Problem has been solved
What I need is if I declare n as 4. I need 4 unique coordinates and push them to array, here is my code that can produce duplicates:
function entriesToDel(n) {
var array = [];
for (let i = 0; i < array.length; i++) {
var row = Math.round(3*Math.random());
var col = Math.round(3*Math.random());
var output = [row,col];
array.push(output);
}
return array;
}
What you can do is simply put a condition based checking for the coords.
e.g. You can do something like,
function entriesToDel(n) {
var array = [];
for (var i = 0; i < n; i++) {
const repeat = () => {
var row = Math.round(3 * Math.random());
var col = Math.round(3 * Math.random());
var coords = [row, col];
let sameEntry = array.find(i => i[0] === row && i[1] === col)
if (sameEntry) {
repeat();
} else {
array.push(coords);
}
}
repeat();
}
console.log(array);
return array;
}Instead of creating the coordinates randomly, shuffle an array with all 3x3 coordinates and pick the first n of them:
function shuffle(a) { // Generic shuffle function
for (let i = a.length - 1; i > 0; i--) {
let j = Math.floor(Math.random() * (i + 1));
let x = a[i];
a[i] = a[j];
a[j] = x;
}
return a;
}
function entriesToDel(n) {
// Generate all possible coordinates in a 3x3 grid
let coord = Array.from({length: 3}, (_, i) =>
Array.from({length: 3}, (_, j) => [i, j])
).flat();
// Shuffle and select
return shuffle(coord).slice(0, n);
}
console.log(entriesToDel(4));You can just change your for loop into a while loop, and just terminate when the array.length is equal to your parameter.
You can use Arrray.some() to check if you already have the coordinate.
eg.
function entriesToDel(n) {
var array = [];
while (array.length < n) {
var row = Math.round(3*Math.random());
var col = Math.round(3*Math.random());
var coords = [row,col];
if (!array.some(([r,c]) =>
r === row && c === col))
array.push(coords);
}
return array;
}
console.log(entriesToDel(4));