Remove duplication algorithm in place and stable (javascript)

Question

In the class today, we were asked to write an algorithm.

Given the array, remove duplicate values:

• It must be stable and should not use the inner loop.
• Should be done in place as best as possible
• No built-in functions (I was allowed to use .push)

After struggling with him for a while, this is what I came up with.

function remove_dupes(arr){
  var seen = {};
  var count = 0;

  for( var i = 0; i < arr.length - count ; i++) {
    arr[i] = arr[i+count];

    if( seen[arr[i]] ) {
      count++;
      arr[i] = arr[i+count];
      i--;
    }

    seen[arr[i]] = true;
  }

  arr.length = arr.length - count;
}

Working jsbin

I have a bit of duplicate code here, and I feel that maybe i--not the best way.

Is there a way to improve this code (without using the built-in functions)?

Answer 1

, , , , :

function remove_dupes(arr){
  var seen = {};
  
  var k = 0;
  for( var i=0; i<arr.length ;i++) {
    if( !seen[arr[i]] ) {
      arr[k++] = arr[i];
      seen[arr[i]] = 'seen';
    }
  }
  
  arr.length = k;
}


var x = [ 1, 2, 1, 4, 5, 3, 'dojo', 4, 6, 6, 7, 7, 6, 7, 5, 6, 6, 6, 6, 7, 'dojo', 11 ];
remove_dupes(x);


document.write(x);

Hide result

, .

Answer 2

:

function remove_dupes(arr){
  var seen = {};
  var dupes_removed = [];

  for( var i = 0; i < arr.length; i++) {
    if (!seen[arr[i]]) {
      dupes_removed.push(arr[i]);
      seen[arr[i]] = true;
    }
  }

  return dupes_removed;
}

- O (n) O (nlogn) time ( JS-- O (1) O (logn) time). , . O (n ^ 2), .

Answer 3

indexOf, , arr,

function remove_dupes(arr){
  var newArr = [];
  for( var i = 0; i < arr.length; i++){ 
    if(newArr.indexOf(arr[i]) === -1){
      newArr.push(arr[i]);
    }
  }
  
  return newArr;
}

var myArr = [2,4,2,4,6,6,6,2,2,1,10,33,3,4,4,4];

console.log(remove_dupes(myArr));

Hide result

Answer 4

Array.prototype.splice, (fiddle - ):

var arr = [1, 54, 5, 3, 1, 5, 20, 1, 54, 54];

var seen = {};

var length = arr.length;

var i = 0;

while (i < length) {
    if (seen[arr[i]] !== undefined) {
        arr.splice(i, 1);
        length--;
    } else {
        seen[arr[i]] = true;
    }

    i++;
}

console.log(arr);

O (n ^ 2), O (n), n .

Answer 5

, JS:

if ( !Array.unique )
{
    Array.prototype.unique = function()
    {
        var tmp = {}, out = [], _i, _n ;
        for( _i = 0, _n = this.length; _i < _n; ++_i )
        if(!tmp[this[_i]]) { tmp[this[_i]] = true; out.push(this[_i]); }
        return out;
    }
}