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javascript function to find the second largest element in an array

I am completing the hackerrank's 10 days of javascript challenge. The question:
write a function to take an array as an argument and then return the second largest element in the array.
I have written the code but my code is returning the largest element and not the second largest as asked.
function getSecondLargest(nums) {
// Complete the function
var largest=nums[0];
for(let i=1;i<nums.length;++i)
{
if(nums[i]>largest)
largest=nums[i];
}
var large=nums[0];
for(let j=1;j<nums.length;++j)
{
if(large<nums[j]&&large<largest)
large=nums[j];
}
return large;
}
When input array nums={2,3,6,6,5} the result is coming 6 while expected output is 5. Please help and point out the errors in the function code below.

should not initialize large with first value var large=nums[0]; because it may appear the biggest value and won't work
should use nums[j]<largest instead of large<largest as mentioned above
I think don't need second loop as all checks can be done in first loop, and you can assign prev largest to large whenever you change it:
function getSecondLargest(nums) {
var largest = nums[0];
var large;
for (let i = 1; i < nums.length; ++i) {
if (nums[i] > largest) {
large = largest;
largest = nums[i];
} else if (nums[i] > large || typeof large === 'undefined') {
large = nums[i]
}
}
return large;
}
console.log(getSecondLargest([5,1-2,3]))
console.log(getSecondLargest([-5,1,-2,3]))

GET SECOND LARGEST
first, I create new array with unique values.
let arr = [...new Set(nums)];
second, sort value using built-in function .sort().
note : by default .sort() always sorts asciibetically, but for some testcase, it doesn't work. So, I put (a, b) => { return a - b } to make sure it will work properly.
arr = arr.sort((a, b) => { return a -b });
third, get the value from arr
let result = arr[arr.length - 2] || arr[0];
finally, return the result
return result
function getSecondLargest(nums) {
let arr = [...new Set(nums)];
//Javascript's array member method .sort( always sorts asciibetically.
arr = arr.sort((a, b) => { return a - b });
let result = arr[arr.length - 2] || arr[0];
return result
}

Just one minor change:
Use nums[j]<largest instead of large<largest in the second for loop
function getSecondLargest(nums) {
// Complete the function
var largest=nums[0];
for(let i=1;i<nums.length;++i)
{
if(nums[i]>largest)
largest=nums[i];
}
var large;
//To ensure that the selected number is not the largest
for(let j=0;j<nums.length;++j)
{
if (nums[j] !== largest){
large = nums[j];
break;
}
}
for(let j=1;j<nums.length;++j)
{
if(large<nums[j]&&nums[j]!=largest)
large=nums[j];
else
console.log(large)
}
return large;
}
var secondLargest = getSecondLargest([6,3,6,6,5]);
console.log("Second largest number", secondLargest);

If you want to avoid using library functions like #ifaruki suggests, this line
if(large<nums[j]&&large<largest)
should read
if (large<nums[j] && nums[j] < largest)
Sorting and picking the second or second-to-last value fails when there are duplicates of the highest value in the input array.

Another easiest logic is to remove duplicates from the array and sort.
let givenArray = [2, 3, 6, 6, 5];
let uniqueArray = [...new Set(givenArray)];
console.log("The second largets element is", uniqueArray.sort()[uniqueArray.length - 2]);

I know you had your question answered, just thought I would provide my solution for any future users looking into this.
You can use reduce to go through the array while remembering the two largest numbers so far.
You just make a simple reduction function:
function twoMax(two_max, candidate)
{
if (candidate > two_max[0]) return [candidate,two_max[0]];
else if (candidate > two_max[1]) return [two_max[0],candidate];
else return two_max;
}
And then you use it for example like this:
let my_array = [0,1,5,7,0,8,12];
let two_largest = my_array.reduce(twoMax,[-Infinity,-Infinity]);
let second_largest = two_largest[1];
This solution doesn't require sorting and goes through the array only once.

