Generate unique random numbers between 1 and 100

Question

How can I generate some unique random numbers between 1 and 100 using JavaScript?

Answer 1

For example: To generate 8 unique random numbers and store them to an array, you can simply do this:

var arr = [];
while(arr.length < 8){
    var r = Math.floor(Math.random() * 100) + 1;
    if(arr.indexOf(r) === -1) arr.push(r);
}
console.log(arr);

Answer 2

1. Populate an array with the numbers 1 through 100.
2. Shuffle it.
3. Take the first 8 elements of the resulting array.

Answer 3

Modern JS Solution using Set (and average case O(n))

const nums = new Set();
while(nums.size !== 8) {
  nums.add(Math.floor(Math.random() * 100) + 1);
}

console.log([...nums]);

Answer 4

Another approach is to generate an 100 items array with ascending numbers and sort it randomly. This leads actually to a really short and (in my opinion) simple snippet.

const numbers = Array(100).fill().map((_, index) => index + 1);
numbers.sort(() => Math.random() - 0.5);
console.log(numbers.slice(0, 8));

Answer 5

Generate permutation of 100 numbers and then choose serially.

Use Knuth Shuffle(aka the Fisher-Yates shuffle) Algorithm.

JavaScript:

  function fisherYates ( myArray,stop_count ) {
  var i = myArray.length;
  if ( i == 0 ) return false;
  int c = 0;
  while ( --i ) {
     var j = Math.floor( Math.random() * ( i + 1 ) );
     var tempi = myArray[i];
     var tempj = myArray[j];
     myArray[i] = tempj;
     myArray[j] = tempi;

     // Edited thanks to Frerich Raabe
     c++;
     if(c == stop_count)return;

   }
}

CODE COPIED FROM LINK.

EDIT:

Improved code:

function fisherYates(myArray,nb_picks)
{
    for (i = myArray.length-1; i > 1  ; i--)
    {
        var r = Math.floor(Math.random()*i);
        var t = myArray[i];
        myArray[i] = myArray[r];
        myArray[r] = t;
    }

    return myArray.slice(0,nb_picks);
}

Potential problem:

Suppose we have array of 100 numbers {e.g. [1,2,3...100]} and we stop swapping after 8 swaps; then most of the times array will look like {1,2,3,76,5,6,7,8,...numbers here will be shuffled ...10}.

Because every number will be swapped with probability 1/100 so prob. of swapping first 8 numbers is 8/100 whereas prob. of swapping other 92 is 92/100.

But if we run algorithm for full array then we are sure (almost)every entry is swapped.

Otherwise we face a question : which 8 numbers to choose?

Answer 6

The above techniques are good if you want to avoid a library, but depending if you would be alright with a library, I would suggest checking out Chance for generating random stuff in JavaScript.

Specifically to solve your question, using Chance it's as easy as:

// One line!
var uniques = chance.unique(chance.natural, 8, {min: 1, max: 100});

// Print it out to the document for this snippet so we can see it in action
document.write(JSON.stringify(uniques));

<script src="http://chancejs.com/chance.min.js"></script>

Disclaimer, as the author of Chance, I am a bit biased ;)

Answer 7

To avoid any long and unreliable shuffles, I'd do the following...

1. Generate an array that contains the number between 1 and 100, in order.
2. Generate a random number between 1 and 100
3. Look up the number at this index in the array and store in your results
4. Remove the elemnt from the array, making it one shorter
5. Repeat from step 2, but use 99 as the upper limit of the random number
6. Repeat from step 2, but use 98 as the upper limit of the random number
7. Repeat from step 2, but use 97 as the upper limit of the random number
8. Repeat from step 2, but use 96 as the upper limit of the random number
9. Repeat from step 2, but use 95 as the upper limit of the random number
10. Repeat from step 2, but use 94 as the upper limit of the random number
11. Repeat from step 2, but use 93 as the upper limit of the random number

Voila - no repeated numbers.

I may post some actual code later, if anybody is interested.

Edit: It's probably the competitive streak in me but, having seen the post by @Alsciende, I couldn't resist posting the code that I promised.

