Get all non-unique values (i.e.: duplicate/more than one occurrence) in an array

I need to check a JavaScript array to see if there are any duplicate values. What's the easiest way to do this? I just need to find what the duplicated values are - I don't actually need their indexes or how many times they are duplicated.

I know I can loop through the array and check all the other values for a match, but it seems like there should be an easier way.

Similar question:

• https://stackoverflow.com/questions/1960473

You could sort the array and then run through it and then see if the next (or previous) index is the same as the current. Assuming your sort algorithm is good, this should be less than O(n²):

const findDuplicates = (arr) => {
  let sorted_arr = arr.slice().sort(); // You can define the comparing function here. 
  // JS by default uses a crappy string compare.
  // (we use slice to clone the array so the
  // original array won't be modified)
  let results = [];
  for (let i = 0; i < sorted_arr.length - 1; i++) {
    if (sorted_arr[i + 1] == sorted_arr[i]) {
      results.push(sorted_arr[i]);
    }
  }
  return results;
}

let duplicatedArray = [9, 9, 111, 2, 3, 4, 4, 5, 7];
console.log(`The duplicates in ${duplicatedArray} are ${findDuplicates(duplicatedArray)}`);

In case, if you are to return as a function for duplicates. This is for similar type of case.

Reference: https://stackoverflow.com/a/57532964/8119511

If you want to elimate the duplicates, try this great solution:

function eliminateDuplicates(arr) {
  var i,
      len = arr.length,
      out = [],
      obj = {};

  for (i = 0; i < len; i++) {
    obj[arr[i]] = 0;
  }
  for (i in obj) {
    out.push(i);
  }
  return out;
}

console.log(eliminateDuplicates([1,6,7,3,6,8,1,3,4,5,1,7,2,6]))

Source: http://dreaminginjavascript.wordpress.com/2008/08/22/eliminating-duplicates/

This is my answer from the duplicate thread (!):

When writing this entry 2014 - all examples were for-loops or jQuery. JavaScript has the perfect tools for this: sort, map and reduce.

Find duplicate items

var names = ['Mike', 'Matt', 'Nancy', 'Adam', 'Jenny', 'Nancy', 'Carl']

const uniq = names
  .map((name) => {
    return {
      count: 1,
      name: name
    };
  })
  .reduce((result, b) => {
    result[b.name] = (result[b.name] || 0) + b.count;

    return result;
  }, {});
const duplicates = Object.keys(uniq).filter((a) => uniq[a] > 1);

console.log(duplicates); // [ 'Nancy' ]

More functional syntax:

@Dmytro-Laptin pointed out some code that can be removed. This is a more compact version of the same code. Using some ES6 tricks and higher-order functions:

const names = ['Mike', 'Matt', 'Nancy', 'Adam', 'Jenny', 'Nancy', 'Carl'];
const count = names =>
  names.reduce((result, value) => ({ ...result,
    [value]: (result[value] || 0) + 1
  }), {}); // don't forget to initialize the accumulator
const duplicates = dict =>
  Object.keys(dict).filter((a) => dict[a] > 1);

console.log(count(names)); // { Mike: 1, Matt: 1, Nancy: 2, Adam: 1, Jenny: 1, Carl: 1 }
console.log(duplicates(count(names))); // [ 'Nancy' ]

UPDATED: Short one-liner to get the duplicates:

[1, 2, 2, 4, 3, 4].filter((e, i, a) => a.indexOf(e) !== i) // [2, 4]

To get the array without duplicates simply invert the condition:

[1, 2, 2, 4, 3, 4].filter((e, i, a) => a.indexOf(e) === i) // [1, 2, 3, 4]

I simply did not think about filter() in my old answer below ;)

---

When all you need is to check that there are no duplicates as asked in this question you can use the every() method:

[1, 2, 3].every((e, i, a) => a.indexOf(e) === i) // true

[1, 2, 1].every((e, i, a) => a.indexOf(e) === i) // false

Note that every() doesn't work for IE 8 and below.

Find duplicate values in an array

This should be one of the shortest ways to actually find duplicate values in an array. As specifically asked for by the OP, this does not remove duplicates but finds them.

var input = [1, 2, 3, 1, 3, 1];

var duplicates = input.reduce(function(acc, el, i, arr) {
  if (arr.indexOf(el) !== i && acc.indexOf(el) < 0) acc.push(el); return acc;
}, []);

document.write(duplicates); // = 1,3 (actual array == [1, 3])

This doesn't need sorting or any third party framework. It also doesn't need manual loops. It works with every value indexOf() (or to be clearer: the strict comparision operator) supports.

Because of reduce() and indexOf() it needs at least IE 9.

You can add this function, or tweak it and add it to Javascript's Array prototype:

Array.prototype.unique = function () {
	var r = new Array();
	o:for(var i = 0, n = this.length; i < n; i++)
	{
		for(var x = 0, y = r.length; x < y; x++)
		{
			if(r[x]==this[i])
			{
                alert('this is a DUPE!');
				continue o;
			}
		}
		r[r.length] = this[i];
	}
	return r;
}

var arr = [1,2,2,3,3,4,5,6,2,3,7,8,5,9];
var unique = arr.unique();
alert(unique);

UPDATED: The following uses an optimized combined strategy. It optimizes primitive lookups to benefit from hash O(1) lookup time (running unique on an array of primitives is O(n)). Object lookups are optimized by tagging objects with a unique id while iterating through so so identifying duplicate objects is also O(1) per item and O(n) for the whole list. The only exception is items that are frozen, but those are rare and a fallback is provided using an array and indexOf.

var unique = function(){
  var hasOwn = {}.hasOwnProperty,
      toString = {}.toString,
      uids = {};

  function uid(){
    var key = Math.random().toString(36).slice(2);
    return key in uids ? uid() : uids[key] = key;
  }

  function unique(array){
    var strings = {}, numbers = {}, others = {},
        tagged = [], failed = [],
        count = 0, i = array.length,
        item, type;

    var id = uid();

    while (i--) {
      item = array[i];
      type = typeof item;
      if (item == null || type !== 'object' && type !== 'function') {
        // primitive
        switch (type) {
          case 'string': strings[item] = true; break;
          case 'number': numbers[item] = true; break;
          default: others[item] = item; break;
        }
      } else {
        // object
        if (!hasOwn.call(item, id)) {
          try {
            item[id] = true;
            tagged[count++] = item;
          } catch (e){
            if (failed.indexOf(item) === -1)
              failed[failed.length] = item;
          }
        }
      }
    }