If you want to avoid using **sort method. I think here's the easiest logic to do that, which will also work in arrays where there's duplicates of largest integer exists.
function getSecondLargest(arr) {
const largest = Math.max.apply(null, arr);
for (let i = 0; i < arr.length; i++) {
if (largest === arr[i]) {
arr[i] = -Infinity;
}
}
return Math.max.apply(null, arr);
}
console.log(getSecondLargest([5, 7, 11, 11, 11])); //7

Javascript - Sum of few numbers in array with reduce() or some other built-in method

So I was wondering, if there is a shorter or different way to get the sum of FEW numbers in an array, not the complete array.
For example if i want to get the sum of first 3 numbers in array, I know I can go trough loop like this:
var arr=[1,2,3,4,5];
var sum=0;
for(var i=0; i<3;i++){
sum+= arr[i];
}
console.log(sum);
But I would like to know if you can somehow use reduce() method? Or even some other built-in method?
I tried reduce like this,few times on my own, but it isn't working:
var arr= [1,2,3,4,5];
arr.reduce(function(total,amount,index){
total+=amount;
if(index===2){
return total;
}
});

arr
.slice(0,3) //get the range
.reduce((a,b) => a + b)//sum up
Or without slice:
arr.reduce((a,b,i) => i < 3 ? a + b : a, 0);

As an example, if you want to skip element by index 2.
var arr = [1,2,3,4,5];
arr.reduce(function(total,amount,index){
if(index === 2) {
return total;
}
return total + amount;
});
In your case you forgot to return new total, when the index is not 2.

A function that takes an array, and two numbers representing the element range. It slices the array, and returns the sum of the numbers in that range using reduce.
let arr= [1,2,3,4,5];
function getRangeSum(arr, from, to) {
return arr.slice(from, to).reduce((p, c) => {
return p + c;
}, 0);
}
console.log(getRangeSum(arr, 2, 4));

Return index of greatest value in an array

I have this:
var arr = [0, 21, 22, 7];
What's the best way to return the index of the highest value into another variable?

This is probably the best way, since it’s reliable and works on old browsers:
function indexOfMax(arr) {
if (arr.length === 0) {
return -1;
}
var max = arr[0];
var maxIndex = 0;
for (var i = 1; i < arr.length; i++) {
if (arr[i] > max) {
maxIndex = i;
max = arr[i];
}
}
return maxIndex;
}
There’s also this one-liner:
let i = arr.indexOf(Math.max(...arr));
It performs twice as many comparisons as necessary and will throw a RangeError on large arrays, though. I’d stick to the function.

In one line and probably faster then arr.indexOf(Math.max.apply(Math, arr)):
var a = [0, 21, 22, 7];
var indexOfMaxValue = a.reduce((iMax, x, i, arr) => x > arr[iMax] ? i : iMax, 0);
document.write("indexOfMaxValue = " + indexOfMaxValue); // prints "indexOfMaxValue = 2"
Where:
iMax - the best index so far (the index of the max element so far, on the first iteration iMax = 0 because the second argument to reduce() is 0, we can't omit the second argument to reduce() in our case)
x - the currently tested element from the array
i - the currently tested index
arr - our array ([0, 21, 22, 7])
About the reduce() method (from "JavaScript: The Definitive Guide" by David Flanagan):
reduce() takes two arguments. The first is the function that performs the reduction operation. The task of this reduction function is to somehow combine or reduce two values into a single value, and to return that reduced value.
Functions used with reduce() are different than the functions used with forEach() and map(). The familiar value, index, and array values are passed as the second, third, and fourth arguments. The first argument is the accumulated result of the reduction so far. On the first call to the function, this first argument is the initial value you passed as the
second argument to reduce(). On subsequent calls, it is the value returned by the previous invocation of the function.
When you invoke reduce() with no initial value, it uses the first element of the array as the initial value. This means that the first call to the reduction function will have the first and second array elements as its
first and second arguments.

