Find the min/max element of an array in JavaScript
JavascriptJavascript Problem Overview
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
Javascript Solutions
Solution 1 - Javascript
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);
};
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 the spread operator which functions similarly to the apply
method:
var min = Math.min( ...arr ),
max = Math.max( ...arr );
Solution 2 - Javascript
var max_of_array = Math.max.apply(Math, array);
For a full discussion see: http://aaroncrane.co.uk/2008/11/javascript_max_api/
Solution 3 - Javascript
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;
};
Solution 4 - Javascript
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)
)
Solution 5 - Javascript
##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.
###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);
Solution 6 - Javascript
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.
Solution 7 - Javascript
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.
Solution 8 - Javascript
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);
});
Solution 9 - Javascript
.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']
Solution 10 - Javascript
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)
Solution 11 - Javascript
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 }
Solution 12 - Javascript
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
Solution 13 - Javascript
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.
Solution 14 - Javascript
One more way to do it:
var arrayMax = Function.prototype.apply.bind(Math.max, null);
Usage:
var max = arrayMax([2, 5, 1]);
Solution 15 - Javascript
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]
Solution 16 - Javascript
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.
Solution 17 - Javascript
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;
}
Solution 18 - Javascript
let array = [267, 306, 108] let longest = Math.max(...array);
Solution 19 - Javascript
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;
};
Solution 20 - Javascript
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);
Solution 21 - Javascript
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]
Solution 22 - Javascript
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).
Solution 23 - Javascript
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
Solution 24 - Javascript
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); });
Solution 25 - Javascript
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);
Solution 26 - Javascript
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)
Solution 27 - Javascript
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 }
Solution 28 - Javascript
> let arr=[20,8,29,76,7,21,9]
Math.max.apply( Math, arr ); // 76
Solution 29 - Javascript
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;
Solution 30 - Javascript
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());
Solution 31 - Javascript
Using Math.max()
or Math.min()
Math.max(10, 20); // 20
Math.min(-10, -20); // -20
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); // 3
var min = Math.min(...arr); // 1
Solution 32 - Javascript
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.
Solution 33 - Javascript
If you use the library sugar.js, you can write arr.min() and arr.max() as you suggest. You can also get min and max values from non-numeric arrays.
> min( map , all = false ) Returns the element in the array with the > lowest value. map may be a function mapping the value to be checked or > a string acting as a shortcut. If all is true, will return all min > values in an array. > > max( map , all = false ) Returns the element in the array with the > greatest value. map may be a function mapping the value to be checked > or a string acting as a shortcut. If all is true, will return all max > values in an array.
Examples:
[1,2,3].min() == 1
['fee','fo','fum'].min('length') == "fo"
['fee','fo','fum'].min('length', true) == ["fo"]
['fee','fo','fum'].min(function(n) { return n.length; }); == "fo"
[{a:3,a:2}].min(function(n) { return n['a']; }) == {"a":2}
['fee','fo','fum'].max('length', true) == ["fee","fum"]
Libraries like Lo-Dash and underscore.js also provide similar powerful min and max functions:
Example from Lo-Dash:
_.max([4, 2, 8, 6]) == 8
var characters = [
{ 'name': 'barney', 'age': 36 },
{ 'name': 'fred', 'age': 40 }
];
_.max(characters, function(chr) { return chr.age; }) == { 'name': 'fred', 'age': 40 }
Solution 34 - Javascript
You can use lodash's methods
_.max([4, 2, 8, 6]);
returns => 8
https://lodash.com/docs/4.17.15#max
_.min([4, 2, 8, 6]);
returns => 2
Solution 35 - Javascript
let arr = [2,5,3,5,6,7,1];
let max = Math.max(...arr); // 7
let min = Math.min(...arr); // 1
Solution 36 - Javascript
Try
let max= a=> a.reduce((m,x)=> m>x ? m:x);
let min= a=> a.reduce((m,x)=> m<x ? m:x);
let max= a=> a.reduce((m,x)=> m>x ? m:x);
let min= a=> a.reduce((m,x)=> m<x ? m:x);
// TEST - pixel buffer
let arr = Array(200*800*4).fill(0);
arr.forEach((x,i)=> arr[i]=100-i%101);
console.log('Max', max(arr));
console.log('Min', min(arr))
For Math.min/max (+apply) we get error:
> Maximum call stack size exceeded (Chrome 74.0.3729.131)
// TEST - pixel buffer
let arr = Array(200*800*4).fill(0);
arr.forEach((x,i)=> arr[i]=100-i%101);
// Exception: Maximum call stack size exceeded
try {
let max1= Math.max(...arr);
} catch(e) { console.error('Math.max :', e.message) }
try {
let max2= Math.max.apply(null, arr);
} catch(e) { console.error('Math.max.apply :', e.message) }
// same for min
Solution 37 - Javascript
If you are using prototype.js framework, then this code will work ok:
arr.min();
arr.max();
Documented here: Javascript prototype framework for max
Solution 38 - Javascript
minHeight = Math.min.apply({},YourArray);
minKey = getCertainKey(YourArray,minHeight);
maxHeight = Math.max.apply({},YourArray);
maxKey = getCertainKey(YourArray,minHeight);
function getCertainKey(array,certainValue){
for(var key in array){
if (array[key]==certainValue)
return key;
}
}
Solution 39 - Javascript
Here's a plain vanilla JS approach.
