Javascript- how to find the minimum of an array using a function?

Question

I have the following function

function min() {
  var array = Array.prototype.slice.call(arguments);
  array = array.length === 1 && isNumeric(array[0].length) ? array[0] : array;

  var min = array[0];
  var i, count;
  for (i = 1, count = array.length; i < count; i++) {
    if (array[i] < min) min = array[i];
  }
  return min;
},

I don't understand why the following line was placed in, what is its purpose?

array = array.length === 1 && isNumeric(array[0].length) ? array[0] : array;

That whole function seems way too complicated when you can just call Math.min.apply(null, arrayOfNumbers) OR Math.min.call(null, num1, num2, num3) — Niet the Dark Absol
– Niet the Dark Absol, Commented May 5, 2016 at 15:02

Sergiu Paraschiv · Accepted Answer · 2016-05-05 15:02:15Z

3

condition ? returnIfTrue : returnIfFalse is called a ternary operator.
array.length === 1 && isNumeric(array[0].length) means "If array has a single element and that first element is itself an Array".
a = ternaryOperator means set a to the result of the ternary operator.

These three together mean that you can call min in two ways:

min(1, 2, 3) or min([1, 2, 3]).

answered May 5, 2016 at 15:02

Sergiu Paraschiv

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Comments

Dark Falcon · Accepted Answer · 2016-05-05 15:01:10Z

2

That line allows passing numbers directly as arguments or passing a single array as the first argument:

min(1,2,3);
min([1,2,3]);

answered May 5, 2016 at 15:01

Dark Falcon

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Comments

Redu · Accepted Answer · 2016-05-05 15:16:30Z

1

I would refactor this function into

var min = (...args) => Array.isArray(args[0]) ? Math.min(...args[0]) : Math.min(...args);

document.write("<pre>" + min([1,2,3,4,-1]) + "</pre>");
document.write("<pre>" + min(1,2,-3,4,-1) + "</pre>");

answered May 5, 2016 at 15:16

Redu

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2 Comments

Sergiu Paraschiv Over a year ago

Arrow functions are a bit of a stretch here. Browser support is not really there yet.

Redu Over a year ago

True.. Safari doesn't support arrows but i just wanted to demonstrate the idea. It's no big deal to implement the same thing in conventional ways. It's also worth to notice that exactly the same ...args syntax work both as rest and spread operator.

adam.v · Accepted Answer · 2016-05-05 15:08:05Z

-1

Your code is confusing, try doing something like this

var min = 1000000;
 var temp = 0;
if ( element < min){
 temp = element; 
 min = element;
 }

answered May 5, 2016 at 15:08

adam.v

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