Another solution of max using reduce:
[1,2,5,0,4].reduce((a,b,i) => a[0] < b ? [b,i] : a, [Number.MIN_VALUE,-1])
//[5,2]
This returns [5e-324, -1] if the array is empty. If you want just the index, put [1] after.
Min via (Change to > and MAX_VALUE):
[1,2,5,0,4].reduce((a,b,i) => a[0] > b ? [b,i] : a, [Number.MAX_VALUE,-1])
//[0, 3]

To complete the work of #VFDan, I benchmarked the 3 methods: the accepted one (custom loop), reduce, and find(max(arr)) on an array of 10000 floats.
Results on chromimum 85 linux (higher is better):
custom loop: 100%
reduce: 94.36%
indexOf(max): 70%
Results on firefox 80 linux (higher is better):
custom loop: 100%
reduce: 96.39%
indexOf(max): 31.16%
Conclusion:
If you need your code to run fast, don't use indexOf(max).
reduce is ok but use the custom loop if you need the best performances.
You can run this benchmark on other browser using this link:
https://jsben.ch/wkd4c

If you are utilizing underscore, you can use this nice short one-liner:
_.indexOf(arr, _.max(arr))
It will first find the value of the largest item in the array, in this case 22. Then it will return the index of where 22 is within the array, in this case 2.

Unless I'm mistaken, I'd say it's to write your own function.
function findIndexOfGreatest(array) {
var greatest;
var indexOfGreatest;
for (var i = 0; i < array.length; i++) {
if (!greatest || array[i] > greatest) {
greatest = array[i];
indexOfGreatest = i;
}
}
return indexOfGreatest;
}

var arr=[0,6,7,7,7];
var largest=[0];
//find the largest num;
for(var i=0;i<arr.length;i++){
var comp=(arr[i]-largest[0])>0;
if(comp){
largest =[];
largest.push(arr[i]);
}
}
alert(largest )//7
//find the index of 'arr'
var arrIndex=[];
for(var i=0;i<arr.length;i++){
var comp=arr[i]-largest[0]==0;
if(comp){
arrIndex.push(i);
}
}
alert(arrIndex);//[2,3,4]

EDIT: Years ago I gave an answer to this that was gross, too specific, and too complicated. So I'm editing it. I favor the functional answers above for their neat factor but not their readability; but if I were more familiar with javascript then I might like them for that, too.
Pseudo code:
Track index that contains largest value. Assume index 0 is largest initially. Compare against current index. Update index with largest value if necessary.
Code:
var mountains = [3, 1, 5, 9, 4];
function largestIndex(array){
var counter = 1;
var max = 0;
for(counter; counter < array.length; counter++){
if(array[max] < array[counter]){
max = counter;
}
}
return max;
}
console.log("index with largest value is: " +largestIndex(mountains));
// index with largest value is: 3

function findIndicesOf(haystack, needle)
{
var indices = [];
var j = 0;
for (var i = 0; i < haystack.length; ++i) {
if (haystack[i] == needle)
indices[j++] = i;
}
return indices;
}
pass array to haystack and Math.max(...array) to needle. This will give all max elements of the array, and it is more extensible (for example, you also need to find min values)

If you create a copy of the array and sort it descending, the first element of the copy will be the largest. Than you can find its index in the original array.
var sorted = [...arr].sort((a,b) => b - a)
arr.indexOf(sorted[0])
Time complexity is O(n) for the copy, O(n*log(n)) for sorting and O(n) for the indexOf.
If you need to do it faster, Ry's answer is O(n).

A minor modification revised from the "reduce" version of #traxium 's solution taking the empty array into consideration:
function indexOfMaxElement(array) {
return array.reduce((iMax, x, i, arr) =>
arr[iMax] === undefined ? i :
x > arr[iMax] ? i : iMax
, -1 // return -1 if empty
);
}

A stable version of this function looks like this:
// not defined for empty array
function max_index(elements) {
var i = 1;
var mi = 0;
while (i < elements.length) {
if (!(elements[i] < elements[mi]))
mi = i;
i += 1;
}
return mi;
}

To find the index of the greatest value in an array, copy the original array into the new array and then sort the original array in decreasing order to get the output [22, 21, 7, 0]; now find the value 22 index in the copyNumbers array using this code copyNumbers.indexOf(numbers[0]);
<script>
const numbers = [0, 21, 22, 7];
const copyNumbers = [];
copyNumbers.push(...numbers);
numbers.sort(function(a, b){
return b - a
});
const index = copyNumbers.indexOf(numbers[0]);
console.log(index);
</script>

Make this
const max = arr.reduce((m, n) => Math.max(m, n)), then the indexes of the max
get index with findIndex
var index = arr.findIndex(i => i === max)

Max value in array, based on a function

I have an array of objects. I want to find the "maximum" of this array based on a function that returns whichever object is bigger when given 2 objects.
function comparison(first, second) {
// ... arbitrary comparison based on properties...
return first; // or second
}
var a = [obj1, obj2, obj3];
var maxObj = ????(comparison);
What do I fill in here? What's elegant and short?

