Find the min/max element of an Array in JavaScript

Question

How can I easily obtain the min or max element of a JavaScript Array?

Example Psuedocode:

let array = [100, 0, 50]

array.min() //=> 0
array.max()         


        

Answer 1

array.sort((a, b) => b - a)[0];

Gives you the maximum value in an array of numbers.

array.sort((a, b) => a - b)[0];

Gives you the minimum value in an array of numbers.

let array = [0,20,45,85,41,5,7,85,90,111];

let maximum = array.sort((a, b) => b - a)[0];
let minimum = array.sort((a, b) => a - b)[0];

console.log(minimum, maximum)

Answer 2

ChaosPandion's solution works if you're using protoype. If not, consider this:

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

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

The above will return NaN if an array value is not an integer so you should build some functionality to avoid that. Otherwise this will work.

Answer 3

Here's one way to get the max value from an array of objects. Create a copy (with slice), then sort the copy in descending order and grab the first item.

var myArray = [
    {"ID": 1, "Cost": 200},
    {"ID": 2, "Cost": 1000},
    {"ID": 3, "Cost": 50},
    {"ID": 4, "Cost": 500}
]

maxsort = myArray.slice(0).sort(function(a, b) { return b.ID - a.ID })[0].ID; 

Answer 4

Aside using the math function max and min, another function to use is the built in function of sort(): here we go

const nums = [12, 67, 58, 30].sort((x, y) => 
x -  y)
let min_val = nums[0]
let max_val = nums[nums.length -1]

Answer 5

A simple solution to find the minimum value over an Array of elements is to use the Array prototype function reduce:

A = [4,3,-9,-2,2,1];
A.reduce((min, val) => val < min ? val : min, A[0]); // returns -9

or using JavaScript's built-in Math.Min() function (thanks @Tenflex):

A.reduce((min,val) => Math.min(min,val), A[0]);

This sets min to A[0], and then checks for A[1]...A[n] whether it is strictly less than the current min. If A[i] < min then min is updated to A[i]. When all array elements has been processed, min is returned as the result.

EDIT: Include position of minimum value:

A = [4,3,-9,-2,2,1];
A.reduce((min, val) => val < min._min ? {_min: val, _idx: min._curr, _curr: min._curr + 1} : {_min: min._min, _idx: min._idx, _curr: min._curr + 1}, {_min: A[0], _idx: 0, _curr: 0}); // returns { _min: -9, _idx: 2, _curr: 6 }

Answer 6

This may suit your purposes.

Array.prototype.min = function(comparer) {

    if (this.length === 0) return null;
    if (this.length === 1) return this[0];

    comparer = (comparer || Math.min);

    var v = this[0];
    for (var i = 1; i < this.length; i++) {
        v = comparer(this[i], v);    
    }

    return v;
}

Array.prototype.max = function(comparer) {

    if (this.length === 0) return null;
    if (this.length === 1) return this[0];

    comparer = (comparer || Math.max);

    var v = this[0];
    for (var i = 1; i < this.length; i++) {
        v = comparer(this[i], v);    
    }

    return v;
}