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 3.2 Final//EN">
<html>
<head>
<title>8 unique random number between 1 and 100</title>
<script type="text/javascript" language="Javascript">
    function pick(n, min, max){
        var values = [], i = max;
        while(i >= min) values.push(i--);
        var results = [];
        var maxIndex = max;
        for(i=1; i <= n; i++){
            maxIndex--;
            var index = Math.floor(maxIndex * Math.random());
            results.push(values[index]);
            values[index] = values[maxIndex];
        }
        return results;
    }
    function go(){
        var running = true;
        do{
            if(!confirm(pick(8, 1, 100).sort(function(a,b){return a - b;}))){
                running = false;
            }
        }while(running)
    }
</script>
</head>

<body>
    <h1>8 unique random number between 1 and 100</h1>
    <p><button onclick="go()">Click me</button> to start generating numbers.</p>
    <p>When the numbers appear, click OK to generate another set, or Cancel to stop.</p>
</body>

Answer 8

I would do this:

function randomInt(min, max) {
    return Math.round(min + Math.random()*(max-min));
}
var index = {}, numbers = [];
for (var i=0; i<8; ++i) {
    var number;
    do {
        number = randomInt(1, 100);
    } while (index.hasOwnProperty("_"+number));
    index["_"+number] = true;
    numbers.push(number);
}
delete index;

Answer 9

This is a very generic function I have written to generate random unique/non-unique integers for an array. Assume the last parameter to be true in this scenario for this answer.

/* Creates an array of random integers between the range specified 
     len = length of the array you want to generate
     min = min value you require
     max = max value you require
     unique = whether you want unique or not (assume 'true' for this answer)
*/
    function _arrayRandom(len, min, max, unique) {
        var len = (len) ? len : 10,
                min = (min !== undefined) ? min : 1,
                max = (max !== undefined) ? max : 100,
                unique = (unique) ? unique : false,
                toReturn = [], tempObj = {}, i = 0;

        if(unique === true) {
            for(; i < len; i++) {
                var randomInt = Math.floor(Math.random() * ((max - min) + min));
                if(tempObj['key_'+ randomInt] === undefined) {
                    tempObj['key_'+ randomInt] = randomInt;
                    toReturn.push(randomInt);
                } else {
                    i--;
                }
            }
        } else {
            for(; i < len; i++) {
                toReturn.push(Math.floor(Math.random() * ((max - min) + min)));
            }
        }

        return toReturn;
    }

Here the 'tempObj' is a very useful obj since every random number generated will directly check in this tempObj if that key already exists, if not, then we reduce the i by one since we need 1 extra run since the current random number already exists.

In your case, run the following

_arrayRandom(8, 1, 100, true);

That's all.

Answer 10

Shuffling the numbers from 1 to 100 is the right basic strategy, but if you need only 8 shuffled numbers, there's no need to shuffle all 100 numbers.

I don't know Javascript very well, but I believe it's easy to create an array of 100 nulls quickly. Then, for 8 rounds, you swap the n'th element of the array (n starting at 0) with a randomly selected element from n+1 through 99. Of course, any elements not populated yet mean that the element would really have been the original index plus 1, so that's trivial to factor in. When you're done with the 8 rounds, the first 8 elements of your array will have your 8 shuffled numbers.

Answer 11

var arr = []
while(arr.length < 8){
  var randomnumber=Math.ceil(Math.random()*100)
  if(arr.indexOf(randomnumber) === -1){arr.push(randomnumber)}  
}
document.write(arr);

shorter than other answers I've seen

Answer 12

Implementing this as a generator makes it pretty nice to work with. Note, this implementation differs from ones that require the entire input array to be shuffled first.

This sample function works lazily, giving you 1 random item per iteration up to N items you ask for. This is nice because if you just want 3 items from a list of 1000, you don't have to touch all 1000 items first.

// sample :: Integer -> [a] -> [a]
const sample = n => function* (xs) {
  let ys = xs.slice(0);
  let len = xs.length;
  while (n > 0 && len > 0) {
    let i = (Math.random() * len) >> 0;
    yield ys.splice(i,1)[0];
    n--; len--;
  }
}

// example inputs
let items = ['a', 'b', 'c', 'd', 'e', 'f', 'g'];
let numbers = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];

// get 3 random items
for (let i of sample(3) (items))
  console.log(i); // f g c

// partial application
const lotto = sample(3);
for (let i of lotto(numbers))
  console.log(i); // 3 8 7

// shuffle an array
const shuffle = xs => Array.from(sample (Infinity) (xs))
console.log(shuffle(items)) // [b c g f d e a]

I chose to implement sample in a way that does not mutate the input array, but you could easily argue that a mutating implementation is favourable.

For example, the shuffle function might wish to mutate the original input array. Or you might wish to sample from the same input at various times, updating the input each time.