Something like this should be quicker than sort (depending on the data):
/*
values: array of values to test.
fn: function that takes two arguements and returns true if the first is bigger.
*/
var maximum = function(values, fn) {
var currentValue, maxValue = values.pop();
while(values.length)
maxValue = fn(maxValue, currentValue = values.pop()) ? maxValue : currentValue;
return maxValue;
}
Examples: http://jsfiddle.net/SaBJ4/2/
Even better, use Array.reduce:
var a = ['abc', 'defg', 'highlkasd', 'ac', 'asdh'];
a.reduce(function(a, b) { return a.length > b.length ? a : b; }); // highlkasd

What's wrong with the obvious approach?
for(var i = 0, max; i < a.length; ++i)
max = typeof max == 'undefined' ? a[i] : comparison(a[i], max);
Wrap that up however you like.
Or you can take advantage of the fact that a = []; x = a[0] leaves you with undefined in x and do it RobG's way:
for(var i = 1, max = a[0]; i < a.length; ++i)
max = comparison(a[i], max);
That nicely avoids a bunch of typeof operators and comparisons that you really don't need.

[obj,obj,obj].sort(comparison)
// aka
var sorted = [obj,obj,obj].sort(function(a,b){
// return 1/0/-1
});
Then either pop the top or bottom element off (however you're sorting) to get the "max" object.
Array.Sort

Find the min/max element of an array in JavaScript

How can I easily obtain the min or max element of a JavaScript array?
Example pseudocode:
let array = [100, 0, 50]
array.min() //=> 0
array.max() //=> 100

How about augmenting the built-in Array object to use Math.max/Math.min instead:
Array.prototype.max = function() {
return Math.max.apply(null, this);
};
Array.prototype.min = function() {
return Math.min.apply(null, this);
};
let p = [35,2,65,7,8,9,12,121,33,99];
console.log(`Max value is: ${p.max()}` +
`\nMin value is: ${p.min()}`);
Here is a JSFiddle.
Augmenting the built-ins can cause collisions with other libraries (some see), so you may be more comfortable with just apply'ing Math.xxx() to your array directly:
var min = Math.min.apply(null, arr),
max = Math.max.apply(null, arr);
Alternately, assuming your browser supports ECMAScript 6, you can use spread syntax which functions similarly to the apply method:
var min = Math.min( ...arr ),
max = Math.max( ...arr );

var max_of_array = Math.max.apply(Math, array);
For a full discussion see:
http://aaroncrane.co.uk/2008/11/javascript_max_api/

Using spread operator (ES6)
Math.max(...array) // The same with "min" => Math.min(...array)
const array = [10, 2, 33, 4, 5];
console.log(
Math.max(...array)
)

For big arrays (~10⁷ elements), Math.min and Math.max both produces the following error in Node.js.
RangeError: Maximum call stack size exceeded
A more robust solution is to not add every element to the call stack, but to instead pass an array:
function arrayMin(arr) {
return arr.reduce(function (p, v) {
return ( p < v ? p : v );
});
}
function arrayMax(arr) {
return arr.reduce(function (p, v) {
return ( p > v ? p : v );
});
}
If you are concerned about speed, the following code is ~3 times faster then Math.max.apply is on my computer. See https://jsben.ch/JPOyL.
function arrayMin(arr) {
var len = arr.length, min = Infinity;
while (len--) {
if (arr[len] < min) {
min = arr[len];
}
}
return min;
};
function arrayMax(arr) {
var len = arr.length, max = -Infinity;
while (len--) {
if (arr[len] > max) {
max = arr[len];
}
}
return max;
};
If your arrays contains strings instead of numbers, you also need to coerce them into numbers. The below code does that, but it slows the code down ~10 times on my machine. See https://jsben.ch/uPipD.
function arrayMin(arr) {
var len = arr.length, min = Infinity;
while (len--) {
if (Number(arr[len]) < min) {
min = Number(arr[len]);
}
}
return min;
};
function arrayMax(arr) {
var len = arr.length, max = -Infinity;
while (len--) {
if (Number(arr[len]) > max) {
max = Number(arr[len]);
}
}
return max;
};

