Javascript for loop behaviour in arrays

Question

I'm currently learning JS and doing some problems on Codewars. I've written a function to solve a problem, but I don't understand why the for loop in JS behaves the way it does in my solution, I'd be glad if anyone could help me. The problem is as follows:

Write an algorithm that takes an array and moves all of the zeros to the end, preserving the order of the other elements.

moveZeros([1, 2, 0, 1, 0, 1, 0, 3, 0, 1]) // returns[1, 2, 1, 1, 3, 1, 0, 0, 0, 0]

My solution is:

var moveZeros = function(arr) {
    let newarr = [];
    for (let i = 0; i < arr.length; i++) {
        if (arr[i] === 0) {
            newarr.push(arr[i])
        } else {
            newarr.unshift(arr[i])
        };
    }
    return newarr;
}

What my function returns is:

[1, 3, 1, 1, 2, 1, 0, 0, 0, 0]

Why does JS put 3 before 2, although its index in the array is 7?

Thanks in advance!

Unshift adds elements to the beginning of an array (developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…). You should use push instead (developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…) — briosheje
– briosheje, Commented Dec 13, 2018 at 12:53

Code Maniac · Accepted Answer · 2018-12-13 12:54:44Z

4

Well unshift() pushes the value in front of array.

Have a look at this example.ti understand how unshift() works.

let a = [];
 a.unshift(1);
 a.unshift(2);
 a.unshift(3);
 a.unshift(4);
console.log(a);

Push add values at the back of array.

look at example below to understand how push works.

let a = [];
     a.push(1);
     a.push(2);
     a.push(3);
     a.push(4);
console.log(a);

var moveZeros = function (arr) {
let newarr = [];
for (let i = 0; i < arr.length; i++) {
if (arr[i] === 0) {newarr.push(arr[i])}
else {newarr.unshift(arr[i])};
} return newarr;
}



let op = moveZeros([1, 3, 1, 1, 2, 1, 0, 0, 0, 0]);
console.log(op);

edited Dec 13, 2018 at 12:54

answered Dec 13, 2018 at 12:50

Code Maniac

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3 Comments

Konstantin Over a year ago

"Well unshift() pushes the value in front of array. Push add values at the back of array." That's right, but for the input [1, 2, 0, 1, 0, 1, 0, 3, 0, 1] my function returns [1, 3, 1, 1, 2, 1, 0, 0, 0, 0] and NOT [1, 2, 1, 1, 3, 1, 0, 0, 0, 0]. It switches [2] at index 1 for [3] which in the input is at index 7 and I don't see why

HMR Over a year ago

@Konstantin It reverses everything that isn't zero, try it with another input.

Code Maniac Over a year ago

Your function is return is that way only. as @HMR said except zero it will reverse everything.

Zack Keyzol · Accepted Answer · 2018-12-13 13:15:01Z

3

The reason is because unshift always adds the number to the beginning of newarr

E.g from above code the newarr will look as below: i=0: newarr = [1] i=1: newarr = [2,1] i=2: newarr = [2,1,0] i=3: newarr = [1,2,1,0] i=4: newarr = [1,2,1,0,0] i=5: newarr = [1,2,1,0,0,0] i=6: newarr = [1,1,2,1,0,0,0] i=7: newarr = [3,1,1,2,1,0,0,0] i=8: newarr = [3,1,1,2,1,0,0,0,0] i=9: newarr = [1,3,1,1,2,1,0,0,0,0]

answered Dec 13, 2018 at 13:15

Zack Keyzol

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Comments

Ahmzyjazzy · Accepted Answer · 2019-05-08 16:39:19Z

using your sample code. You can introduce a temporary array to store all zeros that exist. Then concatenate it with the newarr

var moveZeros = function(arr) {
  let newarr = []; 
  let temp = []; //store all zeros here
  for (let i = 0; i < arr.length; i++) {
    if (arr[i] === 0) {
      temp.push(arr[i])
  } else {
      newarr.push(arr[i])
  };
 }

 return newarr.concat(temp); //this moves all zeros after the array order
}

Also if you want to maintain the function argument i.e the arr as the result variable, you can remove all zeros from it after storing each in the temporaray array. Then concatenate the arr with the temp

var moveZeros = function(arr) {
  let temp = []; //store all zeros here
  for (let i = 0; i < arr.length; i++) {
    if (arr[i] === 0) {
      temp.push(arr[i]);
      arr.splice(i,1); /*delete all zeros after storing them this does not affect array order*/
    } 
  }

  return arr.concat(temp);
}

HMR · Accepted Answer · 2018-12-13 13:32:25Z

1

welcome to Stack Overflow. The following code will put all zero's at the end of the array and keep the rest in the order they were found:

const moveToEnd = (arr) => {
  const [head, tail] = arr.reduce(
    ([head, tail], val) =>
      val === 0
        ? [head, tail.concat(val)]
        : [head.concat(val), tail],
    [[], []],
  );
  return head.concat(tail);
};
console.log(moveToEnd([1, 2, 0, 1, 0, 1, 0, 3, 0, 1]));

answered Dec 13, 2018 at 13:32

HMR

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Comments

Luis Maracara Cruz · Accepted Answer · 2018-12-13 13:47:37Z

-1

You can use sort() function for array ordering:

var sortedArray = [1, 2, 0, 1, 0, 1, 0, 3, 0, 1].sort(function(a,b){
 return b === 0 && (b - a) 
});

the b - a indicates that order descendant.

https://www.w3schools.com/js/js_array_sort.asp

edited Dec 13, 2018 at 13:47

answered Dec 13, 2018 at 12:54

Luis Maracara Cruz

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1 Comment

Andy Over a year ago

..."preserving the order of the other elements"?

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