CODEWITHSUNDEEP
javascript
Asking
Asking

Reputation: 4192

Compare two array using sort() method

I have next code what should compare the arrays;

function compare(arr) {
    const sorted = arr.sort((a,b)=> a -b).join(',');
    const unsorted = arr.join(',')
    console.log(sorted === unsorted) // true
}

compare([1, 16, 7])

Why i get true, or the arrays should be different?

Upvotes: 0

Views: 387

Answers (5)

TheMaster
TheMaster

Reputation: 50654

Array#Sort sorts in place. So, Join the array before sorting: 

function compare(arr) {
    const unsorted = arr.join(',');//moved up
    const sorted = arr.sort((a,b)=> a -b).join(',');
    console.log(sorted === unsorted) // false
}

compare([1, 16, 7])

Upvotes: 0

Tobbin2
Tobbin2

Reputation: 105

Since you´re sorting an array is it pretty much impossible to check if they are equal or not. If you wan't to check if all values that exists are the same as both arrays could you try something like this.

    function compare(arr) {
    const unsorted = arr.slice()
    const sorted = arr.sort((a,b)=> a -b)

    console.log(sorted)
    console.log(unsorted)

    if(checkIfValuesAreEqual(sorted, unsorted))
        console.log("they are equal")
    else
        console.log("they are not equal")
}

checkIfValuesAreEqual = (sorted, unsorted) => {
    if(sorted.length != unsorted.length)
        return false

    let sortedObj = {}
    let unsortedObj = {}

    for(const value of sorted){
        if(value in sortedObj)
            sortedObj[value] += 1
        else
            sortedObj[value] = 1
    }

    for(const value of unsorted){
        if(value in unsortedObj)
            unsortedObj[value] += 1
        else
            unsortedObj[value] = 1    
    }

    const keysSorted = Object.keys(sortedObj)
    const keysUnSorted = Object.keys(unsortedObj)


    if(keysSorted.length != keysUnSorted.length)
        return false


    for(const key of keysSorted){
        if(sortedObj[key] != unsortedObj[key])
            return false
    }

    return true

}

compare([1, 16, 7])

Had to change some things like unsorted = arr.slice(), Javascript puts everything by reference so unsorted n sorted are the same. To check if both are the same do i need to loop through so i needed them as arrays and not string.

I don't think you can compare two strings if one is sorted and the other is not since some values swap places.

Upvotes: 0

amatyas001
amatyas001

Reputation: 326

According to MDN sort() will modify the original array! MDN Array.prototype.sort()

Upvotes: 0

DSCH
DSCH

Reputation: 2376

The reason is that you are comparing the same string. How is it the same string? Well, .sort() sorts in place - meaning it doesn't return a new sorted array but the the same array just gets sorted. So for unsorted you are joining the same sorted array. You can try switch the order of the assignments and the result should be diferent

Upvotes: 0

User863
User863

Reputation: 20039

Try copying the array using spread opeartor

Note that the array is sorted in place, and no copy is made.

function compare(arr) {
  const sorted = [...arr].sort((a, b) => a - b).join(',');
  const unsorted = arr.join(',')
  console.log(sorted === unsorted) // true
}

compare([1, 16, 7])

Upvotes: 2

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