Find the smallest and largest value in an array with JavaScript

Question

I am trying to write an algorithm that finds and smallest and largest value in an array, and the second largest and second smallest.

I tried with the following:

numbers = [2, 4, 9, 2, 0, 16, 24]

var largest = numbers[0];
var smallest = numbers[0];

for (var i = 1; i < numbers.length; i++) {

  if (numbers[i] > largest) {
    largest = numbers[i];
  } else if (numbers[i] < smallest) {
    smallest = numbers[i];
  }

  console.log(largest);
  console.log(smallest);
}

This does not seem to work and just prints out the array...what am I doing wrong?

Your logic's fine for finding the largest and smallest numbers – you just need to move the `console.log` statements outside the loop. — Rick Hitchcock, Jul 07 '16 at 22:21
Oh wow, how did I not see that...how embarrassing. Any suggestions for getting the second largest and the second smallest? — MadCatm2, Jul 07 '16 at 22:25
"Any suggestions for getting the second largest and the second smallest?" --- create a function that would maintain a sorted array of N elements. Then feed it with your `numbers` array. — zerkms, Jul 07 '16 at 22:26
@MadCatm2 Use the same algorithm, just use a new array with the same data and remove the largest and smallest from it. Or sort the array and get index `1` and `n-2` (assuming `n-1` is the last element and `0` is the first). — Spencer Wieczorek, Jul 07 '16 at 22:28
Oh, but that would just give me my number array sorted from smallest to largest no? I just need to return one value that is the second smallest... — MadCatm2, Jul 07 '16 at 22:30

score 9 · Accepted Answer · answered Jul 07 '16 at 22:29

The easiest way to do this would be to sort the array, then return the first two and last two elements.

Using slice() prevents the array itself from being sorted:

var numbers = [2, 4, 9, 2, 0, 16, 24];

var sorted = numbers.slice().sort(function(a, b) {
  return a - b;
});

var smallest = sorted[0],                      
    secondSmallest = sorted[1],                
    secondLargest = sorted[sorted.length - 2], 
    largest  = sorted[sorted.length - 1];

console.log('Smallest: ' + smallest);
console.log('Second Smallest: ' + secondSmallest);
console.log('Second Largest: ' + secondLargest);
console.log('Largest: ' + largest);

score 3 · Answer 2 · answered Jul 07 '16 at 22:22

3

Move your console.log statements outside of your for loop.

answered Jul 07 '16 at 22:22

Chester Millisock Jr

240
1
9

score 3 · Answer 3 · answered Nov 18 '18 at 01:09

You can use the spread operator to pass each array element as an argument to Math.min() or Math.max().

Using this method, you can find the smallest and largest number in an array:

const numbers = [2, 4, 9, 2, 0, 16, 24];

const smallest_number = Math.min(...numbers);
const largest_number = Math.max(...numbers);

console.log('Smallest Value:', smallest_number); // Smallest Value: 0
console.log('Largest Value:', largest_number);   // Largest Value: 24

score 2 · Answer 4 · answered Jul 22 '20 at 04:23

2

const numbers = [2, 4, 9, 2, 0, 16, 24];

//we can use reduce to find the largest and smallest number

const largest = numbers.reduce((previousValue, currentValue) =>
            previousValue > currentValue ? previousValue : currentValue
        );

const smallest = numbers.reduce((previousValue, currentValue) =>
            previousValue < currentValue ? previousValue : currentValue
        );

answered Jul 22 '20 at 04:23

Ram Prajapati

21
2

2

Please add some description with your answer. – SayAz Jul 22 '20 at 04:55

score 1 · Answer 5 · edited May 23 '17 at 11:53

As Roatin Marth (edited by Mr. Llama) said in this post: https://stackoverflow.com/a/6102340/6074388 ,

You can make the arrays use math.min and math.max like below

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

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

If this causes any problems you can also use

var min = Math.min.apply(null, largest),
max = Math.max.apply(null, largest);

var min = Math.min.apply(null, numbers),
max = Math.max.apply(null, numbers);

It won't help with finding the second smallest and second largest numbers, but I find this to be a simpler solution for finding the largest and smallest numbers.

