How to find non repeated numbers in an Array using JavaScript?

Question

Dear all I'm trying to find non repeated value in an array using javascript.I have written some code but it not working properly ..can you guys tell me where is the problem.thanks.

var arr = [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3];
var n = arr.length;
var result = '';

function nonrep() {
  for (var i = 0; i < n; i++) {
    var j;
    for (j = 0; j < n; j++)
      if (i != j && arr[i] == arr[j]) {
        result = arr[i];
        break;
      }
    if (j == n)
      return arr[i];
  }
  return result;
}
console.log(nonrep())

Possible duplicate of [Get all unique values in an array (remove duplicates)](https://stackoverflow.com/questions/1960473/get-all-unique-values-in-an-array-remove-duplicates) — Tholle, Jun 14 '18 at 10:01
What is your expected output? Do you need the unique values or values appear only once in the array? — Eddie, Jun 14 '18 at 10:02
`var result = '';` why do you initialize `result` to be a string? You need an array, don't you? — Yury Tarabanko, Jun 14 '18 at 10:18

Manu · Answer 1 · 2020-08-12T16:59:08.137

There is possibly a more neat approach to this solution, but this works as expected by filtering the array and compare it's current value with the array items itself (expect current item's index value).

const sampleArray = [1,2,3,7,2,1,3];
const getNonDuplicatedValues = (arr) => 
    arr.filter((item,index) => {
      arr.splice(index,1)
      const unique = !arr.includes(item)
      arr.splice(index,0,item)
      return unique
  })

console.log("Non duplicated values: " , ...getNonDuplicatedValues(sampleArray))

Nina Scholz · Answer 2 · 2018-06-14T11:00:09.123

Some changes:

Move all variable declarations inside of the function.
Use a function parameter for the handed over array, keep the function pure.
Declare all needed variables at top of the function in advance.
Take an array as result array unique.
Check i and j and if equal continue the (inner) loop.
Check the value at i and j and exit the (inner) loop, because a duplicate is found.
Take the check at the end of the inner loop and check the index j with the length of the array l, and if equal push the value to unique.
Use a single return statement with unique array at the end of the outer loop.

function getUnique(array) {
    var l = array.length,
        i, j,
        unique = [];

    for (i = 0; i < l; i++) {
        for (j = 0; j < l; j++) {
            if (i === j) {
                continue;
            }
            if (array[i] === array[j]) {
                break;
            }
        }
        if (j === l) {
            unique.push(array[i]);
        }
    }
    return unique;
}

console.log(getUnique([-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3]));

Another solution could be to check if indexOf and lastIndexOf returns the same value. Then you found a unique value.

var array = [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3],
    unique = array.filter((v, i) => array.indexOf(v) === array.lastIndexOf(v));

console.log(unique);

Although as always you bring to light fascinating concepts, this time I will have to point the O(2N^2) complexity is redundant. Just an observation though. — Attersson, Jun 14 '18 at 10:28

Dhruvin Modi · Answer 3 · 2023-02-04T10:41:51.830

2

 const arr = [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3];
    
    
    const non_repeating = arr.filter(num => arr.indexOf(num) === arr.lastIndexOf(num))
    
    console.log(non_repeating)

edited Feb 04 '23 at 10:41

answered Feb 04 '23 at 10:15

Dhruvin Modi

21
3

Nenad Vracar · Answer 4 · 2018-06-14T10:18:21.040

You could first use reduce to get one object with count for each number element and then filter on Object.keys to return array of non-repeating numbers.

var arr=[-1,2,5,6,2,9,-1,6,5,-1,3];
var obj = arr.reduce((r, e) => (r[e] = (r[e] || 0) + 1, r), {});
var uniq = Object.keys(obj).filter(e => obj[e] == 1).map(Number)

console.log(uniq)

Solution with for loop.

var arr = [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3];

var uniq = [];
for (var i = 0; i < arr.length; i++) {
  for (var j = 0; j < arr.length; j++) {
    if (arr[i] == arr[j] && i != j) break;
    else if (j == arr.length - 1) uniq.push(arr[i])
  }
}

console.log(uniq)

score 1 · Answer 5 · answered Feb 06 '22 at 13:36

Another simple approach

var arr = [1,1,2,3,3,4,4,5];

let duplicateArr = [];
var repeatorCheck = (item) => {
  const currentItemCount = arr.filter(val => val=== item).length;
  if(currentItemCount > 1) duplicateArr.push(item);
  return currentItemCount;
}
var result = arr.filter((item,index) => {
  var itemRepeaterCheck = !duplicateArr.includes(item) && repeatorCheck(item);
  if(itemRepeaterCheck === 1){
    return item;
  }
});
console.log(result);

score 1 · Answer 6 · answered Sep 09 '22 at 04:16

let arr = [1, 2, 1, 3, 3, 5];

    function nonRepeatableNo(arr) {
        let val = []
        for (let i = 0; i < arr.length; i++) {
            let count = 0;
            for (let j = 0; j < arr.length; j++) {
                if (arr[i] === arr[j]) {
                    count += 1
                }
            }
            if (count === 1) {
                val.push(arr[i])
            }
        }
        console.log(val)
    }
    nonRepeatableNo(arr)

score 0 · Answer 7 · edited Jun 20 '20 at 09:12

0

Filtering only unique elements according to OP request:

