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

Question

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?

Answer 1

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)

Answer 2

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);

Answer 3

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]);

Answer 4

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

Answer 5

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;
}

Answer 6

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

Answer 7

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];
  }
}

Answer 8

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);

Answer 9

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;

Answer 10

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)

Answer 11

Use:

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

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

console.log(largest);

Answer 12

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

Answer 13

Use Array.reduce:

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

Answer 14

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.

Answer 15

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.

Answer 16

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);

Answer 17

Simple one liner

[].sort().pop()

Answer 18

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);

Answer 19

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.

Answer 20

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;
}

Answer 21

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
}

Answer 22

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]);

Answer 23

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;
}

Answer 24

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);

Answer 25

Run this:

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

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

Answer 26

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);

Answer 27

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

Answer 28

The easiest syntax, with the new spread operator:

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

Source : Mozilla MDN

Answer 29

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

Answer 30

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() );