    // remove the tags
    while (count--)
      delete tagged[count][id];

    tagged = tagged.concat(failed);
    count = tagged.length;

    // append primitives to results
    for (i in strings)
      if (hasOwn.call(strings, i))
        tagged[count++] = i;

    for (i in numbers)
      if (hasOwn.call(numbers, i))
        tagged[count++] = +i;

    for (i in others)
      if (hasOwn.call(others, i))
        tagged[count++] = others[i];

    return tagged;
  }

  return unique;
}();

If you have ES6 Collections available, then there is a much simpler and significantly faster version. (shim for IE9+ and other browsers here: https://github.com/Benvie/ES6-Harmony-Collections-Shim)

function unique(array){
  var seen = new Set;
  return array.filter(function(item){
    if (!seen.has(item)) {
      seen.add(item);
      return true;
    }
  });
}

var a = ["a","a","b","c","c"];

a.filter(function(value,index,self){ return (self.indexOf(value) !== index )})

This should get you what you want, Just the duplicates.

function find_duplicates(arr) {
  var len=arr.length,
      out=[],
      counts={};

  for (var i=0;i<len;i++) {
    var item = arr[i];
    counts[item] = counts[item] >= 1 ? counts[item] + 1 : 1;
    if (counts[item] === 2) {
      out.push(item);
    }
  }
    
  return out;
}

find_duplicates(['one',2,3,4,4,4,5,6,7,7,7,'pig','one']); // -> ['one',4,7] in no particular order.

Find non-unique values from 3 arrays (or more):

ES2015

//          🚩🚩  🚩   🚩             🚩 
var arr =  [1,2,2,3,3,4,5,6,2,3,7,8,5,22],
    arr2 = [1,2,511,12,50],
    arr3 = [22,0],
    merged,
    nonUnique;

// Combine all the arrays to a single one
merged = arr.concat(arr2, arr3)

// create a new (dirty) Array with only the non-unique items
nonUnique = merged.filter((item,i) => merged.includes(item, i+1))

// Cleanup - remove duplicate & empty items items 
nonUnique = [...new Set(nonUnique)]

console.log(nonUnique)

---

PRE-ES2015:

In the below example I chose to superimpose a unique method on top of the Array prototype, allowing access from everywhere and has more "declarative" syntax. I do not recommend this approach on large projects, since it might very well collide with another method with the same custom name.

Array.prototype.unique = function () {
    var arr = this.sort(), i=arr.length; // input must be sorted for this to work
    while(i--)
      arr[i] === arr[i-1] && arr.splice(i,1) // remove duplicate item
    return arr
}

Array.prototype.nonunique = function () {
    var arr = this.sort(), i=arr.length, res = []; // input must be sorted for this to work
    while(i--)
      arr[i] === arr[i-1] && (res.indexOf(arr[i]) == -1) && res.push(arr[i]) 
    return res
}

//          🚩🚩  🚩    🚩            🚩 
var arr =  [1,2,2,3,3,4,5,6,2,3,7,8,5,22],
    arr2 = [1,2,511,12,50],
    arr3 = [22,0],
    // merge all arrays & call custom Array Prototype - "unique"
    unique = arr.concat(arr2, arr3).unique(),
    nonunique = arr.concat(arr2, arr3).nonunique()

console.log(unique)     // [1,12,2,22,3,4,5,50,511,6,7,8]
console.log(nonunique)  // [1,12,2,22,3,4,5,50,511,6,7,8]

using underscore.js

function hasDuplicate(arr){
    return (arr.length != _.uniq(arr).length);
}

Here's the simplest solution I could think of:

const arr = [-1, 2, 2, 2, 0, 0, 0, 500, -1, 'a', 'a', 'a']

const filtered = arr.filter((el, index) => arr.indexOf(el) !== index)
// => filtered = [ 2, 2, 0, 0, -1, 'a', 'a' ]

const duplicates = [...new Set(filtered)]

console.log(duplicates)
// => [ 2, 0, -1, 'a' ]

That's it.

Note:

1. It works with any numbers including 0, strings and negative numbers e.g. -1 - Related question: https://stackoverflow.com/questions/1960473

2. The original array arr is preserved (filter returns the new array instead of modifying the original)

3. The filtered array contains all duplicates; it can also contain more than 1 same value (e.g. our filtered array here is [ 2, 2, 0, 0, -1, 'a', 'a' ])

4. If you want to get only values that are duplicated (you don't want to have multiple duplicates with the same value) you can use [...new Set(filtered)] (ES6 has an object Set which can store only unique values)

Hope this helps.

The simplest and quickest way is to use the Set object:

const numbers = [1, 2, 3, 2, 4, 5, 5, 6];

const set = new Set(numbers);

const duplicates = numbers.filter(item => {
    if (set.has(item)) {
        set.delete(item);
    } else {
        return item;
    }
});

console.log(duplicates);
// [ 2, 5 ]

Here is mine simple and one line solution.

It searches not unique elements first, then makes found array unique with the use of Set.

So we have array of duplicates in the end.

var array = [1, 2, 2, 3, 3, 4, 5, 6, 2, 3, 7, 8, 5, 22, 1, 2, 511, 12, 50, 22];

console.log([...new Set(
  array.filter((value, index, self) => self.indexOf(value) !== index))]
);

This is my proposal (ES6):

let a = [1, 2, 3, 4, 2, 2, 4, 1, 5, 6]
let b = [...new Set(a.sort().filter((o, i) => o !== undefined && a[i + 1] !== undefined && o === a[i + 1]))]

// b is now [1, 2, 4]

one liner simple way

var arr = [9,1,2,4,3,4,9]
console.log(arr.filter((ele,indx)=>indx!==arr.indexOf(ele))) //get the duplicates
console.log(arr.filter((ele,indx)=>indx===arr.indexOf(ele))) //remove the duplicates

var a = [324,3,32,5,52,2100,1,20,2,3,3,2,2,2,1,1,1].sort();
a.filter(function(v,i,o){return i&&v!==o[i-1]?v:0;});

or when added to the prototyp.chain of Array

//copy and paste: without error handling
Array.prototype.unique = 
   function(){return this.sort().filter(function(v,i,o){return i&&v!==o[i-1]?v:0;});}