tl;dr
// For regular arrays:
var max = Math.max(...arrayOfNumbers);
// For arrays with tens of thousands of items:
let max = testArray[0];
for (let i = 1; i < testArrayLength; ++i) {
if (testArray[i] > max) {
max = testArray[i];
}
}
MDN solution
The official MDN docs on Math.max() already covers this issue:
The following function uses Function.prototype.apply() to find the maximum element in a numeric array. getMaxOfArray([1, 2, 3]) is equivalent to Math.max(1, 2, 3), but you can use getMaxOfArray() on programmatically constructed arrays of any size.
function getMaxOfArray(numArray) {
return Math.max.apply(null, numArray);
}
Or with the new spread operator, getting the maximum of an array becomes a lot easier.
var arr = [1, 2, 3];
var max = Math.max(...arr);
Maximum size of an array
According to MDN the apply and spread solutions had a limitation of 65536 that came from the limit of the maximum number of arguments:
But beware: in using apply this way, you run the risk of exceeding the JavaScript engine's argument length limit. The consequences of applying a function with too many arguments (think more than tens of thousands of arguments) vary across engines (JavaScriptCore has hard-coded argument limit of 65536), because the limit (indeed even the nature of any excessively-large-stack behavior) is unspecified. Some engines will throw an exception. More perniciously, others will arbitrarily limit the number of arguments actually passed to the applied function. To illustrate this latter case: if such an engine had a limit of four arguments (actual limits are of course significantly higher), it would be as if the arguments 5, 6, 2, 3 had been passed to apply in the examples above, rather than the full array.
They even provide a hybrid solution which doesn't really have good performance compared to other solutions. See performance test below for more.
In 2019 the actual limit is the maximum size of the call stack. For modern Chromium based desktop browsers this means that when it comes to finding min/max with apply or spread, practically the maximum size for numbers only arrays is ~120000. Above this, there will be a stack overflow and the following error will be thrown:
RangeError: Maximum call stack size exceeded
With the script below (based on this blog post), by catching that error you can calculate the limit for your specific environment.
Warning! Running this script takes time and depending on the performance of your system it might slow or crash your browser/system!
let testArray = Array.from({length: 10000}, () => Math.floor(Math.random() * 2000000));
for (i = 10000; i < 1000000; ++i) {
testArray.push(Math.floor(Math.random() * 2000000));
try {
Math.max.apply(null, testArray);
} catch (e) {
console.log(i);
break;
}
}
Performance on large arrays
Based on the test in EscapeNetscape's comment I created some benchmarks that tests 5 different methods on a random number only array with 100000 items.
In 2019, the results show that the standard loop (which BTW doesn't have the size limitation) is the fastest everywhere. apply and spread comes closely after it, then much later MDN's hybrid solution then reduce as the slowest.
Almost all tests gave the same results, except for one where spread somewhy ended up being the slowest.
If you step up your array to have 1 million items, things start to break and you are left with the standard loop as a fast solution and reduce as a slower.
JSPerf benchmark
JSBen benchmark
JSBench.me benchmark
Benchmark source code
var testArrayLength = 100000
var testArray = Array.from({length: testArrayLength}, () => Math.floor(Math.random() * 2000000));
// ES6 spread
Math.min(...testArray);
Math.max(...testArray);
// reduce
testArray.reduce(function(a, b) {
return Math.max(a, b);
});
testArray.reduce(function(a, b) {
return Math.min(a, b);
});
// apply
Math.min.apply(Math, testArray);
Math.max.apply(Math, testArray);
// standard loop
let max = testArray[0];
for (let i = 1; i < testArrayLength; ++i) {
if (testArray[i] > max) {
max = testArray[i];
}
}
let min = testArray[0];
for (let i = 1; i < testArrayLength; ++i) {
if (testArray[i] < min) {
min = testArray[i];
}
}
// MDN hibrid soltuion
// Source: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Function/apply#Using_apply_and_built-in_functions
function minOfArray(arr) {
var min = Infinity;
var QUANTUM = 32768;
for (var i = 0, len = arr.length; i < len; i += QUANTUM) {
var submin = Math.min.apply(null, arr.slice(i, Math.min(i + QUANTUM, len)));
min = Math.min(submin, min);
}
return min;
}
minOfArray(testArray);
function maxOfArray(arr) {
var max = -Infinity;
var QUANTUM = 32768;
for (var i = 0, len = arr.length; i < len; i += QUANTUM) {
var submax = Math.max.apply(null, arr.slice(i, Math.max(i + QUANTUM, len)));
max = Math.max(submax, max);
}
return max;
}
maxOfArray(testArray);

