How to stop a loop after finding n elements javascript

Question

I am writing a quick bit of code which find the first three even numbers in a random length of random numbers in an array.

  const getEven = (array) => {
  const evenArr = [];
  for (let index = 0; index < array.length; index++) {
    if (index % 2 != 0) {
      evenArr.push(array[index]);
    }
  }
  console.log(evenArr);
};

getEven([1, 2, 3, 4, 5, 6, 7, 8, 9]);
Output = [ 2, 4, 6, 8 ]

The question I have is, once it finds the first three even elements the loop continues to run, how can I stop it after finding the first three elements and is there a more efficient way of doing this?

I could use a filter but decided to go traditional and wrote a for-loop.

coolDog · Accepted Answer · 2022-04-18 09:13:09Z

2

Try this

 if (index % 2 != 0 && evenArr.length<3)

answered Apr 18, 2022 at 9:13

coolDog

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1 Comment

Zack Amin Over a year ago

This is really clean, and I never even thought of it! thanks.

James Jr. · Accepted Answer · 2022-04-18 09:08:39Z

1

Well, you may just add break statement with your desired condition:

 const getEven = (array) => {
  const evenArr = [];
  for (let index = 0; index < array.length; index++) {
    if (index % 2 !== 0) {
      evenArr.push(array[index]);
    }
    
    if (evenArr.length === 3) {
      break;
    }
  }
  console.log(evenArr);
};

getEven([1, 2, 3, 4, 5, 6, 7, 8, 9]);
Output = [ 2, 4, 6, 8 ]

answered Apr 18, 2022 at 9:08

James Jr.

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1 Comment

Zack Amin Over a year ago

Amazing, thanks Sergey. I did not know you could use a break with conditions. Great to learn something new!

Ville Venäläinen · Accepted Answer · 2022-04-18 09:09:56Z

1

With your current approach, the easiest change would be to add break -statement to your loop:

  for (let index = 0; index < array.length; index++) {
    if (index % 2 != 0) {
      evenArr.push(array[index]);
      if (arrray.length >= 3) {
        break; // <= this will end the loop
      }
    }    
  }

It is also good to mention that your algorithm is collecting odd numbers not even numbers currently. SHould be index % 2 === 0.

answered Apr 18, 2022 at 9:09

Ville Venäläinen

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2 Comments

Zack Amin Over a year ago

Thanks, great answer. i.e. Even numbers are those numbers that can be divided into two equal groups or pairs and are exactly divisible by 2. For example, 2, 4, 6, 8, 10, cuemath.com/numbers/even-numbers

Ville Venäläinen Over a year ago

Exactly my point. The code is adding values to the array when they are not even. ndex % 2 != 0 is true when the value is odd value. 2 % 2 = 0 // even 3 % 2 = 1 // odd ==> != 0 4 % 2 = 0 // event

Ilijanovic · Accepted Answer · 2022-04-18 09:11:24Z

1

Well you can use break

   if (evenArr.length === 3) {
      break;
   }

or return

   if (evenArr.length === 3) {
      return;
   }

or better:

   if (evenArr.length === 3) {
      return evenArr
   }

Return not just stops the loop but the whole function

const getEven = (array) => {
  const evenArr = [];
  for (let index = 0; index < array.length; index++) {
    if (index % 2 != 0) evenArr.push(array[index]);
    if(evenArr.length === 3) return evenArr;
  }
  return evenArr
};

answered Apr 18, 2022 at 9:11

Ilijanovic

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