See here: https://gist.github.com/1305056

Fast and elegant way using es6 object destructuring and reduce

It runs in O(n) (1 iteration over the array) and doesn't repeat values that appear more than 2 times

const arr = ['hi', 'hi', 'hi', 'bye', 'bye', 'asd']
const {
  dup
} = arr.reduce(
  (acc, curr) => {
    acc.items[curr] = acc.items[curr] ? acc.items[curr] += 1 : 1
    if (acc.items[curr] === 2) acc.dup.push(curr)
    return acc
  }, {
    items: {},
    dup: []
  },
)

console.log(dup)
// ['hi', 'bye']

Shortest vanilla JS:

[1,1,2,2,2,3].filter((v,i,a) => a.indexOf(v) !== i) // [1, 2, 2]

You can use filter method and indexOf() to get all the duplicate values

function duplicate(arr) {
    return duplicateArray = arr.filter((item, index) => arr.indexOf(item) !== index) 
}

arr.indexOf(item) will always return the first index at which a given element can be found

Here is a very light and easy way:

var codes = dc_1.split(',');
var i = codes.length;
while (i--) {
  if (codes.indexOf(codes[i]) != i) {
    codes.splice(i,1);
  }
}

With ES6 (or using Babel or Typescipt) you can simply do:

var duplicates = myArray.filter(i => myArray.filter(ii => ii === i).length > 1);

https://es6console.com/j58euhbt/

I have just figured out a simple way to achieve this using an Array filter

    var list = [9, 9, 111, 2, 3, 4, 4, 5, 7];
    
    // Filter 1: to find all duplicates elements
    var duplicates = list.filter(function(value,index,self) {
       return self.indexOf(value) !== self.lastIndexOf(value) && self.indexOf(value) === index;
    });
    
    console.log(duplicates);

Simple code with ES6 syntax (return sorted array of duplicates):

let duplicates = a => {d=[]; a.sort((a,b) => a-b).reduce((a,b)=>{a==b&&!d.includes(a)&&d.push(a); return b}); return d};

How to use:

duplicates([1,2,3,10,10,2,3,3,10]);

This answer might also be helpful, it leverages js reduce operator/method to remove duplicates from array.

const result = [1, 2, 2, 3, 3, 3, 3].reduce((x, y) => x.includes(y) ? x : [...x, y], []);

console.log(result);

Higher ranked answers have a few inherent issues including the use of legacy javascript, incorrect ordering or with only support for 2 duplicated items.

Here's a modern solution which fixes those problems:

const arrayNonUniq = array => {
	if (!Array.isArray(array)) {
		throw new TypeError("An array must be provided!")
	}

	return array.filter((value, index) => array.indexOf(value) === index && array.lastIndexOf(value) !== index)
}

arrayNonUniq([1, 1, 2, 3, 3])
//=> [1, 3]

arrayNonUniq(["foo", "foo", "bar", "foo"])
//=> ['foo']

You can also use the npm package array-non-uniq.

The following function (a variation of the eliminateDuplicates function already mentioned) seems to do the trick, returning test2,1,7,5 for the input ["test", "test2", "test2", 1, 1, 1, 2, 3, 4, 5, 6, 7, 7, 10, 22, 43, 1, 5, 8]

Note that the problem is stranger in JavaScript than in most other languages, because a JavaScript array can hold just about anything. Note that solutions that use sorting might need to provide an appropriate sorting function--I haven't tried that route yet.

This particular implementation works for (at least) strings and numbers.

function findDuplicates(arr) {
	var i,
		len=arr.length,
		out=[],
		obj={};

	for (i=0;i<len;i++) {
		if (obj[arr[i]] != null) {
			if (!obj[arr[i]]) {
				out.push(arr[i]);
				obj[arr[i]] = 1;
			}
		} else {
			obj[arr[i]] = 0;			
		}
	}
	return out;
}

ES5 only (i.e., it needs a filter() polyfill for IE8 and below):

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

arrayToFilter.
    sort().
    filter( function(me,i,arr){
       return (i===0) || ( me !== arr[i-1] );
    });

This is probably one of the fastest way to remove permanently the duplicates from an array 10x times faster than the most functions here.& 78x faster in safari

function toUnique(a,b,c){//array,placeholder,placeholder
 b=a.length;
 while(c=--b)while(c--)a[b]!==a[c]||a.splice(c,1)
}
var array=[1,2,3,4,5,6,7,8,9,0,1,2,1];
toUnique(array);
console.log(array);

1. Test: http://jsperf.com/wgu
2. Demo: http://jsfiddle.net/46S7g/
3. More: https://stackoverflow.com/a/25082874/2450730

if you can't read the code above ask, read a javascript book or here are some explainations about shorter code. https://stackoverflow.com/a/21353032/2450730

EDIT As stated in the comments this function does return an array with uniques, the question however asks to find the duplicates. in that case a simple modification to this function allows to push the duplicates into an array, then using the previous function toUnique removes the duplicates of the duplicates.

function theDuplicates(a,b,c,d){//array,placeholder,placeholder
 b=a.length,d=[];
 while(c=--b)while(c--)a[b]!==a[c]||d.push(a.splice(c,1))
}
var array=[1,2,3,4,5,6,7,8,9,0,1,2,1];

toUnique(theDuplicates(array));

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

function returnDuplicates(arr) {
  return arr.reduce(function(dupes, val, i) {
    if (arr.indexOf(val) !== i && dupes.indexOf(val) === -1) {
      dupes.push(val);
    }
    return dupes;
  }, []);
}

alert(returnDuplicates(arr));

This function avoids the sorting step and uses the reduce() method to push duplicates to a new array if it doesn't already exist in it.

Using "includes" to test if the element already exists.

var arr = [1, 1, 4, 5, 5], darr = [], duplicates = [];

for(var i = 0; i < arr.length; i++){
  if(darr.includes(arr[i]) && !duplicates.includes(arr[i]))
    duplicates.push(arr[i])
  else
    darr.push(arr[i]);
}

console.log(duplicates);

<h3>Array with duplicates</h3>
<p>[1, 1, 4, 5, 5]</p>
<h3>Array with distinct elements</h3>
<p>[1, 4, 5]</p>
<h3>duplicate values are</h3>
<p>[1, 5]</p>

ES6 offers the Set data structure which is basically an array that doesn't accept duplicates. With the Set data structure, there's a very easy way to find duplicates in an array (using only one loop).

