Sort 2D array of pairs in JavaScript

January 11, 2023

I need to sort 2D array of pairs in JavaScript.

const a = [
  [1, 1],
  [1, 2],
  [1, 3],
  [5, 6]
]
const b = [
  [5, 6],
  [1, 2],
  [1, 3],
  [1, 1]
]
b.sort((c, d) => d[0] + d[1] - c[0] + c[1]);
console.log(b)

function compare(a, b) {
  if (a.length != b.length) return false;
  for (let i = 0; i < b.length; i++)
    for (let j = 0; j < b[i].length; j++)
      if (a[i][j] !== b[i][j]) return false;
  return true;
}
console.log(compare(a, b))

I want to sort array b to be equal to array a. Of course real life array is much longer. This is just an example.

I have written function for comparing, but sort function doesn’t work properly. Could you please help me to fix this?

>Solution :

The issue isn’t your compare function, but instead your .sort() call.

First, d[0] + d[1] - c[0] + c[1] is backwards. You should subtract the elements of d from c, not the other way around. Like this: c[0] + c[1] - d[0] + d[1].

Second, you have an order of operations error. You need to subtract the elements of d from c, but your current code subtracts one element of d and adds the other. You need to distribute the negative sign, just like this: c[0] + c[1] - d[0] - d[1]

const a = [
  [1, 1],
  [1, 2],
  [1, 3],
  [5, 6]
]
const b = [
  [5, 6],
  [1, 2],
  [1, 3],
  [1, 1]
]
b.sort((c, d) => c[0] + c[1] - d[0] - d[1]);
console.log(b)

function compare(a, b) {
  if (a.length != b.length) return false;
  for (let i = 0; i < b.length; i++)
    for (let j = 0; j < b[i].length; j++)
      if (a[i][j] !== b[i][j]) return false;
  return true;
}
console.log(compare(a, b))