I have a simple JavaScript Array object containing a few numbers.

```
[267, 306, 108]
```

Is there a function that would find the largest number in this array?

```
Array.max = function( array ){
return Math.max.apply( Math, array );
};
```

**Warning**: since the maximum number of arguments is as low as 65535 on some VMs, use a for loop if you're not certain the array is that small.

You can use the apply function, to call Math.max:

```
var array = [267, 306, 108];
var largest = Math.max.apply(Math, array); // 306
```

How it works?

The apply function is used to call another function, with a given context and arguments, provided as an array. The min and max functions can take an arbitrary number of input arguments: Math.max(val1, val2, ..., valN)

So if we call:

```
Math.min.apply(Math, [1,2,3,4]);
```

The apply function will execute:

```
Math.min(1,2,3,4);
```

Note that the first parameter, the context, is not important for these functions since they are static, they will work regardless of what is passed as the context.

The easiest syntax, with the new spread operator:

```
var arr = [1, 2, 3];
var max = Math.max(...arr);
```

Source : Mozilla MDN

I'm no JS expert, but I wanted to see how these methods stack up, so this was good practice for me. I don't know if this is technically the right way to performance test these, but I just ran them one right after another, as you can see in my code.

Sorting and getting the 0th value is by far the worst method (and it modifies the order of your array, which may not be desirable). For the others, the difference is negligible unless you're talking millions of indices.

**Average results of five runs with a 100,000-index array of random numbers:**

- reduce took
**4.0392ms**to run - Math.max.apply took
**3.3742ms**to run - sorting and getting the 0th value took
**67.4724ms**to run - Math.max within reduce() took
**6.5804ms**to run - custom findmax function took
**1.6102ms**to run

```
var performance = window.performance
function findmax(array)
{
var max = 0,
a = array.length,
counter
for (counter=0;counter<a;counter++)
{
if (array[counter] > max)
{
max = array[counter]
}
}
return max
}
function findBiggestNumber(num) {
var counts = []
var i
for (i = 0; i < num; i++) {
counts.push(Math.random())
}
var a, b
a = performance.now()
var biggest = counts.reduce(function(highest, count){
return highest > count ? highest : count
}, 0)
b = performance.now()
console.log('reduce took ' + (b - a) + ' ms to run')
a = performance.now()
var biggest2 = Math.max.apply(Math, counts)
b = performance.now()
console.log('Math.max.apply took ' + (b - a) + ' ms to run')
a = performance.now()
var biggest3 = counts.sort(function(a,b){return b-a;})[0]
b = performance.now()
console.log('sorting and getting the 0th value took ' + (b - a) + ' ms to run')
a = performance.now()
var biggest4 = counts.reduce(function(highest, count){
return Math.max(highest,count)
}, 0)
b = performance.now()
console.log('Math.max within reduce() took ' + (b - a) + ' ms to run')
a = performance.now()
var biggest5 = findmax(counts)
b = performance.now()
console.log('custom findmax function took ' + (b - a) + ' ms to run')
console.log(biggest + '-' + biggest2 + '-' + biggest3 + '-' + biggest4 + '-' + biggest5)
}
findBiggestNumber(1E5)
```

I've found that for bigger arrays (~100k elements), it actually pays to simply iterate the array with a humble `for`

loop, performing ~30% better than `Math.max.apply()`

:

```
function mymax(a)
{
var m = -Infinity, i = 0, n = a.length;
for (; i != n; ++i) {
if (a[i] > m) {
m = a[i];
}
}
return m;
}
```

You could sort the array in descending order and get the first item:

```
[267, 306, 108].sort(function(a,b){return b-a;})[0]
```

How about this:

```
var arr = [1,2,3,4];
var largest = arr.reduce(function(x,y){
return (x > y) ? x : y;
});
console.log(largest);
```

how about using Array.reduce ?

```
[0,1,2,3,4].reduce(function(previousValue, currentValue){
return Math.max(previousValue,currentValue);
});
```

Almost all of the answers use `Math.max.apply()`

which is nice and dandy but has limitations.

Function arguments are placed onto stack which has a downside - a limit. So if your array is bigger than limit it will fail with `RangeError: Maximum call stack size exceeded.`

To find a call stack size I used this code:

```
var ar = [];
for (var i = 1; i < 100*99999; i++) {
ar.push(1);
try {
var max = Math.max.apply(Math, ar);
} catch(e) {
console.log('Limit reached: '+i+' error is: '+e);
break;
}
}
```

It proved to be biggest on FireFox on my machine - **591519**. This means that if you array contains more than **591519** items, `Math.max.apply()`

will result in **RangeError**.

Best solution for this problem is iterative way(credit: https://developer.mozilla.org/):

```
max = -Infinity, min = +Infinity;
for (var i = 0; i < numbers.length; i++) {
if (numbers[i] > max)
max = numbers[i];
if (numbers[i] < min)
min = numbers[i];
}
```

I have written about this question on my blog here.