Here's my code

function findDuplicate(arr) {
var set = new Set();
var duplicates = new Set();
  for (let i = 0; i< arr.length; i++) {
     var size = set.size;
     set.add(arr[i]);
     if (set.size === size) {
         duplicates.add(arr[i]);
     }
  }
 return duplicates;
}

Following logic will be easier and faster

// @Param:data:Array that is the source 
// @Return : Array that have the duplicate entries
findDuplicates(data: Array<any>): Array<any> {
        return Array.from(new Set(data)).filter((value) => data.indexOf(value) !== data.lastIndexOf(value));
      }

Advantages :

1. Single line :-P
2. All inbuilt data structure helping in improving the efficiency
3. Faster

Description of Logic :

1. Converting to set to remove all duplicates
2. Iterating through the set values
3. With each set value check in the source array for the condition "values first index is not equal to the last index" == > Then inferred as duplicate else it is 'unique'

Note: map() and filter() methods are efficient and faster.

Just to add some theory to the above.

Finding duplicates has a lower bound of O(n*log(n) in the comparison model. SO theoretically, you cannot do any better than first sorting then going through the list sequentially removing any duplicates you find.

If you want to find the duplicates in linear (O(n)) expected time, you could hash each element of the list; if there is a collision, remove/label it as a duplicate, and continue.

I prefer the function way of doing this.

function removeDuplicates(links) {
    return _.reduce(links, function(list, elem) { 
        if (list.indexOf(elem) == -1) {
            list.push(elem);
        }   
        return list;
    }, []);
}

This uses underscore, but Array has a reduce function, too

var input = ['a', 'b', 'a', 'c', 'c'],
    duplicates = [],
    i, j;
for (i = 0, j = input.length; i < j; i++) {
  if (duplicates.indexOf(input[i]) === -1 && input.indexOf(input[i], i+1) !== -1) {
    duplicates.push(input[i]);
  }
}

console.log(duplicates);

I think the below is the easiest and fastest O(n) way to accomplish exactly what you asked:

function getDuplicates( arr ) {
  var i, value;
  var all = {};
  var duplicates = [];

  for( i=0; i<arr.length; i++ ) {
    value = arr[i];
    if( all[value] ) {
      duplicates.push( value );
      all[value] = false;
    } else if( typeof all[value] == "undefined" ) {
      all[value] = true;
    }
  }

  return duplicates;
}

Or for ES5 or greater:

function getDuplicates( arr ) {
  var all = {};
  return arr.reduce(function( duplicates, value ) {
    if( all[value] ) {
      duplicates.push(value);
      all[value] = false;
    } else if( typeof all[value] == "undefined" ) {
      all[value] = true;
    }
    return duplicates;
  }, []);
}

Modifying @RaphaelMontanaro's solution, borrowing from @Nosredna's blog, here is what you could do if you just want to identify the duplicate elements from your array.

function identifyDuplicatesFromArray(arr) {
        var i;
        var len = arr.length;
        var obj = {};
        var duplicates = [];

        for (i = 0; i < len; i++) {

            if (!obj[arr[i]]) {

                obj[arr[i]] = {};

            }

            else
            {
                duplicates.push(arr[i]);
            }

        }
        return duplicates;
    }

Thanks for the elegant solution, @Nosredna!

I did not like most answers.

Why? Too complicated, too much code, inefficient code and many do not answer the question, which is to find the duplicates (and not to give an array without the duplicates).

Next function returns all duplicates:

function GetDuplicates(arr) {
  var i, out=[], obj={};
  for (i=0; i < arr.length; i++) 
    obj[arr[i]] == undefined ? obj[arr[i]] ++ : out.push(arr[i]);
  return out;
}  

Because most of the time it is of no use to return ALL duplicates, but just to tell which duplicate values exist. In that case you return an array with unique duplicates ;-)

function GetDuplicates(arr) {
  var i, out=[], obj={};
  for (i=0; i < arr.length; i++)
    obj[arr[i]] == undefined ? obj[arr[i]] ++ : out.push(arr[i]);
  return GetUnique(out);
}

function GetUnique(arr) {
  return $.grep(arr, function(elem, index) {
    return index == $.inArray(elem, arr);
  });
}

Maybe somebody else thinks the same.

To solve the above in O(n) time complexity (without sorting).

var arr = [9, 9, 111, 2, 3, 4, 4, 5, 7];

var obj={};

for(var i=0;i<arr.length;i++){
    if(!obj[arr[i]]){
        obj[arr[i]]=1;
    } else {
        obj[arr[i]]=obj[arr[i]]+1;
    }
}
var result=[]
for(var key in obj){
    if(obj[key]>1){
        result.push(Number(key)) // change this to result.push(key) to find duplicate strings in an array
    }
}

console.log(result)

function GetDuplicates(arr) {
	var i = 0, m = [];
	return arr.filter(function (n) {
		return !m[n] * ~arr.indexOf(n, m[n] = ++i);
	});
}

I feel like the simplest solution would to just use indexOf

full example of pushing only unique elements to an array.

var arr = ['a','b','c','d','a','b','c','d'];
var newA = [];
for(var i = 0; i < arr.length; i++){
    if(newA.indexOf(arr[i]) === -1){
        newA.push(arr[i]);
    }
 }

Magic

a.filter(( t={}, e=>!(1-(t[e]=++t[e]|0)) )) 

O(n) performance; we assume your array is in a and it contains elements that can be cast .toString() in unique way (which is done implicity by JS in t[e]) e.g numbers=[4,5,4], strings=["aa","bb","aa"], arraysNum=[[1,2,3], [43,2,3],[1,2,3]]. Explanation here, unique values here

var a1 = [[2, 17], [2, 17], [2, 17], [1, 12], [5, 9], [1, 12], [6, 2], [1, 12]];
var a2 = ['Mike', 'Adam','Matt', 'Nancy', 'Adam', 'Jenny', 'Nancy', 'Carl'];
var a3 = [5,6,4,9,2,3,5,3,4,1,5,4,9];

let nd = (a) => a.filter((t={},e=>!(1-(t[e]=++t[e]|0)))) 


// Print
let c= x => console.log(JSON.stringify(x));             
c( nd(a1) );
c( nd(a2) );
c( nd(a3) );

There is a really simple way to solve this. If you use the newish 'Set' javascript command. Set can take an array as input and output a new 'Set' that only contains unique values. Then by comparing the length of the array and the 'size' property of the set you can see if they differ. If they differ it must be due to a duplicate entry.

var array1 = ['value1','value2','value3','value1']; // contains duplicates
var array2 = ['value1','value2','value3','value4']; // unique values

console.log('array1 contains duplicates = ' + containsDuplicates(array1));
console.log('array2 contains duplicates = ' + containsDuplicates(array2));


function containsDuplicates(passedArray) {
  let mySet = new Set(passedArray);
  if (mySet.size !== passedArray.length) {
    return true;
  }
  return false;
}

If you run the above snippet you will get this output.

array1 contains duplicates = true

array2 contains duplicates = false

Yet another way by using underscore. Numbers is the source array and dupes has possible duplicate values.

var itemcounts = _.countBy(numbers, function (n) { return n; });
var dupes = _.reduce(itemcounts, function (memo, item, idx) {
    if (item > 1)
        memo.push(idx);
    return memo;
}, []);

Similar to a few other answers, but I used forEach() to make it a bit prettier:

function find_duplicates(data) {

    var track = {};
    var duplicates = [];

    data.forEach(function (item) {
        !track[item] ? track[item] = true : duplicates.push(item);
    });

    return duplicates;
}

If a value is duplicated more than once, all its duplicates are returned, like so:

find_duplicates(['foo', 'foo', 'bar', 'bar', 'bar']);
// returns ['foo', 'bar', 'bar']

This might be what you want, otherwise you just have to follow with an "unique" filtering.

