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fastest way to detect if duplicate entry exists in javascript array?

var arr = ['test0','test2','test0'];


Like the above,there are two identical entries with value "test0",how to check it most efficiently?

8 Answers
8

If you sort the array, the duplicates are next to each other so that they are easy to find:

arr.sort();
var last = arr[0];
for (var i=1; i<arr.length; i++)
if (arr[i] == last) alert('Duplicate : '+last);
last = arr[i];



===

This will do the job on any array and is probably about as optimized as possible for handling the general case (finding a duplicate in any possible array). For more specific cases (e.g. arrays containing only strings) you could do better than this.

function hasDuplicate(arr)
var i = arr.length, j, val;

while (i--)
val = arr[i];
j = i;
while (j--)
if (arr[j] === val)
return true;



return false;



Loop stops when found first duplicate:

function has_duplicates(arr)

var x = , len = arr.length;
for (var i = 0; i < len; i++)
if (x[arr[i]])
return true;

x[arr[i]] = true;

return false;




Edit (fix 'toString' issue):

function has_duplicates(arr)

var x = , len = arr.length;
for (var i = 0; i < len; i++)
if (x[arr[i]] === true)
return true;

x[arr[i]] = true;

return false;




this will correct for case has_duplicates(['toString']); etc..

has_duplicates(['toString'])

true

[ a: 1, b: 2 ]

x

There are lots of answers, here but not all of them "feel" nice...

If you are using lodash:

function containsDuplicates(array)
return _.uniq(array).length !== array.length;



If you can use ES6 Sets, it simply becomes:

function containsDuplicates(array)
return array.length !== new Set(array).length



With vanilla javascript:

function containsDuplicates(array)
return array
.sort()
.some(function (item, i, items)
return item === items[i + 1]
)



However, sometimes you may want to check if the items are duplicated on a certain field.

This is how I'd handle that:

containsDuplicates([country: 'AU', country: 'UK', country: 'AU'], 'country')

function containsDuplicates(array, attribute)
return array
.map(function (item) return item[attribute] )
.sort()
.some(function (item, i, items)
return item === items[i + 1]
)



var index = myArray.indexOf(strElement);
if (index < 0)
myArray.push(strElement);
console.log("Added Into Array" + strElement);
else
console.log("Already Exists at " + index);



Sorting is O(n log n) and not O(n). Building a hash map is O(n). It costs more memory than an in-place sort but you asked for the "fastest." (I'm positive this can be optimized but it is optimal up to a constant factor.)

function hasDuplicate(arr)
var hash = ;
var hasDuplicate = false;
arr.forEach(function(val)
if (hash[val])
hasDuplicate = true;
return;

hash[val] = true;
);
return hasDuplicate;



You can convert the array to to a Set instance, then convert to an array and check if the length is same before and after the conversion.

Set

const hasDuplicates = (array) =>
const arr = ['test0','test2','test0'];
const set1 = new Set(array);
const uniqueArray = [...set1];

return array.length !== uniqueArray.length;
;

console.log(`Has duplicates : $hasDuplicates(['test0','test2','test0'])`);
console.log(`Has duplicates : $hasDuplicates(['test0','test2','test3'])`);

Assuming all you want is to detect how many duplicates of 'test0' are in the array. I guess an easy way to do that is to use the join method to transform the array in a string, and then use the match method.

var arr= ['test0','test2','test0'];
var str = arr.join();

console.log(str) //"test0,test2,test0"

var duplicates = str.match(/test0/g);
var duplicateNumber = duplicates.length;

console.log(duplicateNumber); //2


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