// sample :: Integer -> [a] -> [a]
const sample = n => function* (xs) {
  let len = xs.length;
  while (n > 0 && len > 0) {
    let i = (Math.random() * len) >> 0;
    yield xs.splice(i,1)[0];
    n--; len--;
  }
}

// deal :: [Card] -> [Card]
const deal = xs => Array.from(sample (2) (xs));

// setup a deck of cards (13 in this case)
// cards :: [Card]
let cards = 'A234567890JQK'.split('');

// deal 6 players 2 cards each
// players :: [[Card]]
let players = Array.from(Array(6), $=> deal(cards))

console.log(players);
// [K, J], [6, 0], [2, 8], [Q, 7], [5, 4], [9, A]

// `cards` has been mutated. only 1 card remains in the deck
console.log(cards);
// [3]

sample is no longer a pure function because of the array input mutation, but in certain circumstances (demonstrated above) it might make more sense.

---

Another reason I chose a generator instead of a function that just returns an array is because you may want to continue sampling until some specific condition.

Perhaps I want the first prime number from a list of 1,000,000 random numbers.

• "How many should I sample?" – you don't have to specify
• "Do I have to find all the primes first and then select a random prime?" – Nope.

Because we're working with a generator, this task is trivial

const randomPrimeNumber = listOfNumbers => {
  for (let x of sample(Infinity) (listOfNumbers)) {
    if (isPrime(x))
      return x;
  }
  return NaN;
}

This will continuously sample 1 random number at a time, x, check if it's prime, then return x if it is. If the list of numbers is exhausted before a prime is found, NaN is returned.

---

Note:

This answer was originally shared on another question that was closed as a duplicate of this one. Because it's very different from the other solutions provided here, I've decided to share it here as well

Answer 13

var numbers = [];

for (let i = 0; i < 8; i++) {
  let a = true,
      n;
  while(a) {
    n = Math.floor(Math.random() * 100) + 1;
    a = numbers.includes(n);
  }
  numbers.push(n);
}

console.log(numbers);

Answer 14

for arrays with holes like this [,2,,4,,6,7,,] because my problem was to fill these holes. So I modified it as per my need :)

the following modified solution worked for me :)

var arr = [,2,,4,,6,7,,]; //example
while(arr.length < 9){
  var randomnumber=Math.floor(Math.random()*9+1);
  var found=false;
  for(var i=0;i<arr.length;i++){
    if(arr[i]==randomnumber){found=true;break;}
  }

  if(!found)
    for(k=0;k<9;k++)
    {if(!arr[k]) //if it's empty  !!MODIFICATION
      {arr[k]=randomnumber; break;}}
}

alert(arr); //outputs on the screen

Answer 15

The best earlier answer is the answer by sje397. You will get as good random numbers as you can get, as quick as possible.

My solution is very similar to his solution. However, sometimes you want the random numbers in random order, and that is why I decided to post an answer. In addition, I provide a general function.

function selectKOutOfN(k, n) {
  if (k>n) throw "k>n";
  var selection = [];
  var sorted = [];
  for (var i = 0; i < k; i++) {
    var rand = Math.floor(Math.random()*(n - i));
    for (var j = 0; j < i; j++) {
      if (sorted[j]<=rand)
        rand++;
      else
        break;
    }
    selection.push(rand);
    sorted.splice(j, 0, rand);
  }
  return selection;
}

alert(selectKOutOfN(8, 100));

Answer 16

Same permutation algorithm as The Machine Charmer, but with a prototyped implementation. Better suited to large number of picks. Uses js 1.7 destructuring assignment if available.

// swaps elements at index i and j in array this
// swapping is easy on js 1.7 (feature detection)
Array.prototype.swap = (function () {
    var i=0, j=1;
    try { [i,j]=[j,i]; }
    catch (e) {}
    if(i) {
        return function(i,j) {
            [this[i],this[j]] = [this[j],this[i]];
            return this;
        }
    } else {
        return function(i,j) {
            var temp = this[i];
            this[i] = this[j];
            this[j] = temp;
            return this;
        }
    }
})();


// shuffles array this
Array.prototype.shuffle = function() {
    for(var i=this.length; i>1; i--) {
        this.swap(i-1, Math.floor(i*Math.random()));
    }
    return this;
}

// returns n unique random numbers between min and max
function pick(n, min, max) {
    var a = [], i = max;
    while(i >= min) a.push(i--);
    return a.shuffle().slice(0,n);
}

pick(8,1,100);

Edit: An other proposition, better suited to small number of picks, based on belugabob's answer. To guarantee uniqueness, we remove the picked numbers from the array.