function getMinArrayVal(seq){
var minVal = seq[0];
for(var i = 0; i<seq.length-1; i++){
if(minVal < seq[i+1]){
continue;
} else {
minVal = seq[i+1];
}
}
return minVal;
}
Solution 40 - Javascript
You can use Array.sort but you'll have to write a simple number sorting function since the default is alphabetic.
Then you can grab arr[0]
and arr[arr.length-1]
to get min and max.
Solution 41 - Javascript
create a simple object
var myArray = new Array();
myArray = [10,12,14,100];
var getMaxHeight = {
hight : function( array ){ return Math.max.apply( Math, array );
}
getMaxHeight.hight(myArray);
Solution 42 - Javascript
I like Linus's reduce() approach, especially for large arrays. But as long as you know you need both min and the max, why iterate over the array twice?
Array.prototype.minmax = function () {
return this.reduce(function (p, v) {
return [(p[0] < v ? p[0] : v), (p[1] > v ? p[1] : v)];
}, [this[0], this[0]]);
}
Of course, if you prefer the iterative approach, you can do that too:
Array.prototype.minmax = function () {
var mn = this[0], mx = this[0];
this.forEach(function (v) {
if (v < mn) mn = v;
if (v > mx) mx = v;
});
return [mn, mx];
};
Solution 43 - Javascript
Below script worked for me in ndoejs:
var numbers = [1, 2, 3, 4];
console.log('Value:: ' + Math.max.apply(null, numbers) ); // 4
Solution 44 - Javascript
well I would like to do this in the below way
const findMaxAndMin = (arr) => {
if (arr.length <= 0) return -1;
let min = arr[0];
let max = arr[0];
arr.forEach((n) => {
n > max ? (max = n) : false;
n < min ? (min = n) : false;
});
return [min, max];
};
Solution 45 - Javascript
If you need performance then this is the best way for small arrays:
var min = 99999;
var max = 0;
for(var i = 0; i < v.length; i++)
{
if(v[i] < min)
{
min = v[i];
}
if(v[i] >= max)
{
max = v[i];
}
}
Solution 46 - Javascript
Insert the numbers seperated by a comma and click on the event you want to call ie Get the Max or min number.
function maximumNumber() {
var numberValue = document.myForm.number.value.split(",");
var numberArray = [];
for (var i = 0, len = numberValue.length; i < len; i += 1) {
numberArray.push(+numberValue[i]);
var largestNumber = numberArray.reduce(function (x, y) {
return (x > y) ? x : y;
});
}
document.getElementById("numberOutput").value = largestNumber;
}
function minimumNumber() {
var numberValue = document.myForm.number.value.split(",");
var numberArray = [];
for (var i = 0, len = numberValue.length; i < len; i += 1) {
numberArray.push(+numberValue[i]);
var smallestNumber = numberArray.reduce(function (x, y) {
return (x < y) ? x : y;
});
}
document.getElementById("numberOutput").value = smallestNumber;
}
function restrictCharacters(evt) {
evt = (evt) ? evt : window.event;
var charCode = (evt.which) ? evt.which : evt.keyCode;
if (((charCode >= '48') && (charCode <= '57')) || (charCode == '44')) {
return true;
}
else {
return false;
}
}
<div>
<form name="myForm">
<table>
<tr>
<td>Insert Number</td>
<td><input type="text" name="number" id="number" onkeypress="return restrictCharacters(event);" /></td>
<td><input type="button" value="Maximum" onclick="maximumNumber();" /></td>
<td><input type="button" value="Minimum" onclick="minimumNumber();"/></td>
<td><input type="text" id="numberOutput" name="numberOutput" /></td>
</tr>
</table>
</form>
</div>
Solution 47 - Javascript
if you have complex object, you could use sort....such as: if I want get item which contains MAX/MIN value of below objs.
var objs= [
{name:"Apple",value:3},
{name:"Love",value:32},
{name:"Cheese",value:1},
{name:"Pork",value:77},
{name:"Xmas",value:99}
];
I will do a sort:
objs.sort(function(a, b){return a.value-b.value});
Then:
objs[0]
is the MIN, objs[objs.length-1]
is the MAX.
