How might I find the largest number contained in a JavaScript array?

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?

Resig to the rescue:

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 does it work?

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 not a JavaScript 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 4.0392 ms to run
• Math.max.apply took 3.3742 ms to run
• sorting and getting the 0th value took 67.4724 ms to run
• Math.max within reduce() took 6.5804 ms to run
• custom findmax function took 1.6102 ms 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;
}

Benchmark results

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

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

Use:

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

var largest = arr.reduce(function(x,y) {
    return (x > y) ? x : y;
});

console.log(largest);

Use 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 it has limitations.

Function arguments are placed onto the stack which has a downside - a limit. So if your array is bigger than the 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.

The 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.

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;
}

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);

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));

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

Simple one liner

[].sort().pop()

Yes, of course there exists Math.max.apply(null,[23,45,67,-45]) and the result is to return 67.

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 JavaScript, but I think this method would be good:

var array = [34, 23, 57, 983, 198];
var score = 0;

for(var i = 0; i = array.length; i++) {
  if(array[ i ] > score) {
    score = array[i];
  }
}

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);

var nums = [1,4,5,3,1,4,7,8,6,2,1,4];
nums.sort();
nums.reverse();
alert(nums[0]);

Simplest Way:

var nums = [1,4,5,3,1,4,7,8,6,2,1,4]; nums.sort(); nums.reverse(); alert(nums[0]);

Find the largest number in a multidimensional array

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;
}

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;
}

Should be quite simple:

var countArray = [1,2,3,4,5,1,3,51,35,1,357,2,34,1,3,5,6];

var highestCount = 0;
for(var i=0; i<=countArray.length; i++){	
	if(countArray[i]>=highestCount){
  	highestCount = countArray[i]
  }
}

console.log("Highest Count is " + highestCount);

highest and smallest value using sort with arrow function

var highest=[267, 306, 108,700,490,678,355,399,500,800].sort((a,b)=>{return b-a;})[0]
console.log(highest)
var smallest=[267, 306, 108,700,490,678,355,399,500,800].sort((a,b)=>{return a-b;})[0]
console.log(smallest)

let array = [267, 306, 108]
let longest = Math.max(...array);