// removes n random elements from array this
// and returns them
Array.prototype.pick = function(n) {
    if(!n || !this.length) return [];
    var i = Math.floor(this.length*Math.random());
    return this.splice(i,1).concat(this.pick(n-1));
}

// returns n unique random numbers between min and max
function pick(n, min, max) {
    var a = [], i = max;
    while(i >= min) a.push(i--);
    return a.pick(n);
}

pick(8,1,100);

Answer 17

Here is my ES6 version I cobbled together. I'm sure it can be a little more consolidated.

function randomArray(i, min, max) {
  min = Math.ceil(min);
  max = Math.floor(max);
  
  let arr = Array.from({length: i}, () => Math.floor(Math.random()* (max - min)) + min);
  
  return arr.sort();
 }
 
 let uniqueItems = [...new Set(randomArray(8, 0, 100))]
 console.log(uniqueItems);

Answer 18

if you need more unique you must generate a array(1..100).

var arr=[];
function generateRandoms(){
for(var i=1;i<=100;i++) arr.push(i);
}
function extractUniqueRandom()
{
   if (arr.length==0) generateRandoms();
   var randIndex=Math.floor(arr.length*Math.random());
   var result=arr[randIndex];
   arr.splice(randIndex,1);
   return result;

}
function extractUniqueRandomArray(n)
{
   var resultArr=[];
   for(var i=0;i<n;i++) resultArr.push(extractUniqueRandom());
   return resultArr;
}

above code is faster:
extractUniqueRandomArray(50)=> [2, 79, 38, 59, 63, 42, 52, 22, 78, 50, 39, 77, 1, 88, 40, 23, 48, 84, 91, 49, 4, 54, 93, 36, 100, 82, 62, 41, 89, 12, 24, 31, 86, 92, 64, 75, 70, 61, 67, 98, 76, 80, 56, 90, 83, 44, 43, 47, 7, 53]

Answer 19

Adding another better version of same code (accepted answer) with JavaScript 1.6 indexOf function. Do not need to loop thru whole array every time you are checking the duplicate.

var arr = []
while(arr.length < 8){
  var randomnumber=Math.ceil(Math.random()*100)
  var found=false;
    if(arr.indexOf(randomnumber) > -1){found=true;}
  if(!found)arr[arr.length]=randomnumber;
}

Older version of Javascript can still use the version at top

PS: Tried suggesting an update to the wiki but it was rejected. I still think it may be useful for others.

Answer 20

This is my personal solution :

<script>

var i, k;
var numbers = new Array();
k = Math.floor((Math.random()*8));
numbers[0]=k;
    for (var j=1;j<8;j++){
        k = Math.floor((Math.random()*8));
i=0;
while (i < numbers.length){
if (numbers[i] == k){
    k = Math.floor((Math.random()*8));
    i=0;
}else {i++;}
}
numbers[j]=k;
    }
    for (var j=0;j<8;j++){
alert (numbers[j]);
    }
</script>

It randomly generates 8 unique array values (between 0 and 7), then displays them using an alert box.

Answer 21

How about using object properties as a hash table? This way your best scenario is to only randomize 8 times. It would only be effective if you want a small part of the range of numbers. It's also much less memory intensive than Fisher-Yates because you don't have to allocate space for an array.

var ht={}, i=rands=8;
while ( i>0 || keys(ht).length<rands) ht[Math.ceil(Math.random()*100)]=i--;
alert(keys(ht));

I then found out that Object.keys(obj) is an ECMAScript 5 feature so the above is pretty much useless on the internets right now. Fear not, because I made it ECMAScript 3 compatible by adding a keys function like this.

if (typeof keys == "undefined") 
{ 
  var keys = function(obj) 
  {
    props=[];
    for (k in ht) if (ht.hasOwnProperty(k)) props.push(k);
    return props;
  }
}

Answer 22

function getUniqueRandomNos() {
    var indexedArrayOfRandomNo = [];
    for (var i = 0; i < 100; i++) {
        var randNo = Math.random();
        indexedArrayOfRandomNo.push([i, randNo]);
    }
    indexedArrayOfRandomNo.sort(function (arr1, arr2) {
        return arr1[1] - arr2[1]
    });
    var uniqueRandNoArray = [];
    for (i = 0; i < 8; i++) {
        uniqueRandNoArray.push(indexedArrayOfRandomNo[i][0]);
    }
    return uniqueRandNoArray;
}

I think this method is different from methods given in most of the answers, so I thought I might add an answer here (though the question was asked 4 years ago).