If you're paranoid like me about using Math.max.apply (which could cause errors when given large arrays according to MDN), try this:
function arrayMax(array) {
return array.reduce(function(a, b) {
return Math.max(a, b);
});
}
function arrayMin(array) {
return array.reduce(function(a, b) {
return Math.min(a, b);
});
}
Or, in ES6:
function arrayMax(array) {
return array.reduce((a, b) => Math.max(a, b));
}
function arrayMin(array) {
return array.reduce((a, b) => Math.min(a, b));
}
The anonymous functions are unfortunately necessary (instead of using Math.max.bind(Math) because reduce doesn't just pass a and b to its function, but also i and a reference to the array itself, so we have to ensure we don't try to call max on those as well.

Alternative Methods
The Math.min and Math.max are great methods to get the minimum and maximum item out of a collection of items, however it's important to be aware of some cavities that can comes with it.
Using them with an array that contains large number of items (more than ~10⁷ items, depends on the user's browser) most likely will crash and give the following error message:
const arr = Array.from(Array(1000000).keys());
Math.min(arr);
Math.max(arr);
Uncaught RangeError: Maximum call stack size exceeded
UPDATE
Latest browsers might return NaN instead. That might be a better way to handle errors, however it doesn't solve the problem just yet.
Instead, consider using something like so:
function maxValue(arr) {
return arr.reduce((max, val) => max > val ? max : val)
}
Or with better run-time:
function maxValue(arr) {
let max = arr[0];
for (let val of arr) {
if (val > max) {
max = val;
}
}
return max;
}
Or to get both Min and Max:
function getMinMax(arr) {
return arr.reduce(({min, max}, v) => ({
min: min < v ? min : v,
max: max > v ? max : v,
}), { min: arr[0], max: arr[0] });
}
Or with even better run-time*:
function getMinMax(arr) {
let min = arr[0];
let max = arr[0];
let i = arr.length;
while (i--) {
min = arr[i] < min ? arr[i] : min;
max = arr[i] > max ? arr[i] : max;
}
return { min, max };
}
* Tested with 1,000,000 items:
Just for a reference, the 1st function run-time (on my machine) was 15.84ms vs 2nd function with only 4.32ms.

Two ways are shorter and easy:
let arr = [2, 6, 1, 0]
Way 1:
let max = Math.max.apply(null, arr)
Way 2:
let max = arr.reduce(function(a, b) {
return Math.max(a, b);
});

.apply is often used when the intention is to invoke a variadic function with a list of argument values, e.g.
The Math.max([value1[,value2, ...]]) function returns the largest of zero or more numbers.
Math.max(10, 20); // 20
Math.max(-10, -20); // -10
Math.max(-10, 20); // 20
The Math.max() method doesn't allow you to pass in an array. If you have a list of values of which you need to get the largest, you would normally call this function using Function.prototype.apply(), e.g.
Math.max.apply(null, [10, 20]); // 20
Math.max.apply(null, [-10, -20]); // -10
Math.max.apply(null, [-10, 20]); // 20
However, as of the ECMAScript 6 you can use the spread operator:
The spread operator allows an expression to be expanded in places where multiple arguments (for function calls) or multiple elements (for array literals) are expected.
Using the spread operator, the above can be rewritten as such:
Math.max(...[10, 20]); // 20
Math.max(...[-10, -20]); // -10
Math.max(...[-10, 20]); // 20
When calling a function using the variadic operator, you can even add additional values, e.g.
Math.max(...[10, 20], 50); // 50
Math.max(...[-10, -20], 50); // 50
Bonus:
Spread operator enables you to use the array literal syntax to create new arrays in situations where in ES5 you would need to fall back to imperative code, using a combination of push, splice, etc.
let foo = ['b', 'c'];
let bar = ['a', ...foo, 'd', 'e']; // ['a', 'b', 'c', 'd', 'e']

