Merge sort based on number of values in arrays

Question

I would like to understand how to use merge sort to sort arrays based on the number of values in each array.

let test = [
{
name: "a",
numbers: [1,2]
},{
name: "b",
numbers: [1,2,3]
},{
name: "c",
numbers: [5]

The 'a' has 2 numbers, the 'b' has 3 numbers and the 'c' has 1 number. So it should be sorte as follows, from high to low: b, a, c.

Have you tried something? I mean, do you have a sample code that performs merge sort? — GOTO 0
– GOTO 0, Commented Dec 25, 2021 at 22:11

Nikita Skrebets · Accepted Answer · 2021-12-27 18:07:24Z

5

Just use the array sort with a compare function.
The function should substract the lengths of the numbers property of objects.
More info: Array.sort() on MDN

test.sort((a, b) => b.numbers.length - a.numbers.length)

edited Dec 27, 2021 at 18:07

answered Dec 25, 2021 at 21:10

Nikita Skrebets

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8 Comments

EnTil Over a year ago

How about if it was like this: let test = [ { name: "abc", numbers: [1,2] },{ name: "def", numbers: [1,2,3] },{ name: "ghi", numbers: [5]} And it should be from high to low: def, abc, ghi...?

Nikita Skrebets Over a year ago

@EnTil test.sort((a, b) => b.numbers.length - a.numbers.length).map(o => o.name)

EnTil Over a year ago

It does not work with arrays with other names than a, b and c...

Nikita Skrebets Over a year ago

@EnTil Please post an example of such array

Nikita Skrebets Over a year ago

Because this works

[ { name: "abc", numbers: [1,2] },{ name: "def", numbers: [1,2,3] },{ name: "ghi", numbers: [5]}].sort((a, b) => b.numbers.length - a.numbers.length).map(o => o.name)

, It gives ['def', 'abc', 'ghi']

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