CODEWITHSUNDEEP
javascriptarrays
Anil kumar
Anil kumar

Reputation: 21

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

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())

Upvotes: 0

Views: 22851

Answers (16)

Santanu Ghosh
Santanu Ghosh

Reputation: 1

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);

Upvotes: 0

Dhruvin Modi
Dhruvin Modi

Reputation: 31

 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)

Upvotes: 3

Mahesh Solanke
Mahesh Solanke

Reputation: 1

//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);

Upvotes: 0

Gaurav
Gaurav

Reputation: 11

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)

Upvotes: 1

Ramkumar G
Ramkumar G

Reputation: 461

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]

Upvotes: 0

Iamwafiq
Iamwafiq

Reputation: 1

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)

Upvotes: 0

shabarinath
shabarinath

Reputation: 593

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);

Upvotes: 1

krishnakirit04
krishnakirit04

Reputation: 166

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);

Upvotes: 0

Manu
Manu

Reputation: 193

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))

Upvotes: 4

Ali Awais
Ali Awais

Reputation: 61

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), [])

Upvotes: 0

Hemanta
Hemanta

Reputation: 11

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)

Upvotes: 0

Tested
Tested

Reputation: 789

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

Upvotes: 0

Nina Scholz
Nina Scholz

Reputation: 386634

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);

Upvotes: 4

Attersson
Attersson

Reputation: 4866

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())

Upvotes: 0

Nenad Vracar
Nenad Vracar

Reputation: 122047

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)

Upvotes: 1

ZER0
ZER0

Reputation: 25322

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 )
);

Upvotes: -1

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