Array nested and conditions

Question

i'm learning javascript and trying to solve some exercises. The one I'm into wants me to remove all the duplicates from an array. By far this is my code:

const removeDuplicateds = (arr=undefined)=>{
let count=0;
for (let i=0; i <arr.length ; i++){
    for (let j=1; j<arr.length; j++){
        if (arr[i]=== arr[j]){
            count= count +1;
        }
    }
}
return console.log(count);

removeDuplicateds([1,2,3,4,1,1]);

So I wanted to validate this structure by using a counter (the real practice would be to use an arr.splice(j,1) instead of the count=count+1). So it returns 12 times, which means it entered 12 times in the if condition, which is impossible because there should be only 3 times (three 1's in the arr). Any ideas of what's happening? thanks!

Does this answer your question? [Get all unique values in a JavaScript array (remove duplicates)](https://stackoverflow.com/questions/1960473/get-all-unique-values-in-a-javascript-array-remove-duplicates) — ponury-kostek, Jul 27 '21 at 18:18
Why would it be only three times? The code nests entire-array iterations, so `i[0] === j[4]` *and* `j[5]`. When you get to `i[4]` it'll equal `j[4]` and `j[5]` again, and so on. FWIW this is a good exercise to "play computer" with pencil and paper on, or step through the code, or put in some console logging to understand what's happening. — Dave Newton, Jul 27 '21 at 18:20
There is no point in providing a default value of `undefined` to the `arr` argument of the function. First of all, without a default value `arr` is `undefined` by default if the function is invoked without arguments (i.e. `removeDuplicates()`). Then, if `arr` is `undefined` then the code throws an error when it reaches `arr.length`. If you want to have a default value for `arr` (even if it does not make any sense for this function) then use an empty array (`(arr = []) => { ... }`). — axiac, Jul 27 '21 at 18:26

score 1 · Accepted Answer · answered Jul 27 '21 at 18:20

1

Despite this solution being not the most effective one, the easiest way to fix the code would be

let count = 0;
for (let i = 0; i < arr.length - 1; i++){
    for (let j = i + 1; j < arr.length; j++){
        if (arr[i] === arr[j]){
            count++;
       }
    }
}

Your original code finds too many duplicates because despite pairs (1, 3) and (3, 1) being identical, your code treats them as different.

answered Jul 27 '21 at 18:20

saferif

187
10

(Noting that this would return `2`, I think, but it's not clear if the OP is trying to count all instances of a duplicated entry, or count the number of duplicates of an entry.) – Dave Newton Jul 27 '21 at 18:26
With this validation, I wanted to count how many times one number gets repeated on an array – Aiden Jul 27 '21 at 19:07

score 0 · Answer 2 · edited Jul 28 '21 at 15:02

The problem is logic. You are executing the first cycle 6 times, while the second cycle is executed 5 times for each iteration of the previous one. There will be cycles that will enter the if in more than one opportunity try the following:

 const removeDuplicateds = (arr=undefined)=>{
    let result = arr.filter((item,index)=>{
      return arr.indexOf(item) === index;
    })
    return result;
 }

Array nested and conditions

2 Answers2