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math/max

```
const inputArray = [ 1, 3, 4, 9, 16, 2, 20, 18];
const maxNumber = Math.max(...inputArray);
console.log(maxNumber);
```

Finding max and min value the easy and manual way. This code is much faster than `Math.max.apply`

; I have tried up to 1000k numbers in array.

```
function findmax(array)
{
var max = 0;
var a = array.length;
for (counter=0;counter<a;counter++)
{
if (array[counter] > max)
{
max = array[counter];
}
}
return max;
}
function findmin(array)
{
var min = array[0];
var a = array.length;
for (counter=0;counter<a;counter++)
{
if (array[counter] < min)
{
min = array[counter];
}
}
return min;
}
```

Yes of course exist: `Math.max.apply(null,[23,45,67,-45])`

and the result return `67`

;

To find the largest number in an array you just need to use `Math.max(...arrayName);`

, it works like this:

```
let myArr = [1, 2, 3, 4, 5, 6];
console.log(Math.max(...myArr));
```

So it basically gets the largest number in the array and outputs it to the console. You put the triple dots cause without it you're basically writing `Math.max([1, 2, 3, 4, 5, 6]);`

which is wrong.

Don't forget that the wrap can be done with `Function.prototype.bind`

, giving you an "all-native" *function*.

```
var aMax = Math.max.apply.bind(Math.max, Math);
aMax([1, 2, 3, 4, 5]); // 5
```

You could also extend `Array`

to have this function and make it part of every array.

```
Array.prototype.max = function(){return Math.max.apply( Math, this )};
myArray = [1,2,3];
console.log( myArray.max() );
```

You can also use forEach:

```
var maximum = Number.MIN_SAFE_INTEGER;
var array = [-3, -2, 217, 9, -8, 46];
array.forEach(function(value){
if(value > maximum) {
maximum = value;
}
});
console.log(maximum); // 217
```

Using - `Array.prototype.reduce()`

is cool!

`[267, 306, 108].reduce((acc,val)=> (acc>val)?acc:val)`

where **acc = accumulator** and **val = current value**;

```
var a = [267, 306, 108].reduce((acc,val)=> (acc>val)?acc:val);
console.log(a);
```

You can try this,

```
var arr = [267,306,108];
var largestNum = 0;
for(i=0;i<arr.length;i++) {
if(arr[i]>largest){
var largest = arr[i];
}
}
console.log(largest);
```

I just started with JS but I think this method would be good:

```
var array = [34, 23, 57, 983, 198];<br>
var score = 0;
for(var i = 0; i = array.length; i++) {
if(array[ i ] > score) {
score = array[i];
}
}
```

Simple one liner

```
[].sort().pop()
```

```
var max = [];
for(var i=0; arr.length>i; i++ ){
var arra = arr[i];
var largest = Math.max.apply(Math, arra);
max.push(largest);
}
return max;
```

Run this:

```
Array.prototype.max = function(){
return Math.max.apply( Math, this );
};
```

And now try `[3,10,2].max()`

returns `10`

**Find Max and Min value using Bubble Sort**

```
var arr = [267, 306, 108];
for(i=0, k=0; i<arr.length; i++) {
for(j=0; j<i; j++) {
if(arr[i]>arr[j]) {
k = arr[i];
arr[i] = arr[j];
arr[j] = k;
}
}
}
console.log('largest Number: '+ arr[0]);
console.log('Smallest Number: '+ arr[arr.length-1]);
```

Try this

```
function largestNum(arr) {
var currentLongest = arr[0]
for (var i=0; i< arr.length; i++){
if (arr[i] > currentLongest){
currentLongest = arr[i]
}
}
return currentLongest
}
```

As per @Quasimondo's comment, which seems to have been largely missed, the below seems to have the best performance as shown here: https://jsperf.com/finding-maximum-element-in-an-array. Note that while for the array in the question, performance may not have a significant effect, for large arrays performance becomes more important, and again as noted using `Math.max()`

doesn't even work if the array length is more than 65535. See also this answer.

```
function largestNum(arr) {
var d = data;
var m = d[d.length - 1];
for (var i = d.length - 1; --i > -1;) {
if (d[i] > m) m = d[i];
}
return m;
}
```

A recursive approach on how to do it using ternary operators

```
const findMax = (arr, max, i) => arr.length === i ? max :
findMax(arr, arr[i] > max ? arr[i] : max, ++i)
const arr = [5, 34, 2, 1, 6, 7, 9, 3];
const max = findMax(arr, arr[0], 0)
console.log(max);
```

One `for/of`

loop solution:

```
const numbers = [2, 4, 6, 8, 80, 56, 10];
const findMax = (...numbers) => {
let currentMax = numbers[0]; // 2
for (const number of numbers) {
if (number > currentMax) {
console.log(number, currentMax);
currentMax = number;
}
}
console.log('Largest ', currentMax);
return currentMax;
};
findMax(...numbers);
```

**My solution to return largest numbers in arrays.**

```
const largestOfFour = arr => {
let arr2 = [];
arr.map(e => {
let numStart = -Infinity;
e.forEach(num => {
if (num > numStart) {
numStart = num;
}
})
arr2.push(numStart);
})
return arr2;
}
```

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