What's the best way in JavaScript to test if a given parameter is a square number?

Question

I created a function that will test to see if a given parameter is a square number.

Read about square numbers here: https://en.wikipedia.org/?title=Square_number

If the number is a square number, it returns true and otherwise false. Negative numbers also return false.

Examples:

isSquare(-12) // => false
isSquare( 5) // => false
isSquare( 9) // => true
isSquare(25) // => true
isSquare(27) // => false

Right now, I am using this method: http://jsfiddle.net/marcusdei/ujtc82dq/5/

But, is there a shorter more cleaner way to get the job done?

What exactly is wrong with the referenced method? Doesn't go much shorter than that... (and it's pretty clean too IMO) — Daniel, Jun 18 '15 at 15:35
yes, I just find this one even shorter, not sure about cleaner though: http://jsfiddle.net/marcusdei/ujtc82dq/7/ — Marco V, Jun 18 '15 at 15:38

score 53 · Accepted Answer · answered Jun 18 '15 at 15:08

53

Try this:

var isSquare = function (n) {
    return n > 0 && Math.sqrt(n) % 1 === 0;
};

Check if number is positive
Check if sqrt is complete number i.e. integer

Demo

answered Jun 18 '15 at 15:08

Tushar

85,780
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1

I don't think you need to check if the value is > 0. Also 0 is technically a square number. In your case it would evaluate to false. – mpn Aug 24 '16 at 14:51
@mpn For `0` as perfect square, you're right. For negative numbers, one function call can be saved. – Tushar Aug 24 '16 at 15:11
5

`n >= 0 && ...` - to include 0 – Sergiu Dec 13 '18 at 23:50

score 9 · Answer 2 · edited Apr 25 '18 at 03:44

9

I would definitely go for:

var isSquare = function (n) {
    return Math.sqrt(n) % 1 === 0;
};

PS: 0 is a square number for those who wonder

Demo

edited Apr 25 '18 at 03:44

Andrew Myers

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answered Apr 17 '18 at 14:55

Stephane Moreau

91
1
1

score 5 · Answer 3 · answered May 09 '20 at 15:29

//1st 
var isPerfectSquare = function(num) {
   return Math.sqrt(num) % 1 === 0;
}

//2nd: loop through all the number from 1 to num 
var isPerfectSquare = function(num) {
    
    for(let i=1; i <= num ; i++){
        let d = i * i;
        if(d === num){
            return true
        }
    }
}

// Optimize solution: Binary Search 
var isPerfectSquare = function(num) {

    if(num ==1)return true
    let left = 2;
    let right = Math.floor(num/2);
    while(left <= right){
        let middle = Math.floor((left + right)/2)
        let sqr = middle * middle;
        if(sqr == num){
            return true
        }else{
            if(sqr > num){
              right = middle -1
            }else{
                left = middle + 1
            }
        }  
    }
    
    return false
};

CodeSignal tests state that the 1st method is the fastest! (note: 0 is full square also) — abd0991, Jan 17 '23 at 16:34

score 4 · Answer 4 · edited Oct 15 '20 at 01:03

4

I went that route:

var isSquare = (n) => n === 0 ? true : n > 0 && Math.sqrt(n) % 1 === 0;

console.log(isSquare(25));
console.log(isSquare(10));
console.log(isSquare(16));

edited Oct 15 '20 at 01:03

Pramod Mali

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17
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answered Oct 14 '20 at 03:38

Fred Cal

41
1

score 2 · Answer 5 · answered Mar 31 '20 at 00:43

I think that this one is a shorter and a cleaner option:

  var isSquare = function(n) {

  return Number.isInteger(Math.sqrt(n));
};

isSquare(25); //true

for even shorter and cleaner than that you could use:

var isSquare = n => Number.isInteger(Math.sqrt(n));

isSquare(25);//true

score 1 · Answer 6 · answered Nov 16 '19 at 18:24

It's a bit trickier if you're using the new BigInt in JavaScript:

// integer square root function (stolen from the interwebs)
function sqrt(n) {
  let a = 1n;
  let b = (n >> 5n) + 8n;
  while (b >= a) {
    let mid = (a + b) >> 1n;
    if (mid * mid > n) {
      b = mid -= 1n;
    } else {
      a = mid += 1n;
    }
  }
  return a -= 1n;
}

sqrt(25n) === 5n
sqrt(26n) === 5n
...
sqrt(35n) === 5n

The best and fastest way I've found (so far) to determine if n is a square is:

function isSquare(n) {
   return n%sqrt(n) === 0n
}

But there's gotta be a faster way for BigInt operations.

score 0 · Answer 7 · answered Mar 18 '20 at 12:24

0

Isn't this (Math.sqrt(number) % 1 === 0) basically enough? it just checks if the sqrt of the number is a whole number, if so, then it's a perfect square.

Obviously, depending on what you want to do with that information, it may require extra code.

answered Mar 18 '20 at 12:24

AnonymousContribute

1

Hi AnonymousContribute, isn't your answer the same as Stephane's answer: https://stackoverflow.com/a/49881377/5488275 ?If so, please upvote this answer instead of adding your own :) – Nander Speerstra Mar 18 '20 at 12:46

What's the best way in JavaScript to test if a given parameter is a square number?

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