You do it by extending the Array type:
Array.max = function( array ){
return Math.max.apply( Math, array );
};
Array.min = function( array ){
return Math.min.apply( Math, array );
};
Boosted from here (by John Resig)

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 }

For a concise, modern solution, one can perform a reduce operation over the array, keeping track of the current minimum and maximum values, so the array is only iterated over once (which is optimal). Destructuring assignment is used here for succinctness.
let array = [100, 0, 50];
let [min, max] = array.reduce(([prevMin,prevMax], curr)=>
[Math.min(prevMin, curr), Math.max(prevMax, curr)], [Infinity, -Infinity]);
console.log("Min:", min);
console.log("Max:", max);
To only find either the minimum or maximum, we can use perform a reduce operation in much the same way, but we only need to keep track of the previous optimal value. This method is better than using apply as it will not cause errors when the array is too large for the stack.
const arr = [-1, 9, 3, -6, 35];
//Only find minimum
const min = arr.reduce((a,b)=>Math.min(a,b), Infinity);
console.log("Min:", min);//-6
//Only find maximum
const max = arr.reduce((a,b)=>Math.max(a,b), -Infinity);
console.log("Max:", max);//35

Others have already given some solutions in which they augment Array.prototype. All I want in this answer is to clarify whether it should be Math.min.apply( Math, array ) or Math.min.apply( null, array ). So what context should be used, Math or null?
When passing null as a context to apply, then the context will default to the global object (the window object in the case of browsers). Passing the Math object as the context would be the correct solution, but it won't hurt passing null either. Here's an example when null might cause trouble, when decorating the Math.max function:
// decorate Math.max
(function (oldMax) {
Math.max = function () {
this.foo(); // call Math.foo, or at least that's what we want
return oldMax.apply(this, arguments);
};
})(Math.max);
Math.foo = function () {
print("foo");
};
Array.prototype.max = function() {
return Math.max.apply(null, this); // <-- passing null as the context
};
var max = [1, 2, 3].max();
print(max);
The above will throw an exception because this.foo will be evaluated as window.foo, which is undefined. If we replace null with Math, things will work as expected and the string "foo" will be printed to the screen (I tested this using Mozilla Rhino).
You can pretty much assume that nobody has decorated Math.max so, passing null will work without problems.

One more way to do it:
var arrayMax = Function.prototype.apply.bind(Math.max, null);
Usage:
var max = arrayMax([2, 5, 1]);

I am surprised not one mentiond the reduce function.
var arr = [1, 10, 5, 11, 2]
var b = arr.reduce(function(previous,current){
return previous > current ? previous:current
});
b => 11
arr => [1, 10, 5, 11, 2]

https://developer.mozilla.org/ru/docs/Web/JavaScript/Reference/Global_Objects/Math/max
function getMaxOfArray(numArray) {
return Math.max.apply(null, numArray);
}
var arr = [100, 0, 50];
console.log(getMaxOfArray(arr))
this worked for me.

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;
}

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

I thought I'd share my simple and easy to understand solution.
For the min:
var arr = [3, 4, 12, 1, 0, 5];
var min = arr[0];
for (var k = 1; k < arr.length; k++) {
if (arr[k] < min) {
min = arr[k];
}
}
console.log("Min is: " + min);
And for the max:
var arr = [3, 4, 12, 1, 0, 5];
var max = arr[0];
for (var k = 1; k < arr.length; k++) {
if (arr[k] > max) {
max = arr[k];
}
}
console.log("Max is: " + max);

For big arrays (~10⁷ elements), Math.min and Math.max procuces a RangeError (Maximum call stack size exceeded) in node.js.
For big arrays, a quick & dirty solution is:
Array.prototype.min = function() {
var r = this[0];
this.forEach(function(v,i,a){if (v<r) r=v;});
return r;
};

For an array containing objects instead of numbers:
arr = [
{ name: 'a', value: 5 },
{ name: 'b', value: 3 },
{ name: 'c', value: 4 }
]
You can use reduce to get the element with the smallest value (min)
arr.reduce((a, b) => a.value < b.value ? a : b)
// { name: 'b', value: 3 }
or the largest value (max)
arr.reduce((a, b) => a.value > b.value ? a : b)
// { name: 'a', value: 5 }