The quickest way to solve is is actually with a flag

var values = [4,2,3,1,4]

// solution
const checkDuplicate = list => {
  var hasDuplicate = false;
  list.sort().sort((a, b) => {
    if (a === b) hasDuplicate = true
  })
  return hasDuplicate
}

console.log(checkDuplicate(values))

This can also be solved using Set().

A value in the Set may only occur once; it is unique in the Set's collection.

    Array.prototype.hasDuplicates = function () {
        if (arr.length !== (new Set(arr).size)) {
            return true;
        }
        return false;
    }

More information on Sets: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Set

Note: Sets are not fully supported in IE.

This is one of the simple ES5 solution I could think of -

function duplicates(arr) {
  var duplicatesArr = [],
      uniqueObj = {};

  for (var i = 0; i < arr.length; i++) {
    if( uniqueObj.hasOwnProperty(arr[i]) && duplicatesArr.indexOf( arr[i] ) === -1) {
      duplicatesArr.push( arr[i] );
    }
    else {
      uniqueObj[ arr[i] ] = true;
    }
  }

  return duplicatesArr;
}
/* Input Arr: [1,1,2,2,2,1,3,4,5,3] */
/* OutPut Arr: [1,2,3] */

//find duplicates:
//sort, then reduce - concat values equal previous element, skip others

//input
var a = [1, 2, 3, 1, 2, 1, 2]

//short version:
var duplicates = a.sort().reduce((d, v, i, a) => i && v === a[i - 1] ? d.concat(v) : d, [])
console.log(duplicates); //[1, 1, 2, 2]

//readable version:
var duplicates = a.sort().reduce((output, element, index, input) => {
  if ((index > 0) && (element === input[index - 1]))
    return output.concat(element)
  return output
}, [])
console.log(duplicates); //[1, 1, 2, 2]

1. Printing duplicate values

 var arr = [1,2,3,4,13,2,3,4,3,4];

    // non_unique Printing 
    function nonUnique(arr){
    var result = [];
    for(var i =0;i<arr.length;i++){
        if(arr.indexOf(arr[i],i+1) > -1){
            result.push(arr[i]);
        }
    }
    console.log(result);
    }nonUnique(arr);

    // unique Printing
    function uniqueDuplicateVal(arr){
       var result = [];
       for(var i =0;i<arr.length;i++){
        if(arr.indexOf(arr[i],i+1) > -1){
          if(result.indexOf(arr[i]) === -1]){
             result.push(arr[i]);
          }
        }
       }    
    }
    uniqueDuplicateVal(arr)

This should be one of the shortest and easiest ways to actually find duplicate values in an array.

var arr = [1,2,3,4,5,6,7,8,1,2,3,4,5,3,3,4];
var data = arr.filter(function(item,index,arr){
  return arr.indexOf(item) != arr.lastIndexOf(item) && arr.indexOf(item) == index;
})

console.log(data );

This is most efficient way i can think of as doesn't include Array.indexOf() or Array.lastIndexOf() which have complexity of O(n) and using inside any loop of complexity O(n) will make complete complexity O(n^2).

My first loop have complexity of O(n/2) or O((n/2) + 1), as complexity of search in hash is O(1). The second loop worst complexity when there's no duplicate in array is O(n) and best complexity when every element have a duplicate is O(n/2).

function duplicates(arr) {
  let duplicates = [],
      d = {},
      i = 0,
      j = arr.length - 1;

  // Complexity O(n/2)
  while (i <= j) {
    if (i === j)
      d[arr[i]] ? d[arr[i]] += 1 : d[arr[i]] = 1;  // Complexity O(1)
    else {
      d[arr[i]] ? d[arr[i]] += 1 : d[arr[i]] = 1;  // Complexity O(1)
      d[arr[j]] ? d[arr[j]] += 1 : d[arr[j]] = 1;  // Complexity O(1)
    }

    ++i;
    --j;
  }

  // Worst complexity O(n), best complexity O(n/2)
  for (let k in d) {
    if (d[k] > 1)
      duplicates.push(k);
  }

  return duplicates;

}

console.log(duplicates([5,6,4,9,2,3,5,3,4,1,5,4,9]));
console.log(duplicates([2,3,4,5,4,3,4]));
console.log(duplicates([4,5,2,9]));
console.log(duplicates([4,5,2,9,2,5,9,4]));

This is a single loop approach with a hash table for counting the elements and filter the array if the count is 2, because it returns the first found duplicate.

Advantage:

• single loop
• uses an object for counting in a closure

var array = [5, 0, 2, 1, 2, 3, 3, 4, 4, 8, 6, 7, 9, 4], duplicates = array.filter((h => v => (h[v] = (h[v] || 0) + 1) === 2)({}));

console.log(duplicates);

You can make use of the sort, filter and sets to do that.

var numbers = [1,2,3,4,5,6,7,8,1,2,3,4,5,3,3,4];
var numbersSorted = numbers.sort();
let result = numbers.filter((e, i) => numbers[i] == numbers[i+1]);
result = [...new Set(result)];
console.log(result);

Based on @bluemoon but shorter, returns all duplicates exactly once!

function checkDuplicateKeys(arr) {
    const counts = {}
    return arr.filter((item) => {
        counts[item] = counts[item] || 1
        if (counts[item]++ === 2) return true
    })
}

// [1,2,2,2,2,2,2] => [1,2]
// ['dog', 'dog', 'cat'] => ['dog']

const names = [
  "Alex",
  "Matt",
  12,
  "You",
  "Me",
  12,
  "Carol",
  "Bike",
  "Carol",
];

const count = (names) =>
  names.reduce((a, b) => ({ ...a, [b]: (a[b] || 0) + 1 }), {});

let obj = count(names);
let objectKeys = Object.keys(obj);

let repetitiveElements = [];
let answer = objectKeys.map((value) => {
  if (obj[value] > 1) {
    return repetitiveElements.push(value);
  }
});
console.log(repetitiveElements); 

There are so many answers already, but unfortunately some are too long, some are short but too cryptic for me while others are beyond my scope of knowledge... I really like this solution that I've come up with, though. Hope it's still helpful for some!

