Resolving simple references in a spreadsheet structure. Is this a linked list problem? - arrays

Question

I have this job test task where I am need some help.
I have to somehow dynamically access the value in the data stucture below, through references. To this point I used loops and indexOf() in an ordered alphabet and numbers. indexOf("B") in alphabet is [[ 1 ]], indexOf(1) is [0] for the two dimensional array.
Here "B1" is referencing "C1", "C1" references "D1" .. to "H1" and when "H1" points to the text "Last", all of the formulas should turn into the same value - the text "Last".
But how? Been breaking my head for a couple of hours and remembered linked lists, but so far have no clue how to implement them. How can I achieve this?
{
data: [
[
{ reference: "B1" },
{ reference: "C1" },
{ reference: "D1" },
{ reference: "E1" },
{ reference: "F1" },
{ reference: "G1" },
{ reference: "H1" },
{ text: "Last" },
]
],
id: "job-20"
}

Answer 1

It looks like you need to work with a spreadsheet type of data structure, and need to recalculate cells based on the formulas they have.
You can use recursion for that.
The base case happens when you arrive at a cell that has no reference. In that case the text of that cell can be returned to the caller. And that caller will assign that text to the cell it was looking at, and also returns that to their caller...etc.
function get(reference) { // return the cell object at the given reference
return data[reference.slice(1)-1][reference.charCodeAt() - "A".charCodeAt()];
}
function resolve(cell, visited=new Set) {
// Have we found a cell with just text? Then return that to caller
if (!("reference" in cell)) return cell.text;
// Did we already meet this cell before? Then we are running in circles!
if (visited.has(cell)) return "Circular Reference!";
// Make a recursive call to know what the text should be
cell.text = resolve(get(cell.reference), visited.add(cell));
delete cell.reference;
return cell.text; // Return this value also to the caller
}
function calculate(data) {
// Get to each cell object...
for (let row of data) {
for (let cell of row) {
resolve(cell); // And update its text property by following references
}
}
}
let data = [
[
{reference: "B1"}, {reference: "C1"}, {reference: "D1"}, {reference: "E1"},
{reference: "F1"}, {reference: "G1"}, {reference: "H1"}, {text: "Last"},
]
];
calculate(data);
console.log(data);
It should be noted that in a real spreadsheet the references (formulas) are not deleted during such a recalculation. The cell will have both the text and the reference, so that when a cell's text is updated, all other dependent cells get their own text updated accordingly again.