We generate 100 random numbers, and tag each of them with numbers from 1 to 100. Then we sort these tagged random numbers, and the tags get shuffled randomly. Alternatively, as needed in this question, one could do away with just finding top 8 of the tagged random numbers. Finding top 8 items is cheaper than sorting the whole array.

One must note here, that the sorting algorithm influences this algorithm. If the sorting algorithm used is stable, there is slight bias in favor of smaller numbers. Ideally, we would want the sorting algorithm to be unstable and not even biased towards stability (or instability) to produce an answer with perfectly uniform probability distribution.

Answer 23

This can handle generating upto 20 digit UNIQUE random number

JS

 var generatedNumbers = [];

    function generateRandomNumber(precision) { // input --> number precision in integer 
        if (precision <= 20) {
            var randomNum = Math.round(Math.random().toFixed(precision) * Math.pow(10, precision));
            if (generatedNumbers.indexOf(randomNum) > -1) {
                if (generatedNumbers.length == Math.pow(10, precision))
                    return "Generated all values with this precision";
                    return generateRandomNumber(precision);
            } else {
                generatedNumbers.push(randomNum);
                return randomNum;
            }
        } else
           return "Number Precision shoould not exceed 20";
    }
    generateRandomNumber(1);

jsFiddle

Answer 24

This solution uses the hash which is much more performant O(1) than checking if the resides in the array. It has extra safe checks too. Hope it helps.

function uniqueArray(minRange, maxRange, arrayLength) {
  var arrayLength = (arrayLength) ? arrayLength : 10
  var minRange = (minRange !== undefined) ? minRange : 1
  var maxRange = (maxRange !== undefined) ? maxRange : 100
  var numberOfItemsInArray = 0
  var hash = {}
  var array = []

  if ( arrayLength > (maxRange - minRange) ) throw new Error('Cannot generate unique array: Array length too high')

  while(numberOfItemsInArray < arrayLength){
    // var randomNumber = Math.floor(Math.random() * (maxRange - minRange + 1) + minRange)
    // following line used for performance benefits
    var randomNumber = (Math.random() * (maxRange - minRange + 1) + minRange) << 0

    if (!hash[randomNumber]) {
      hash[randomNumber] = true
      array.push(randomNumber)
      numberOfItemsInArray++
    }
  }
  return array
}
document.write(uniqueArray(1, 100, 8))

Answer 25

You can also do it with a one liner like this:

[...((add, set) => add(set, add))((set, add) => set.size < 8 ? add(set.add(Math.floor(Math.random()*100) + 1), add) : set, new Set())]

Answer 26

getRandom (min, max) {
  return Math.floor(Math.random() * (max - min)) + min
}

getNRandom (min, max, n) {
  const numbers = []
  if (min > max) {
    return new Error('Max is gt min')
  }

  if (min === max) {
    return [min]
  }

  if ((max - min) >= n) {
    while (numbers.length < n) {
      let rand = this.getRandom(min, max + 1)
      if (numbers.indexOf(rand) === -1) {
        numbers.push(rand)
      }
    }
  }

  if ((max - min) < n) {
    for (let i = min; i <= max; i++) {
      numbers.push(i)
    }
  }
  return numbers
}

Answer 27

Using a Set is your fastest option. Here is a generic function for getting a unique random that uses a callback generator. Now it's fast and reusable.

// Get a unique 'anything'
let unique = new Set()

function getUnique(generator) {
  let number = generator()
  while (!unique.add(number)) {
    number = generator()
  }
  return number;
}

// The generator.  Return anything, not just numbers.
const between_1_100 = () => 1 + Math.floor(Math.random() * 100)

// Test it
for (var i = 0; i < 8; i++) {
  const aNumber = getUnique(between_1_100)
}
// Dump the 'stored numbers'
console.log(Array.from(unique))

Answer 28

This is a implementation of Fisher Yates/Durstenfeld Shuffle, but without actual creation of a array thus reducing space complexity or memory needed, when the pick size is small compared to the number of elements available.

To pick 8 numbers from 100, it is not necessary to create a array of 100 elements.