Solution 48 - Javascript
To prevent "max" and "min" to be listed in a "for ... in" loop:
Object.defineProperty(Array.prototype, "max", {
enumerable: false,
configurable: false,
writable: false,
value: function() {
return Math.max.apply(null, this);
}
});
Object.defineProperty(Array.prototype, "min", {
enumerable: false,
configurable: false,
writable: false,
value: function() {
return Math.min.apply(null, this);
}
});
Usage:
var x = [10,23,44,21,5];
x.max(); //44
x.min(); //5
Solution 49 - Javascript
You may not want to add methods to the Array prototype, which may conflict with other libraries.
I've seen a lot of examples use forEach which, I wouldn't recommend for large arrays due to its poor performance vs a for loop. https://coderwall.com/p/kvzbpa/don-t-use-array-foreach-use-for-instead
Also Math.max(Math, [1,2,3]);
Always gives me NaN?
function minArray(a) {
var min=a[0]; for(var i=0,j=a.length;i<j;i++){min=a[i]<min?a[i]:min;}
return min;
}
function maxArray(a) {
var max=a[0]; for(var i=0,j=a.length;i<j;i++){max=a[i]>max?a[i]:max;}
return max;
}
minArray([1,2,3]); // returns 1
If you have an array of objects the minArray() function example below will accept 2 parameters, the first is the array and the second is the key name for the object key value to compare. The function in this case would return the index of the array that has the smallest given key value.
function minArray(a, key) {
var min, i, j, index=0;
if(!key) {
min=a[0];
for(i=0,j=a.length;i<j;i++){min=a[i]<min?a[i]:min;}
return min;
}
min=a[0][key];
for(i=0,j=a.length;i<j;i++){
if(a[i][key]<min) {
min = a[i][key];
index = i;
}
}
return index;
}
var a = [{fee: 9}, {fee: 2}, {fee: 5}];
minArray(a, "fee"); // returns 1, as 1 is the proper array index for the 2nd array element.
Solution 50 - Javascript
linear, almost-purely-functional-approach
var min=[0, 29, 25].map((function(max) {max=-Infinity; return function(e) {return max=Math.max(max, e);}})())[0]
More examples:
Finding out min value
function getMin(arr) {
return (ar || [0, 29, 25]).
map((function(max) {max=-Infinity; return function(e) {return max=Math.max(max, e);}})())[0];
}
or using Array.map
method with variable closuring
function getMin(arrObjs) {
return (arrObjs || [{val: 0}, {val: 29}, {val: 25}]).
map((function(max) {max=-Infinity; return function(e) {return max=(max.val>e.val?max:e);}})())[0];
}
Finding out max value
function getMax(arr) {
return (ar || [0, 29, 25]).
map((function(v) {v=Infinity; return function(e) {return v=Math.min(v, e);}})())[0];
}
for array of objects
function getMax(arrObjs) {
return (arrObjs || [{val: 0}, {val: 29}, {val: 25}]).
map((function(v) {v=-Infinity; return function(e) {return v=(v.val<e.val?v:e);}})())[0];
}
Solution 51 - Javascript
A recursive solution to the problem
const findMinMax = (arr, max, min, i) => arr.length === i ? {
min,
max
} :
findMinMax(
arr,
arr[i] > max ? arr[i] : max,
arr[i] < min ? arr[i] : min,
++i)
const arr = [5, 34, 2, 1, 6, 7, 9, 3];
const max = findMinMax(arr, arr[0], arr[1], 0)
console.log(max);
Solution 52 - Javascript
Another solution
let arr = [1,10,25,15,31,5,7,101];
let sortedArr = arr.sort((a, b) => a - b)
let min = sortedArr[0];
let max = sortedArr[arr.length-1]
console.log(`min => ${min}. Max => ${max}`)
Solution 53 - Javascript
Here is one more example. Calculate the Max/Min value from an array with lodash.
let array = [100, 0, 50];
var func = _.over(Math.max, Math.min);
var [max, min] = func(...array);
// => [100, 0]
console.log(max);
console.log(min);
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.js"></script>
Solution 54 - Javascript
I managed to solve my problem this way:
var strDiv = "4,8,5,1"
var arrayDivs = strDiv.split(",")
var str = "";
for (i=0;i<arrayDivs.length;i++)
{
if (i<arrayDivs.length-1)
{
str = str + eval('arrayDivs['+i+']')+',';
}
else if (i==arrayDivs.length-1)
{
str = str + eval('arrayDivs['+i+']');
}
}
str = 'Math.max(' + str + ')';
var numMax = eval(str);
I hope I have helped.
Best regards.