This uses for loops, as requested. It returns an array containing only elements appearing once in the original array.

var arr = [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3];
var n = arr.length;
var result = [];

function nonrep() {
  for (var i = 0; i < n; i++) {
    for (var j=0 ; j < n; j++)
      if (i!=j && arr[i]==arr[j])
        break;
    if(j==n)
        result.push(arr[i]);
  }
  return result;
}
console.log(nonrep())

edited Jun 20 '20 at 09:12

Community

1
1

answered Jun 14 '18 at 10:08

Attersson

4,755
1
15
29

i need solution in for loop and also above code not working – Anil kumar Jun 14 '18 at 10:12
@Anilkumar I edited my answer and used for loops, implementing the filtering you requested. – Attersson Jun 14 '18 at 10:24

score 0 · Answer 8 · answered Sep 13 '18 at 08:19

var arr1 = [45, 4,16,25,45,4,16, 9,7, 16, 25];
 var arr=arr1.sort();
  console.log(arr);
 var str=[];
 arr.filter(function(value){

if( arr.indexOf(value) === arr.lastIndexOf(value))
{ str.push(value);
console.log("ntttttttttttttnnn" +str)

}// how this works ===============A
 })

O/P

7,9

score 0 · Answer 9 · answered Jul 21 '19 at 15:17

Please try the below code snippet.

var arr = [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3];
    var uniqArr = [];
    for (var i = 0; i < arr.length; i++) {
      for (var j = 0; j < arr.length; j++) {
        if (arr[i] == arr[j] && i != j) break;
        else if (j == arr.length - 1){
          uniqArr.push(arr[i])
        } 
      }
    }

    console.log(uniqArr)

score 0 · Answer 10 · answered Mar 11 '20 at 12:46

0

this ES6 code worked for me :

a.map(c=>a.filter(b=>c==b)).filter(e=>e.length<2).reduce((total, cur)=> total.concat(cur), [])

answered Mar 11 '20 at 12:46

Ali Awais

31
2

score 0 · Answer 11 · answered Aug 18 '20 at 13:31

Here is a working method with loops.

  var arr = [-1,2,5,6,2,9,-1,6,5,-1,3];
  var len = arr.length;
  const result = arr
        .filter(value=>{
              var count=0;
              for(var i=0;i<len;i++)
              {
                  if(arr[i]===value)
                    count++;
              }
              return count===1;
          })
  console.log(result);

score 0 · Answer 12 · answered Mar 15 '22 at 19:29

const sampleArr = [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3];

function getUnique(arr){
  const result=[]
  const obj={}
  for(let i=0;i<arr.length;i++){
    if(!obj[arr[i]]){
      obj[arr[i]]=true
      result.push(arr[i])
    }else{
      const index= result.indexOf(arr[i])
      if(index!==-1){
        result.splice(result.indexOf(arr[i]),1)
      }
    }
  }
  return result
}

const uniqueArr= getUnique(sampleArr)
console.log(uniqueArr)

Ramkumar G · Answer 13 · 2022-08-29T06:23:42.500

0

Here is the solution..

var x = [1,1,2,3,2,4]
var res = []
x.map(d => {
  if(res.includes(d)) {
    // remove value from array
    res = res.filter((a) => a!=d)
  } else {
    // add value to array
    res.push(d)
  }
})
console.log(res) // [3,4]

edited Aug 29 '22 at 06:23

answered Aug 29 '22 at 05:24

Ramkumar G

415
1
8

Mahesh Solanke · Answer 14 · 2023-02-02T18:27:10.480

0

//without using any filter also with minimum complexity

const array = [1 , 2, 3, 4, 2, 3, 1, 6, 8,1,1 ];
const unique = new Set(); 
const repetedTreses = new Set();
for(let i=0; i<array.length; i++) {
    if(!unique.has(array[i]) && !repetedTreses.has(array[i])){
        unique.add(array[i]);
    }else{
        repetedTreses.add(array[i]);
        unique.delete(array[i]);
    }
}
let uniqueElements=[...unique];
console.log(uniqueElements);

edited Feb 02 '23 at 18:27

answered Feb 02 '23 at 18:22

Mahesh Solanke

1
1

Hi! please read [answer] add some description to your answer – Greg-- Feb 04 '23 at 17:34

score 0 · Answer 15 · answered Jul 01 '23 at 17:01

0

var arr = [1, 2, 4, 5, 6, 3, 6, 4];

const getNonDuplicatedValues = () => {
  let nondup = [];
  arr.forEach((e) => {
    if (nondup.includes(e)) {
      let index = nondup.indexOf(e);
      nondup.splice(index, 1);
    } else {
      nondup.push(e);
    }
  });
  return nondup;
};

const Nondup = getNonDuplicatedValues();

console.log("Nondup", Nondup);

answered Jul 01 '23 at 17:01

Santanu Ghosh

1
1

As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Jul 03 '23 at 10:49

score -1 · Answer 16 · answered Jun 14 '18 at 10:10

-1

You can use filter and indexOf for that:

console.log(
  [-1, 2, 5, 6, 2, 9, -1, 6, 5, -1, 3].filter((v, i, a) => a.indexOf(v, i + 1) === -1 )
);

answered Jun 14 '18 at 10:10

ZER0

24,846
5
51
54

How to find non repeated numbers in an Array using JavaScript?

16 Answers16

Filtering only unique elements according to OP request:

Linked

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