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The challenge to make a shape area calculation in CodeSignal

Camilo Martinez on September 21, 2018

Languages: [🇪🇸] Español - [🇺🇸] English Since college, I was not encouraged to do programming challenges, until I got the invitation to CodeSign...

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Frederik 👨‍💻➡️🌐 Creemers • Sep 22 '18

I think think there's an even easier solution:

shapeArea = function(n){
    return 1 + (n-1)*4
}

Xerghos • Dec 6 '18

I think this is more appropriate:

function shapeArea(n) {
return (1 + (n-1)*n) *2 - 1
}

TrungSuSi97 • Jun 25 '19

Why do you think that formula ?

Axel Espinosa • Jun 28 '19

Did you solve it?

Axel Espinosa • Jun 28 '19

Hey buddy, can you explain your solution?

Camilo Martinez • Sep 22 '18

Yeah, that's exactly the beauty that I talk about code fights. You can learn from other solution.

kikeflowers • Mar 20 '21 • Edited

I found this solution and just replaced the variables

function shapeArea(n) {
return 1 + 2 * n * (n-1)
//1 + 2 * 4 * (3) = 3*4=12*2=24+1=25
//1 + 2 * 3 * (2) = 2*3=6*2=12+1=13
//1 + 2 * 2 * (1) = 1*2=2*2=4+1=5
//1 + 2 * 1 * (0) = 0*1*=0*2=0+1=1
}

Lara Potjewyd • Jan 5 '19 • Edited

Hi! Would it be possible for someone to explain what's happening in the code? I'm a beginner to both code and maths. Any resources and tutorials would be appreciated too :)

Camilo Martinez • Jan 6 '19 • Edited

There is no tutorial about this specific ploblem. You need learn other concepts like recursion and loops.

This function shapeArea is not important. It's used to measure the amount of time.

Relevant things on this exercise is x.loopShapeArea, x.recursionShapeArea and x.mathShapeArea. Three of them solves the problem, but math is more efficient than others.

stxenhammer • Feb 20 '20

But where did you get the 4 from & multiplying it ?

For example when I can recite something for 60 times in a min so if i recite for 100 mins i just do 60 * 100 to find out how many times I recite a line.
What was ur reasoning behind using the number 4 ? I am not clear on how I would go on solving this on my own. @

Camilo Martinez • Feb 21 '20 • Edited

I'm using a technic learned in a subject Numerical Methods. You need reproduce step by steep and try to find something familiar between each interaction.

A square have 4 sides, that's why you need multiply the value by four.

But each side merge a point with other in the other side (red circle), you can't count both. That's why yo need subtract 4, one for each corner.

That's how I found this formula: n * 4 - 4

On first and second iteration is difficult to see it... but next iterations can give you an idea about the solution.

Hope it helps.

Rudolf Arthur Frans Gutlich • Jun 11 '21

the simpler way I think is:


return Math.pow(n,2) + Math.pow(n-1,2)

Just rotate the grid view by 45º and check the patterns

Akshay Dalvi • Mar 30 '19

return n*n + (n-1)*(n-1);

Eric Curtin • Nov 16 '18

It's more like:

int shapeArea(int n) {
  int area = 1;
  while (n > 1) {
    area += (n-- - 1) * 4;
  }
  return area;
}

hodasalah • Sep 14 '23 • Edited

I tried both solution

function solution(n) {
    return 2*n*(n-1) +1;
    //return n*2+(n-1)*2
}

kikeflowers • Mar 20 '21

function shapeArea(n) {
var res = 1;
for(let i = 1; i < n; i++){
res = res + (i*4)
}
return res;
}

Mark Kelly • Feb 20 '21

return n * ((n - 1) * 2) + 1;

Carlos Ayala • Jul 16 '20

n*(n*2*-1)-int((n*2-1/2)

Radwa Alaa • Mar 6 '22

this is the shortest valid solution :

int output = 1;
for (int i = 1; i <= n; i++) {
if (i != 1) {
output = output + 4 * (i - 1);
}
}

return output;

hodasalah • Sep 14 '23

i tried both solution
function solution(n) {
return 2*n*(n-1) +1;
//return n*2+(n-1)*2
}