score 1 · Answer 6 · answered Feb 18 '19 at 08:04

numbers = [2, 4, 9, 2, 0, 16, 24]
var largest = numbers[0];
var smallest = numbers[0];
for (var i = 0; i < numbers.length; i++){
    var temp = numbers[i];
    if (temp > largest)
    {
        largest = numbers[i];
    }
    if (temp < smallest)
    {
        smallest = numbers[i];
    }
}
console.log(largest);
console.log(smallest);

score 0 · Answer 7 · answered Jul 10 '19 at 23:43

0

var numbers = [2, 4, 9, 2, 0, 16, 24];
var largest = -Infinity;
var smallest = Infinity;

for (var i = 0; i < numbers.length; i++) {
    if (numbers[i] > largest) {
        largest = numbers[i];
    }
}

for (var i = 0; i < numbers.length; i++) {
    if (numbers[i] < smallest) {
        smallest = numbers[i];
    }
}

console.log(largest);
console.log(smallest);

answered Jul 10 '19 at 23:43

maja

1
1

please add a description to your code snippet to explain how it solves the problem – vs97 Jul 11 '19 at 00:04

score 0 · Answer 8 · answered Jul 22 '20 at 08:08

You can use the reduce function for this!

var numbers = [45, 4, 9, 16, 25, 100, 43];

var max = numbers.reduce((total, value, index, array) => {
    if(index === 0) return array[0];
    return total > value ? total : value;
});
console.log(max);

var min = numbers.reduce((total, value, index, array) => {
    if(index === 0) return array[0];
    return total < value ? total : value;
});
console.log(min);

You can learn how the reduce function works here. It is really useful when you have to derive a result from all the values of an array.

I'm also using the ternary operator for less code.

score 0 · Answer 9 · answered May 28 '21 at 04:39

0

let arr = [1,2,3,4,5]


const bigandsmallNum = function(){
    const smallNum = Math.min(...arr)
    const largeNum = Math.max(...arr)
    console.log(smallNum,largeNum)

 }
console.log(bigandsmallNum())

answered May 28 '21 at 04:39

Bala kumar

1

score 0 · Answer 10 · answered Dec 02 '21 at 20:40

You can use reduce to return an object with largest and smallest:

const array = [22, 3, 56 , 7, 14, 100, 2, 44]

const findLargestAndSmallest = (arr) => arr.reduce((acc, cur) => {
  acc = acc.largest > cur ? 
   {...acc, largest: acc.largest} : 
   {...acc, largest: cur}
  acc = acc.smallest < cur ? 
   {...acc, smallest: acc.smallest} :
   {...acc, smallest: cur}
  return acc
}, {})

console.log(findLargestAndSmallest(array)) // { largest: 100, smallest: 2 }

// or destruct it: 

const {largest, smallest} = findLargestAndSmallest(array)
console.log(largest) // 100
console.log(smallest) // 2

score 0 · Answer 11 · answered Apr 12 '22 at 00:38

0

const x = [1,2,3,4];

const y = Math.min(...x);

const z =Math.max(...x)

answered Apr 12 '22 at 00:38

ahmed alsaedi

1
1

score 0 · Answer 12 · answered Jul 22 '22 at 19:26

How you can get the first and second largest number in array I hope this would be helpful for everyone thanks

// first method

const getLargestNumber = (nums) => {
    let largest = [0];
    for (let i = 0; i <= nums.length; i++) {
      if (largest <= nums[i]) {
        largest = nums[i]
      }
    }
    return largest;
}

const array = [0, 2, 3, 6, 6, 5]

console.log("first method", getLargestNumber(array))

// second method

const maxNumber = Math.max(...array);

console.log("Second Method", maxNumber)

// third method

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

console.log("Third Method", getMaxNumber)

const getSecondLargest = (nums) => {
    let largest = Math.max(...nums), newArr = [];
    for (let i = 0; i <= nums.length; i++) {
      if ((nums[i] !== undefined) && !(largest <= nums[i])) {
        newArr.push(nums[i]);
      }
    }
    return Math.max(...newArr);
}

const array = [0, 2, 3, 6, 6, 5]

console.log(getSecondLargest(array))

Find the smallest and largest value in an array with JavaScript

12 Answers12