 elements”, that would have terrible performance. Instead, every time it gets full, it is recreated to twice the size it was previously. Specifying the size can help you a bit, but probably not noticeably. \$\endgroup\$
    – svick
    Oct 4, 2012 at 9:25
  • \$\begingroup\$ My mistake. It starts from 0, then 4, 8, 16, 32 and so on. stackoverflow.com/questions/1762817/default-capacity-of-list \$\endgroup\$
    – Kuba
    Oct 4, 2012 at 9:42

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