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 pseudocode:

let array = [100, 0, 50]

array.min() //=> 0
array.max() //=> 100

Answer 1

The best way with Math.min(...array) and Math.max(...array)

Answer 2

Min/Max Alternative Methods

---

The Math.min and Math.max are great methods to get the minimum and maximum items out of a collection of items. However, it's important to be aware of some cavities that can come with it.

Using them with an array containing a large number of items (more than ~10⁷ items, depending 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

---

Some browsers might return a NaN value instead. It might be a better way to handle errors, but it doesn't solve the problem just yet.

Instead, for very large numbers, consider using something like this:

function maxValue(arr) {
  return arr.reduce((max, val) => max > val ? max : val)
}
const arr = Array.from(Array(1000000).keys());
console.log(maxValue(arr)); // 999999

Better run-time:

function maxValue(arr) {
  let max = arr[0];

  for (let val of arr) {
    if (val > max) {
      max = val;
    }
  }
  return max;
}
const arr = Array.from(Array(1000000).keys());
console.log(maxValue(arr)); // 999999

---

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] });
}
const arr = Array.from(Array(1000000).keys());
console.log(getMinMax(arr)); // {"min":0,"max":999999}

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 };
}
const arr = Array.from(Array(1000000).keys());
console.log(getMinMax(arr)); // {"min":0,"max":999999}

* Tested with 1,000,000 items:
Just for reference, the 1st function run-time (on my machine) was 15.84ms vs. 2nd function with only 4.32ms.

Answer 3

Finding the Max and Min elements of an array in JavaScript.

There are several approaches you can use:

Using Math.min() and Math.max()

let array = [100, 0, 50];
Math.min(...array); // 0
Math.max(...array); // 100

Using Sorting

let array = [100, 0, 50];
arraySorted = array.toSorted((a, b) => a - b); // [0, 50, 100];
arraySorted.at(0);  // 0
arraySorted.at(-1); // 100

Using simple for-loop

let array = [100, 0, 50];
let maxNumber = array[0];
let minNumber = array[0];
for (let i = 1; i < array.length; i++) {
  if (array[i] > maxNumber) {
    maxNumber = array[i];
  }
  if (array[i] < minNumber) {
    minNumber = array[i];
  }
}

Answer 4

In this day and age (in 2022), the most efficient way to get min + max from an array is to do it in a single iteration, via reduce.

• In JavaScript:

const arr = [3, 0, -2, 5, 9, 4];

const i = arr.reduce((p, c) => {
    p.min = c < p.min ? c : p.min ?? c;
    p.max = c > p.max ? c : p.max ?? c;
    return p;
}, {min: undefined, max: undefined});

console.log(i); //=> { min: -2, max: 9 }

And when the input has no data, it will output {min: undefined, max: undefined}.

In TypeScript, you would just add type casting, so the return type is inferred as {min: number | undefined, max: number | undefined}, and not as {min: any, max: any}:

const arr = [3, 0, -2, 5, 9, 4];

const i = arr.reduce((p, c) => {
    p.min = c < p.min! ? c : p.min ?? c;
    p.max = c > p.max! ? c : p.max ?? c;
    return p;
}, {min: undefined, max: undefined} as { min: number | undefined, max: number | undefined });

console.log(i); //=> { min: -2, max: 9 }

UPDATE

Following kiran goud comment, here's an alternative that uses arrays instead of objects:

const i = arr.reduce((p, c) => {
    p[0] = c < p[0] ? c : p[0] ?? c;
    p[1] = c > p[1] ? c : p[1] ?? c;
    return p;
}, [undefined, undefined]);

console.log(i); //=> [-2, 9]

Answer 5

To add to the many good answers here, here is a typescript version that can handle lists where some values are undefined.

How it can be used:

const testDates = [
  undefined,
  new Date('July 30, 1986'),
  new Date('July 31, 1986'),
  new Date('August 1, 1986'),
]
const max: Date|undefined = arrayMax(testDates); // Fri Aug 01 1986
const min: Date|undefined = arrayMin(testDates); // Min: Wed Jul 30 1986
const test: Date = arrayMin(testDates); // Static type error
const anotherTest: undefined = arrayMin(testDates); // Static type error

The definitions (the notEmpty definition is from this post):

function arrayMax<T>(values?: (T | null | undefined)[]): T | undefined {
    const nonEmptyValues = filterEmpty(values);
    if (nonEmptyValues.length === 0) {
        return undefined;
    }
    return nonEmptyValues.reduce((a, b) => (a >= b ? a : b), nonEmptyValues[0]);
}

function arrayMin<T>(values?: (T | null | undefined)[]): T | undefined {
    const nonEmptyValues = filterEmpty(values);
    if (nonEmptyValues.length === 0) {
        return undefined;
    }
    return nonEmptyValues.reduce((a, b) => (a <= b ? a : b), nonEmptyValues[0]);
}

function filterEmpty<T>(values?: (T | null | undefined)[] | null): T[] {
    return values?.filter(notEmpty) ?? [];
}

function notEmpty<T>(value: T | null | undefined): value is T {
    if (value === null || value === undefined) return false;
    const testDummy: T = value;
    return true;
}

I didn't use the Math.max function as suggested in the documentation because this way I can use this function with any comparable objects (if you know how to type this let me know so I can better define T).

Answer 6

For learning purpose, you can do it by using variables and for loop without using built-in functions.

// Input sample data to the function
var arr = [-1, 0, 3, 100, 99, 2, 99];
// Just to show the result
console.log(findMinMax(arr));

function findMinMax(arr) {
  let arraySize = arr.length;
  if (arraySize > 0) {
    var MaxNumber = MinNumber = arr[0];
    for (var i = 0; i <= arraySize; i++) {
      if (arr[i] > MaxNumber) {
        MaxNumber = arr[i];
      }else if(arr[i] < MinNumber) {
        MinNumber = arr[i];
      }
    }
    var minMax = [MinNumber,MaxNumber];
    return minMax;
  } else {
    return 0;
  }
}

Answer 7

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 );

Answer 8

Alternative Solns

class SmallestIntegerFinder {
  findSmallestInt(args) {
    return args.reduce((min,item)=>{ return (min<item ? min : item)});
  }
}

class SmallestIntegerFinder {
  findSmallestInt(args) {
    return Math.min(...args)
  }
}

class SmallestIntegerFinder {
  findSmallestInt(args) {
    return Math.min.apply(null, args);
  }
}

class SmallestIntegerFinder {
  findSmallestInt(args) {
    args.sort(function(a, b) {
    return a - b; } )
    return args[0];
  }
}

Answer 9

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;
};

Answer 10

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}`)

Answer 11

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)
)

Answer 12

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];
};

Answer 13

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 14

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

Answer 15

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

Answer 16

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

Answer 17

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 }

Answer 18

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 19

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);
});

Answer 20

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

https://lodash.com/docs/4.17.15#min

Answer 21

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 22

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);

Answer 23

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;
}

Answer 24

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

Answer 25

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);

Answer 26

let arr = [2,5,3,5,6,7,1];

let max = Math.max(...arr); // 7
let min = Math.min(...arr); // 1

Answer 27

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>

Answer 28

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>

Answer 29

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); });

Answer 30

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.