stack size - Page 106 - Developer IT

how to increase the size of the jquery grid row..

- by kumar

how to increase row size using jquery.. I have column like this. { name: 'Comments', index: 'Comments', editable: true, editype: 'textarea', editoptions: { rows: "2", cols: "10"} } Thanks

How can I use client-side JavaScript to restrict the size of file uploads?

- by cometta

Is there a JavaScript library that can be used to check the size of a file before allowing it to be uploaded to a Java servlet?

How to find same-value rectangular areas of a given size in a matrix most efficiently?

- by neo

My problem is very simple but I haven't found an efficient implementation yet. Suppose there is a matrix A like this: 0 0 0 0 0 0 0 4 4 2 2 2 0 0 4 4 2 2 2 0 0 0 0 2 2 2 1 1 0 0 0 0 0 1 1 Now I want to find all starting positions of rectangular areas in this matrix which have a given size. An area is a subset of A where all numbers are the same. Let's say width=2 and height=3. There are 3 areas which have this size: 2 2 2 2 0 0 2 2 2 2 0 0 2 2 2 2 0 0 The result of the function call would be a list of starting positions (x,y starting with 0) of those areas. List((2,1),(3,1),(5,0)) The following is my current implementation. "Areas" are called "surfaces" here. case class Dimension2D(width: Int, height: Int) case class Position2D(x: Int, y: Int) def findFlatSurfaces(matrix: Array[Array[Int]], surfaceSize: Dimension2D): List[Position2D] = { val matrixWidth = matrix.length val matrixHeight = matrix(0).length var resultPositions: List[Position2D] = Nil for (y <- 0 to matrixHeight - surfaceSize.height) { var x = 0 while (x <= matrixWidth - surfaceSize.width) { val topLeft = matrix(x)(y) val topRight = matrix(x + surfaceSize.width - 1)(y) val bottomLeft = matrix(x)(y + surfaceSize.height - 1) val bottomRight = matrix(x + surfaceSize.width - 1)(y + surfaceSize.height - 1) // investigate further if corners are equal if (topLeft == bottomLeft && topLeft == topRight && topLeft == bottomRight) { breakable { for (sx <- x until x + surfaceSize.width; sy <- y until y + surfaceSize.height) { if (matrix(sx)(sy) != topLeft) { x = if (x == sx) sx + 1 else sx break } } // found one! resultPositions ::= Position2D(x, y) x += 1 } } else if (topRight != bottomRight) { // can skip x a bit as there won't be a valid match in current row in this area x += surfaceSize.width } else { x += 1 } } } return resultPositions } I already tried to include some optimizations in it but I am sure that there are far better solutions. Is there a matlab function existing for it which I could port? I'm also wondering whether this problem has its own name as I didn't exactly know what to google for. Thanks for thinking about it! I'm excited to see your proposals or solutions :)

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