CODEWITHSUNDEEP
javascriptarrays
MGS
MGS

Reputation: 23

Sort the numbers so that the even numbers were ahead

I have array with 20 random numbers (from -10 to 10) , I need to sort them . The even numbers must be in front of array. like let arr = [-2,3,6,-12,9,2,-4,-11,-8] must become arr = [-12,-8,-2,2,4,6,-11,3,9] here is my code:

let array = Array(20).fill().map(() => Math.round(Math.random() * 20) - 10);

console.log(array);

function moveEvenToFront(array){
    let temp=0;
    let a=0;
    for(let i=0;i<array.length;i++){

        if(array[i] % 2 == 0){

            for (let j=i; j>a; j-- ){

                temp = array[j-1];

                array[j-1] = array[j];

                array[j] = temp;

            }
            a++;
        }

    }
    return array;
}

moveEvenToFront(array);

I tried this function , but it doesn't works.

Upvotes: 2

Views: 387

Answers (4)

Paul Muriithi
Paul Muriithi

Reputation: 41

function moveEvenToFront(array) {
    let evenIndex = 0;
    for (let i = 0; i < array.length; i++) {
        if (array[i] % 2 === 0) {
            // Swapping the current element with the element at evenIndex
            let temp = array[i];
            array[i] = array[evenIndex];
            array[evenIndex] = temp;
            evenIndex++;
        }
    }

    return array;
}

Upvotes: 0

Nina Scholz
Nina Scholz

Reputation: 386650

You could use a group for sorting even and odd numbers and then sort by value with a chained approach.

function sort(array) {
    return array.sort((a, b) => Math.abs(a) % 2 - Math.abs(b) % 2 || a - b)
}

console.log(sort(Array.from({ length: 20 }, _ => Math.round(Math.random() * 20) - 10)));
.as-console-wrapper { max-height: 100% !important; top: 0; }

Upvotes: 2

kev
kev

Reputation: 1031

Your best bet is to use Array.prototype.sort()

let array = Array(20).fill().map(() => Math.round(Math.random() * 20) - 10);


    function moveEvenToFront(array) {
        console.log(array);

        const tempEven = array.filter(el => {return el % 2 === 0});
        const tempOdd = array.filter(el => {return el % 2 !== 0});

    const tempEvenSorted = tempEven.sort((a, b) => {
        return a - b;
    });

    const tempOddSorted = tempOdd.sort((a, b) => {
        return a - b;
    });

    return tempEvenSorted.concat(tempOddSorted);

    }

     console.log(moveEvenToFront(array));

A combination of filter() and sort(). First filter the evens and then the odds into different arrays. Then sort them separately and then add them together.

The sort() function takes two arguments which are elements from your original array and must return a negative number, a positive number or zero and based on that creates a new sorted array and returns it.

Upvotes: 0

Pointy
Pointy

Reputation: 413767

You can use the .sort() method and a comparator. The comparator function will have to first check the "evenness" of the two arguments. If they're both even or both odd, it'll then base the comparison result on the value:

array.sort(function(a, b) {
  let aeven = !(a % 2), beven = !(b % 2);
  if (aeven && !beven) return -1;
  if (beven && !aeven) return 1;
  return a - b;
});

A comparator function for the .sort() method is passed two values from the array. The function should return:

  • a negative number if the first should sort before the second;
  • a positive number if the first should sort after the second;
  • zero if they're the same for ordering purposes.

Upvotes: 6

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