Sorting using JavaScript's sort method

Question

I am studying javascript sort method. So, I want to sort using only sort without filter or map or otherMethod function.

const arr = [1,100,3,5,7,3,2,7, -1]

In the above array, I want the numbers greater than 5 to be sorted first, followed by the smaller numbers in ascending order.

Expected results are:
// [5, 7, 7, 100, -1, 1, 2, 3, 3];

Again, I know how to sort using other methods!

for example

const arr1 = input.filter(num => num >= 5).sort((a, b) => a - b)
const arr2 = input.filter(num => num < 5).sort((a,b) => a - b)

[...arr1, ...arr2]

But I hope to solve it using only the sort method!

please help me

arr.sort((a,b) => {
  return (a+5) > b ? 1 : -1
})

but result is : 
// [1, 3, -1, 5, 2, 7, 3, 7, 100]

Your sort callback needs to do the following: if both a and b are both below 5 or above 5, return -1 or 1 depending on which number is greater. If one number of below 5 and the other above 5, return -1 or 1 depending on which number is below/above 5. — deceze
– deceze ♦, Commented Nov 9, 2022 at 1:07

CertainPerformance · Accepted Answer · 2022-11-09 01:09:03Z

3

First return the difference between whether one value or the other is less than 5 - proceed to sort numerically (with subtraction) only if that difference is 0 (in which case both values being compared are 5 or above, or both values are less than 5).

const arr = [1,100,3,5,7,3,2,7, -1];
arr.sort((a, b) => 
  ((b >= 5) - (a >= 5))
  || a - b
);
console.log(arr);

answered Nov 9, 2022 at 1:09

CertainPerformance

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