Count number of coloured cells in column with dynamic number of rows - google-apps-script

Question

At the bottom of each column I'm trying to display how many cells in that column have a particular background colour. This is what I'd like it to do:
Picture of intended result
The problem is it's a Gantt chart and I'll be inserting rows constantly. I can do it with a fixed number of rows using:
=countCellsWithBackgroundColor("#cfe2f3", "B1:B22")
which uses this function I found:
function countCellsWithBackgroundColor(color, rangeSpecification) {
var sheet = SpreadsheetApp.getActiveSpreadsheet();
var range = sheet.getRange(rangeSpecification);
var x = 0;
for (var i = 1; i <= range.getNumRows(); i++) {
for (var j = 1; j <= range.getNumColumns(); j++) {
var cell = range.getCell(i, j);
if(cell.getBackgroundColor() == color)
x++;
}
}
return x;
}
I just can't figure out how to give that function every cell in the column. I get errors like "must be a range".

Answer 1

You could use a more efficient way to achieve the same result by using arrays in your function like below
function countCellsWithBackgroundColor(color, rangeSpecification) {
var sheet = SpreadsheetApp.getActiveSpreadsheet();
var backGroundColors = sheet.getRange(rangeSpecification).getBackgrounds();
var x = 0;
for (var i = 0; i < backGroundColors.length; i++) {
for (var j = 0; j < backGroundColors[0].length; j++) {
if(backGroundColors[i][j] == color){ x++ }
}
}
return x;
}

Answer 2

WORKING EXAMPLE WITH AUTO UPDATE
function countCellsWithBackgroundColor(color, rangeSpecification) {
var sheet = SpreadsheetApp.getActiveSpreadsheet();
var backGroundColors = sheet.getRange(rangeSpecification).getBackgrounds();
var x = 0;
for (var i = 0; i < backGroundColors.length; i++) {
for (var j = 0; j < backGroundColors[0].length; j++) {
if(backGroundColors[i][j] == color){ x++ }
}
}
return x;
}
function onEdit(e)
{
SpreadsheetApp.getActiveSheet().getRange('H26').setValue(Math.random());
}
Now Insert the dummy cell into spreadsheet function for auto update
=countCellsWithBackgroundColor("#0f9d58","A5:A26",H26)