Assuming a array is created,

• From the end of array(100), get random number(rnd) from 1 to 100
• Swap 100 and the random number rnd
• Repeat step 1 with array(99)

If a array is not created, A hashMap may be used to remember the actual swapped positions. When the second random number generated is equal to the one of the previously generated numbers, the map provides the current value in that position rather than the actual value.

const getRandom_ = (start, end) => {
  return Math.floor(Math.random() * (end - start + 1)) + start;
};
const getRealValue_ = (map, rnd) => {
  if (map.has(rnd)) {
    return getRealValue_(map, map.get(rnd));
  } else {
    return rnd;
  }
};
const getRandomNumbers = (n, start, end) => {
  const out = new Map();
  while (n--) {
    const rnd = getRandom_(start, end--);
    out.set(getRealValue_(out, rnd), end + 1);
  }
  return [...out.keys()];
};

console.info(getRandomNumbers(8, 1, 100));
console.info(getRandomNumbers(8, 1, Math.pow(10, 12)));
console.info(getRandomNumbers(800000, 1, Math.pow(10, 15)));

Answer 29

Here is an example of random 5 numbers taken from a range of 0 to 100 (both 0 and 100 included) with no duplication.

let finals = [];
const count = 5; // Considering 5 numbers
const max = 100;

for(let i = 0; i < max; i++){
  const rand = Math.round(Math.random() * max);
  !finals.includes(rand) && finals.push(rand)
}

finals = finals.slice(0, count)

Answer 30

This is how I'd do it in ES6 and without using while. This will never return a 0 either.

function randomUniqueIntegers(total, quantity) {
  
  const numbers = Array(total)
    .fill(null)
    .map((_, i) => i+1);
  
  return numbers
    .map((value) => ({ value, sort: Math.random() }))
    .sort((a, b) => a.sort - b.sort)
    .map(({ value }) => value)
    .slice(0, quantity);
}

randomUniqueIntegers(100, 8) //[ 79, 28, 97, 17, 23, 70, 20, 12 ]

You may also want to do some error handling if there is going to be user input in this.

Answer 31

So many options, but mostly they are with the similar algorithms. I decided to recompile them, and make a benchmark test. You can see a source code. Feel free to make pull request with additional functions or corrections.

I would recommend to use the Set version:

/**
 * Get set of unique numbers from the renge between 0 and {max}
 * @param {number} max The top number of the renge 0..max (will not be included in the result)
 * @param {number} qty The amount of unique numbers from the renge Must be less or equal to {max}
 * @returns {Array<number>} List of unique random numbers from the renge (0 <= random number < max) 
 */
export default function uniqueIndexes(max, qty) {
  const retVal = new Set;
  while (retVal.size < qty) {
    retVal.add(Math.floor(Math.random() * max));
  }
  return Array.from(retVal);
}

This function gives a second result when you need to get small amount of unique random numbers independently of size of the array. And it is still has a good performance when you need a bigger set of unique numbers.

And I strongly do not recommend to use a shuffle algorithm, which several people suggested in this topic. The shuffle algorithm is the worse in terms of performance, it is 25x times slower than one I recommended, and may have some impact if you working with bigger arrays in a high demand servers, or making some web based games. Also its performance become worse exponentially when you have a bigger range of numbers.

In my test example, in the range from 0 to 1000, it runs in 300! times slower than the recommended method.

Here is the result of the benchmark.

• Testing 10000 times (80 unique numbers out of 100)

Shuffle function took 205.500ms --- The worse

Hash function took 48.448ms --- The BEST

Set function took 77.849ms --- reasonable

IndexOf function took 105.924ms

---

• Testing 10000 x (10 out of 100)

Shuffle function took 203.991ms --- The worse

Hash function took 11.478ms

Set function took 8.007ms --- reasonable

IndexOf function took 4.859ms --- The BEST

---

• Testing 10000 x (10 out of 1000)

Shuffle function took 2772.697ms --- wow, no words (but expected)

Hash function took 29.295ms

Set function took 8.435ms --- reasonable

IndexOf function took 4.792ms --- The BEST

Answer 32

var bombout=0;
var checkArr=[];
var arr=[];
while(arr.length < 8 && bombout<100){
  bombout++;
  var randomNumber=Math.ceil(Math.random()*100);
  if(typeof checkArr[randomNumber] == "undefined"){
    checkArr[randomNumber]=1;
    arr.push(randomNumber);
  }
}​

// untested - hence bombout