2D array - checking for column with most "on" values - ios

Question

So I have a 2D array of "1" and "0" values, a value is either on or off, this can generate shapes and I want to check for vertical lines, for example:
[0,0,1,0,1,0,0,0]
[0,0,0,0,1,0,0,0]
[0,1,0,0,1,0,0,0]
[0,0,0,0,1,0,0,0]
[1,0,0,0,1,0,0,0]
[0,0,0,0,1,0,0,0]
that has a vertical line in column 5 so we will return those ones and strip all other results (change the 1's that aren't part of that line to 0's...)
[0,0,0,0,1,0,0,0]
[0,0,0,0,1,0,0,0]
[0,0,0,0,1,0,0,0]
[0,0,0,0,1,0,0,0]
[0,0,0,0,1,0,0,0]
[0,0,0,0,1,0,0,0]
My 2 dimensional array is much more complex, it is about a 300x600 2D array. In-order to visually see the values I generated UIViews with red background and stuck them on the screen in a view the size of my array. This is what the returned image was (the blue arrow was later photoshopped in to indicate the longest vertical line (the values we want to keep)
So what's a good method to find the longest vertical line (of "1" values) in a 2 dimensional array and change all other values to zero. (So that if I render the array in a visual graph format again only this is displayed (other red dots are fading out because they have been changed from "1" values to "0" values.)
I was thinking maybe something along the lines of generating a for-loop that would keep track of all of the "1" values in each column and a set of maybe any given 6 columns that are consecutively next to each other that have the largest amount of "1" values is the area (6 columns wide) where the longest vertical line is most likely located, but I can see a few issues with this, also I don't know how to get the rows that are part of that line after I have the columns.... hmmmm
*Note: I am making my "2 dimensional array" by just have a variable for the fixed number of columns and then I have an array that just has all of the values for all column/row combinations. For example a 3x3 board would be [0,0,0,0,1,0,0,0,0] which I can then understand means this:
[0,0,0]
[0,1,0]
[0,0,0]
Because I know there is always 3 columns per row.
[row1column1,row1column2,row1column3,row2column1,row2column2,row2column3,row3column1,row3column2,row3column3]

Answer 1

Here's some sample code (you'll need to take care of the allocations and some semantics - e.g: assigning one array to another)
int array2d[]; // which is 1d from what I understood - this holds your initial values
int columnCount; // you already have the number of columns, this is the var that's holding it
int maxColumn[], column[]; // these will hold the values of the maximum column and the current looped column
int maxIndex = 0, max=0; // used to determine which of the columns is the longest
int rowCount = array2d.count/columnCount; // find out how many rows there are
// first loop through the entire array, set everything to '0', while storing the values of the longest column (so far) in a different array
for(int i=0; i<columnCount; i++){
int columnSum = 0;
int index = i;
// go through every row of the column i
for(int j=0; j<rowCount; j++, index+=rowCount){
columnSum=columnSum+array2d[index];
column[j]=array2d[index];
array2d[index]=0;
}
if(columnSum>max){
max = columnSum;
maxIndex = i; // the column index
maxColumn = column;
}
}
// we found the longest column (maxIndex), now we need to set its values back to what they were previously
int index = maxIndex;
for(int j=0; j<rowCount; j++, index+=rowCount){
array2d[index]=maxColumn[j];
}
Since you want to find the longest of 6 consecutive columns, I believe that setting columnCount/=6; should work (it'll also make the rowCount=rowCount*6)

Answer 2

Your requirements aren't very precise, which makes me think you need to think harder about what you want, or state it more clearly. Take a look at the attached figure:
It's six pixels wide, so I image you could get this sort of a chain in your image. There's no connected sequence of 1's longer than 4 in any of the columns, yet the chain itself is 18 pixels long. Would you need to deal with a situation like this? If yes, then just summing over individual columns isn't enough; you need a more sophisticated method, like finding "connected components".