Is there a way to find the row of a given cell in an ordered matrix?

Question

For example, given a matrix:

    01|02|03|04|05
    06|07|08|09|10
    11|12|13|14|15

And knowing that the matrix is 5x3, is there a way that if given the value '7', that we can know it is in row 2? The case is that the matrix is always ordered from 1 to n, starting from 0.

Lastly, the matrix is stored linearly in a zero based array.

Answer 1

Let X be your index
-------------------
column = X % width
row    = ceiling(X / width)

Edit: Seems to work now that I made some changes.

Answer 2

If the matrix is stored in a row-major order, then the row indices map to the element index as follows:

 rowIndex = (elementIndex - 1) / numcolumns
 columnIndex = (elementIndex % numcolumns) - 1

This is always integer division -- so no remainders. You will get the row and column indices as starting from 0.

It is left as an exercise to figure out what happens in column-major layout.

Answer 3

row = ceiling(7 / 5)  or   ceiling(position / width)

Answer 4

if it is 0-based:

row: n / width
col: n % width

In your example, you say it is zero-based, but it is actually starting at 1, and you count the top-left element as (row,col)=(1,1), so you'd need to adjust the math:

row: (n-1) / width + 1
col: (n-1) % width + 1

In your case, n=7, width= 5:

row = (7-1)/5 + 1 = 1+1 = 2
col = (7-1)%5 + 1 = 1+1 = 2

Note: I'm using standard programmer's integer math, where "a/b" really means "floor(a/b)".