Even though the original post says he/she doesn't actually need the duplicates' indexes or how many times they are duplicated, I think it's still clearest to have 'em counted.

Codes with notes.

function findDuplicates(array, count = {}) {
  // with count declared in the parameter, initialized as an empty object, 
  // it can store the counts of all elements in array  
  
  // using the forEach loop to iterate through the input array, 
  // also using the conditional ternary operators 
  // (works just like a normal if-else statement, but just a bit cleaner)
  // we can store all occurrences of each element from array in count
  array.forEach(el => count[el] ? count[el]++ : count[el] = 1)
  
  // using Object.keys, we get an array of all keys from count (all numbers) 
  // (sorted as well, though of no specific importance here)
  // using filter to find all elements with a count (value) > 1 (duplicates!)
  return Object.keys(count).filter(key => count[key] > 1);
}

Just the codes (with test cases).

function findDuplicates(array, count = {}) {
  array.forEach(el => count[el] ? count[el]++ : count[el] = 1);
  return Object.keys(count).filter(key => count[key] > 1);
}

let arr1 = [9, 9, 111, 2, 3, 4, 4, 5, 7];
let arr2 = [1,6,7,3,6,8,1,3,4,5,1,7,2,6];
console.log(findDuplicates(arr1)); // => ['4', '9']
console.log(findDuplicates(arr2)); // => ['1', '3', '6', '7']

It's actually a shame that this question has so many wrong answers or answers which need a lot of extra memory like a Set. Clean and simple solution:

function findDuplicates<T>(arr: Array<T>): T[] {
  //If the array has less than 2 elements there are no duplicates
  const n = arr.length;
  if (n < 2)
    return [];
  
  const sorted = arr.sort();
  const result = [];

  //Head
  if (sorted[0] === sorted[1])
    result.push(sorted[0]);

  //Inner (Head :: Inner :: Tail)
  for (let i = 1; i < n-1; i++) {
    const elem = sorted[i];
    if (elem === sorted[i - 1] || elem === sorted[i+1])
      result.push(elem)
  }

  //Tail
  if (sorted[n - 1] == sorted[n - 2])
    result.push(sorted[n - 1]);

  return result;
}

console.dir(findDuplicates(['a', 'a', 'b', 'b']));
console.dir(findDuplicates(['a', 'b']));
console.dir(findDuplicates(['a', 'a', 'a']));
console.dir(findDuplicates(['a']));
console.dir(findDuplicates([]));

The Prototype library has a uniq function, which returns the array without the dupes. That's only half of the work though.

/* The indexOf method of the Array object is useful for comparing array items. IE is the only major browser that does not natively support it, but it is easy to implement: */

Array.prototype.indexOf= Array.prototype.indexOf || function(what, i){
	i= i || 0;
	var L= this.length;
	while(i<L){
		if(this[i]=== what) return i;
		++i;
	}
	return -1;
}

function getarrayduplicates(arg){
	var itm, A= arg.slice(0, arg.length), dups= [];
	while(A.length){
		itm= A.shift();
		if(A.indexOf(itm)!= -1 && dups.indexOf(itm)== -1){
			dups[dups.length]= itm;
		}
	}
	return dups;
}

var a1= [1, 22, 3, 2, 2, 3, 3, 4, 1, 22, 7, 8, 9];

alert(getarrayduplicates(a1));

For very large arrays, it can be faster to remove the duplicates from the array as they are found, so that they will not be looked at again:

function getarrayduplicates(arg){
	var itm, A= arg.slice(0, arg.length), dups= [];
	while(A.length){
		itm= A.shift();
		if(A.indexOf(itm)!= -1){
			dups[dups.length]= itm;
			while(A.indexOf(itm)!= -1){
				A.splice(A.indexOf(itm), 1);
			}
		}
	}
	return dups;
}

From Raphael Montanaro answer, it can improve to use with array/object item as follows.

function eliminateDuplicates(arr) {
  var len = arr.length,
      out = [],
      obj = {};

  for (var key, i=0; i < len; i++) {
    key = JSON.stringify(arr[i]);
    obj[key] = (obj[key]) ? obj[key] + 1 : 1;
  }
  for (var key in obj) {
    out.push(JSON.parse(key));
  }
  return [out, obj];
}

Note: You need to use JSON library for browser that's not supported JSON.

http://jsfiddle.net/vol7ron/gfJ28/

var arr  = ['hello','goodbye','foo','hello','foo','bar',1,2,3,4,5,6,7,8,9,0,1,2,3];
var hash = [];

// build hash
for (var n=arr.length; n--; ){
   if (typeof hash[arr[n]] === 'undefined') hash[arr[n]] = [];
   hash[arr[n]].push(n);
}


// work with compiled hash (not necessary)
var duplicates = [];
for (var key in hash){
    if (hash.hasOwnProperty(key) && hash[key].length > 1){
        duplicates.push(key);
    }
}    
alert(duplicates);

1. The result will be the hash array, which will contain both a unique set of values and the position of those values. So if there are 2 or more positions, we can determine that the value has a duplicate. Thus, every place hash[<value>].length > 1, signifies a duplicate.

2. hash['hello'] will return [0,3] because 'hello' was found in node 0 and 3 in arr[].

Note: the length of [0,3] is what's used to determine if it was a duplicate.

3. Using for(var key in hash){ if (hash.hasOwnProperty(key)){ alert(key); } } will alert each unique value.

Here is the one of methods to avoid duplicates into javascript array...and it supports for strings and numbers...

 var unique = function(origArr) {
	var newArray = [],
		origLen = origArr.length,
		found,
		x = 0; y = 0;
		
	for ( x = 0; x < origLen; x++ ) {
		found = undefined;
		for ( y = 0; y < newArray.length; y++ ) {
			if ( origArr[x] === newArray[y] ) found = true;
		}
		if ( !found) newArray.push( origArr[x] );    
	}
   return newArray;
}

check this fiddle..

I am trying to improve the answer from @swilliams, this will return an array without duplicates.