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]

I had the same problem, I needed to obtain the minimum and maximum values of an array and, to my surprise, there were no built-in functions for arrays. After reading a lot, I decided to test the "top 3" solutions myself:
discrete solution: a FOR loop to check every element of the array against the current max and/or min value;
APPLY solution: sending the array to the Math.max and/or Math.min internal functions using apply(null,array);
REDUCE solution: recursing a check against every element of the array using reduce(function).
The test code was this:
function GetMaxDISCRETE(A)
{ var MaxX=A[0];
for (var X=0;X<A.length;X++)
if (MaxX<A[X])
MaxX=A[X];
return MaxX;
}
function GetMaxAPPLY(A)
{ return Math.max.apply(null,A);
}
function GetMaxREDUCE(A)
{ return A.reduce(function(p,c)
{ return p>c?p:c;
});
}
The array A was filled with 100,000 random integer numbers, each function was executed 10,000 times on Mozilla Firefox 28.0 on an intel Pentium 4 2.99GHz desktop with Windows Vista. The times are in seconds, retrieved by performance.now() function. The results were these, with 3 fractional digits and standard deviation:
Discrete solution: mean=0.161s, sd=0.078
APPLY solution: mean=3.571s, sd=0.487
REDUCE solution: mean=0.350s, sd=0.044
The REDUCE solution was 117% slower than the discrete solution. The APPLY solution was the worse, 2,118% slower than the discrete solution. Besides, as Peter observed, it doesn't work for large arrays (about more than 1,000,000 elements).
Also, to complete the tests, I tested this extended discrete code:
var MaxX=A[0],MinX=A[0];
for (var X=0;X<A.length;X++)
{ if (MaxX<A[X])
MaxX=A[X];
if (MinX>A[X])
MinX=A[X];
}
The timing: mean=0.218s, sd=0.094
So, it is 35% slower than the simple discrete solution, but it retrieves both the maximum and the minimum values at once (any other solution would take at least twice that to retrieve them). Once the OP needed both values, the discrete solution would be the best choice (even as two separate functions, one for calculating maximum and another for calculating minimum, they would outperform the second best, the REDUCE solution).

Iterate through, keeping track as you go.
var min = null;
var max = null;
for (var i = 0, len = arr.length; i < len; ++i)
{
var elem = arr[i];
if (min === null || min > elem) min = elem;
if (max === null || max < elem) max = elem;
}
alert( "min = " + min + ", max = " + max );
This will leave min/max null if there are no elements in the array. Will set min and max in one pass if the array has any elements.
You could also extend Array with a range method using the above to allow reuse and improve on readability. See a working fiddle at http://jsfiddle.net/9C9fU/
Array.prototype.range = function() {
var min = null,
max = null,
i, len;
for (i = 0, len = this.length; i < len; ++i)
{
var elem = this[i];
if (min === null || min > elem) min = elem;
if (max === null || max < elem) max = elem;
}
return { min: min, max: max }
};
Used as
var arr = [3, 9, 22, -7, 44, 18, 7, 9, 15];
var range = arr.range();
console.log(range.min);
console.log(range.max);

You can use the following function anywhere in your project:
function getMin(array){
return Math.min.apply(Math,array);
}
function getMax(array){
return Math.max.apply(Math,array);
}
And then you can call the functions passing the array:
var myArray = [1,2,3,4,5,6,7];
var maximo = getMax(myArray); //return the highest number

The following code works for me :
var valueList = [10,4,17,9,3];
var maxValue = valueList.reduce(function(a, b) { return Math.max(a, b); });
var minValue = valueList.reduce(function(a, b) { return Math.min(a, b); });

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)

let arr=[20,8,29,76,7,21,9]
Math.max.apply( Math, arr ); // 76

Simple stuff, really.
var arr = [10,20,30,40];
arr.max = function() { return Math.max.apply(Math, this); }; //attach max funct
arr.min = function() { return Math.min.apply(Math, this); }; //attach min funct
alert("min: " + arr.min() + " max: " + arr.max());

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;