Minimum and Maximum of an array in JavaScript

Question

I need to sort an array of numbers, so that it returns its minumum and its maximum. The code I've written so far doesn't seem to do anything at all.

function sort(array) {
    arrayNew = [];
    maximum = array[0];
    minimum = array[0];

    for (i = 0; i < array.length; i++) {
        if (maximum < array[i]) {
            maximum = array[i];
        }
    }
    arrayNew.push(maximum);

    for (i = 0; i < array.length; i++) {
        if (minimum > array[i]) {
            minimum = array[i];
        }
    }
    arrayNew.unshift(minumum);
    return arrayNew;
}


var arr1 = [3, 8, 7, 6, 5, -4, 3, 2, 1];
alert(sort(arr1));

You have a misspelling: arrayNew.unshift(minumum);

trnelson
– trnelson

2014-10-22 10:26:01 +00:00
Commented Oct 22, 2014 at 10:26 — trnelson
– trnelson, Commented Oct 22, 2014 at 10:26

Andy · Accepted Answer · 2014-10-22 10:25:08Z

2

In case you didn't know there is a much simpler way to get the min and max values from an array - use Math.min and Math.min with apply:

var min = Math.min.apply(null, arr1);
var max = Math.max.apply(null, arr1);

DEMO

answered Oct 22, 2014 at 10:25

Andy

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2 Comments

Gioacchino Gargiulo Over a year ago

Thanks for your reply, I knew about the Math object, I was just trying this as a "side" exercise of one I was assigned

thefourtheye Over a year ago

Even I thought about this, but then I decided to go with the looping. Because, it finds the max and min in single iteration.

xec · Accepted Answer · 2014-10-23 10:02:25Z

1

minumum is spelled incorrectly. It should have been minimum
Don't use your variables without declaring them with var keyword. Otherwise they will become global properties.

Also, you can have your minimum check also in the same loop as maximum. So, your code can be shortened like this

function sort(array) {
    var maximum = array[0],
        minimum = array[0],
        i;

    for (i = 0; i < array.length; i++) {
        if (maximum < array[i]) {
            maximum = array[i];
        }

        if (minimum > array[i]) {
            minimum = array[i];
        }
    }
    return [minimum, maximum];
}

Now that we know, we can do it like this, lets try to shorten it more with ternary operator

function sort(array) {
    var maximum = array[0],
        minimum = array[0],
        i;

    for (i = 0; i < array.length; i++) {
        maximum = maximum < array[i] ? array[i] : maximum;
        minimum = minimum > array[i] ? array[i] : minimum;
    }
    return [minimum, maximum];
}

edited Oct 23, 2014 at 10:02

xec

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answered Oct 22, 2014 at 10:24

thefourtheye

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1 Comment

Andy Over a year ago

Not for (var i = 0? particularly given your second point? @thefourtheye?

score 1 · Accepted Answer · 2014-10-22 10:44:24Z

1

var arr = [ 3, 8, 7, 6, 5, -4, 31, 2, 21, 20, 1 ].sort(function(a, b) { return a - b }),  // [-4, 1, 2, 3, 5, 6, 7, 8, 20, 21, 31]
    min = arr[0],  // min
    max = arr[arr.length-1];  //max

edited Oct 22, 2014 at 10:44

answered Oct 22, 2014 at 10:26

user235273

2 Comments

caffeinated.tech Over a year ago

javascript's default sort is lexiographic not numerical, that function won't work for numbers above 10. Yes the OP used the same array but they never said they are limited to this set. Best to correct it for posterity.

user235273 Over a year ago

Updated the sort method with a sort function.

thefourtheye · Accepted Answer · 2014-10-23 03:22:18Z

0

arrayNew.unshift(minumum); is wrong variable minimum

correct one : arrayNew.unshift(minimum);

edited Oct 23, 2014 at 3:22

thefourtheye

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answered Oct 22, 2014 at 10:24

Gurkan İlleez

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2 Comments

PM 77-1 Over a year ago

Please double-check your answer. It does not make any sense.

Gurkan İlleez Over a year ago

I checked my answer and he is using wrong variable. minumum does not exists.

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