// arrays for testing
var arr = [9, 9, 111, 2, 3, 4, 4, 5, 7];

// ascending order
var sorted_arr = arr.sort(function(a,b){return a-b;}); 

var arr_length = arr.length;
var results = [];
if(arr_length){
    if(arr_length == 1){
        results = arr;
    }else{
        for (var i = 0; i < arr.length - 1; i++) {
            if (sorted_arr[i + 1] != sorted_arr[i]) {
                results.push(sorted_arr[i]);
            }
            // for last element
            if (i == arr.length - 2){
                results.push(sorted_arr[i+1]);
            }
        }
    }
}

alert(results);

Here is one implemented using sort() and JSON.stringify()

https://gist.github.com/korczis/7598657

function removeDuplicates(vals) {
	var res = [];
	var tmp = vals.sort();
	
	for (var i = 0; i < tmp.length; i++) {
		res.push(tmp[i]);
                	while (JSON.stringify(tmp[i]) == JSON.stringify(tmp[i + 1])) {
			i++;
		}
	}
	
	return res;
}
console.log(removeDuplicates([1,2,3,4,5,4,3,3,2,1,]));

here is a small simple snippet to find unique and duplicate values with out sorting and two loops.

var _unique = function (arr) {
    var h = [], t = [];
    arr.forEach(function (n) {
        if (h.indexOf(n) == -1)
            h.push(n);
        else t.push(n);
    });
    return [h, t];
}
var result = _unique(["test",1,4,2,34,6,21,3,4,"test","prince","th",34]);
console.log("Has duplicate values : " + (result[1].length > 0))  //and you can check count of duplicate values
console.log(result[0]) //unique values
console.log(result[1]) //duplicate values

You can use the following construction:

var arr = [1,2,3,4,5,6,7,8,9,0,5];
var duplicate = arr.filter(function(item, i, arr) {
  return -1 !== arr.indexOf(item, i + 1);
})

using Pure Js

function arr(){
  var a= [1,2,3,4,5,34,2,5,7,8,6,4,3,25,8,34,56,7,8,76,4,23,34,46,575,8564,53,5345657566];
  var b= [];
  b.push(a[0]);
  var z=0;

  for(var i=0; i< a.length; i++){
      for(var j=0; j< b.length; j++){
        if(b[j] == a[i]){
          z = 0;
          break;
        }
        else
          z = 1;
      }
      if(z)
        b.push(a[i]);
    }
  console.log(b);
}

var isUnique = true;      
for (var i= 0; i< targetItems.length; i++) {
        var itemValue = $(targetItems[i]).val();
        if (targetListValues.indexOf(itemValue) >= 0) {
          isUnique = false;
           break;
        }
      targetListValues.push(itemValue);
        if (!isUnique) {
          //raise any error msg
          return false;
        }
      }

Here we output the duplicates just once per dupe.

var arr = [9, 9, 9, 9, 111, 2, 3, 4, 4, 5, 7];
arr.sort(); 

var results = [];
for (var i = 0; i < arr.length - 1; i++) {
    if (arr[i + 1] == arr[i]) {
        results.push(arr[i]);
    }
}

results = Array.from(new Set(results))

console.log(results);

This is more advanced function based on number of occurences.

function getMostDuplicated(arr, count) {
      const result = [];
      arr.forEach((item) => {
        let i = 0;
        arr.forEach((checkTo) => {
          if (item === checkTo) {
            i++;
            if (i === count && result.indexOf(item) === -1) {
              result.push(item);
            }
          }
        });
      });
      return result;
}

arr = [1,1,1,2,5,67,3,2,3,2,3,1,2,3,4,1,4];
result = getMostDuplicated(arr, 5); // result is 1
result = getMostDuplicated(arr, 2); // result 1, 2, 3, 4
console.log(result);

This is how I implemented it with map. It should run in O(n) time and should kinda be easy to gasp.

    var first_array=[1,1,2,3,4,4,5,6];
    var find_dup=new Map;

    for (const iterator of first_array) {
            // if present value++
            if(find_dup.has(iterator)){ 
                find_dup.set(iterator,find_dup.get(iterator)+1); 
            }else{
            // else add it
                find_dup.set(iterator,1);
            }
        }
    console.log(find_dup.get(2));

Then you can find_dup.get(key) to find if it has duplicates (it should give > 1).

Returns duplicates and preserves data type.

With O(4n) performance

const dupes = arr => {
  const map = arr.reduce((map, curr) => {
    return (map.set(curr, (map.get(curr) || 0) + 1), map)
  }, new Map());

  return Array.from(map).filter(([key, val])=> val > 1).map(([key, val]) => key)
}

With O(2n) performance

const dupes = arr => {
  const map = arr.reduce((map, curr) => {
    return (map.set(curr, (map.get(curr) || 0) + 1), map)
  }, new Map());

  const dupes_ = [];
  for (let [key, val] of map.entries()) {
    if (val > 1) dupes_.push(key);
  }
  return dupes_;
}

Simplest way to fetch duplicates/repeated values from array/string :

function getDuplicates(param) {
  var duplicates = {}

  for (var i = 0; i < param.length; i++) {
    var char = param[i]
    if (duplicates[char]) {
      duplicates[char]++
    } else {
      duplicates[char] = 1
    }
  }
  return duplicates
}

console.log(getDuplicates("aeiouaeiou"));
console.log(getDuplicates(["a", "e", "i", "o", "u", "a", "e"]));
console.log(getDuplicates([1, 2, 3, 4, 5, 1, 1, 2, 3]));

This will return duplicates from an Array as an Array of duplicates.

    const duplicates = function(arr) {
      // let try moving in pairs.. maybe that will work
      let dups = new Set(),
          r = []
      arr.sort()
      arr.reduce((pv, cv) => {
        if (pv === cv) {
          dups.add(pv)
        }
        return cv
      })
      for (let m of dups.values()) {
        r.push(m)
      }
      return r
    }
    
    console.log(duplicates([1,3,5,6,7,4,4,5,1,4,6,3,8,9,5,0]))

Very simple way:

function getDuplicateValues(someArray) {
	const duplicateValues = new Set([])
	const check = new Set([])
	someArray.forEach(v => {
		if (check.has(v)) {
			duplicateValues.add(v)
		} else {
			check.add(v)
		}
	})
	return Array.from(duplicateValues);
}

const result = getDuplicateValues(['coffee', 'soda', 'water', 'juice', 'water', 'water', 'coffee'])

repeated_values.textContent = JSON.stringify(result, null, '  ')

<pre id="repeated_values"></pre>

The accepted answer is the most perfect one but as some users has pointed that for cases where an element is repeated more than 2 times it will gives us the array with repeated elements:

This solution covers that scenarios too::

const peoples = [
  {id: 1, name:"Arjun"},
  {id: 2, name:"quinze"},
  {id: 3, name:"catorze"},
  {id: 1, name:"Arjun"},
  {id: 4, name:"dezesseis"},
  {id: 1, name:"Arjun"},
  {id: 2, name:"quinze"},
  {id: 3, name:"catorzee"}
]


function repeated(ppl){

  const newppl = ppl.slice().sort((a,b) => a.id -b.id);

  let rept = [];
  for(let i = 0; i < newppl.length-1 ; i++){
    if (newppl[i+1].id == newppl[i].id){
      rept.push(newppl[i+1]);
    }
  }

  return [...new Set(rept.map(el => el.id))].map(rid => 
    rept.find(el => el.id === rid)
  );

}

repeated(peoples);

This was asked me in an interview, My answer is,

List<int> getDublicates(List<int> x)
{
   List<int> result = new List<int>();
   while (x.Count>0)
   {
      int d = x[0];
      x.Remove(x[0]);
      if (x.Contains(d)) 
      result.Add(d);
   }
   return result;
}

it have good performance

The shortest way to remove duplicates is by using Set and Spread syntax

const remove = (array) => [...new Set(array)];
console.log(remove([1,1,2,2,3]); //1,2,3

[1, 2, 2, 3, 3, 4, 5, 6, 2, 3, 50, 8, 5, 22, 1, 2, 511, 12, 50, 22].reduce(function (total, currentValue, currentIndex, arr) {
    if (total.indexOf(currentValue) === -1 && arr.indexOf(currentValue) !== currentIndex) 
        total.push(currentValue);
    return total;
}, [])

const a = ['a', 'b', 'b']

function findDuplicates(a) {
    return Object.keys(_.pickBy(_.countBy(a), x => x > 1))
}

https://runkit.com/nitzanav/62659f09bd2d5f0008ef46d4

You can use the following code to get the duplicate elements in a given array:

let name = ['satya', 'amit', 'aditya', 'abhay', 'satya', 'amit'];
let dup = [];
let uniq = [];
name.forEach((item, index) => {
  if(!uniq.includes(item)) {
    uniq[index] = item;
  }
  if (name.indexOf(item, index + 1) != -1) {
    dup[index] = item;
  }
})

Here's one without using a temp Array to store the non-duplicate ones:

// simple duplicate removal for non-object types
Array.prototype.removeSimpleDupes = function() {
	var i, cntr = 0, arr = this, len = arr.length;
	
	var uniqueVal = function(val,n,len) { // remove duplicates
		var dupe = false;
			for (i = n; i < len; i++) { 
				if (typeof arr[i]!=="undefined" && val===arr[i]) { arr.splice(i,1); dupe = true; }
			}
		return (dupe) ? arr.length : len;
	};
	
	while (cntr < len) {
		len = uniqueVal(arr[cntr],cntr+1,len);
		cntr++;
	}
	
	return arr;
};

function remove_dups(arrayName){
  var newArray = new Array();

  label:for(var i=0; i<arrayName.length; i++ ){  

     for(var j=0; j<newArray.length;j++ ){
       if(newArray[j]==arrayName[i]){
         continue label;
       }
     }

     newArray[newArray.length] = arrayName[i];

  }

  return newArray;
}

Surprised no one posted this solution.

<!DOCTYPE html>
<html>
<head>
<meta charset=utf-8 />
<title>
</title>
</head>
<body>
  <script>
       var list = [100,33,45,54,9,12,80,100];
       var newObj = {};
       var newArr = [];
        for(var i=0; i<list.length; i++){
          newObj[list[i]] = i;               
        }
        for(var j in newObj){
            newArr.push(j);  
        }
       console.log(newArr);
  </script>
</body>
</html>

var arr = [4,5,1,1,2,3,4,4,7,5,2,6,10,9];
var sorted_arr = arr.sort();
var len = arr.length;
var results = [];
for (var i = 0; i < len; i++) {
  if (sorted_arr[i + 1] !== sorted_arr[i]) {
    results.push(sorted_arr[i]);
  }
}
document.write(results);

var a= [1, 2,2,3,3,4,4,4];
var m=[];
var n = [];
a.forEach(function(e) {
  if(m.indexOf(e)=== -1) {
    m.push(e);
}else if(n.indexOf(e)=== -1){
    n.push(e);
}

});

You can proceed by comparing index:

function getDuplicate(array) {
    return array.filter((value, index) => array.value !== index)
}

We will use Javascript ES6 Functionality to do magic!

var arr = [9, 9, 111, 2, 3, 4, 4, 5, 7];
const filtered = arr.filter((value, index) => {
 return arr.indexOf(value) >= index;
});

console.log(filtered);

https://jsfiddle.net/97Lxupnz/

var arr = ['a','b','c','a'];

arr.filter( (item , index ) => {  
console.log(item , index , arr.indexOf(item) , arr.indexOf( item ) == index);
return index == arr.indexOf(item)
 } );

var array = ['a', 'b', 'c', 'a'];

function unique(array) {
    var unique_arr = [];
    array.forEach(function(i, e) {
        if (unique_arr.indexOf(i)===-1) unique_arr.push(i);
    });
    return unique_arr;
}
console.log(unique(array));

In this post was useful for duplication check if u are using Jquery.

https://stackoverflow.com/questions/4346547/how-to-find-the-duplicates-in-an-array-using-jquery

var unique_values = {}; var list_of_values = []; $('input[name$="recordset"]').     each(function(item) {          if ( ! unique_values[item.value] ) {             unique_values[item.value] = true;             list_of_values.push(item.value);         } else {             // We have duplicate values!         }     });

//program to find the duplicate elements in arraylist

import java.util.ArrayList;
import java.util.Scanner;

public class DistinctEle 
{ 
    public static void main(String args[])
    {
        System.out.println("Enter elements");
        ArrayList<Integer> abc=new ArrayList<Integer>();
        ArrayList<Integer> ab=new ArrayList<Integer>();
        Scanner a=new Scanner(System.in);
        int b;
        for(int i=0;i<=10;i++)
        {
            b=a.nextInt();
            if(!abc.contains(b))
            {
                abc.add(b);
            }
            else
            {
                System.out.println("duplicate elements"+b);
            }
        }
    }
}