CODEWITHSUNDEEP
javascript
Sara Ree
Sara Ree

Reputation: 3543

adding to array from last element to the first one

I want some help to modify or create a function to do a simple task;

Here is my function:

function arra(sum, length){
  var array = [];
  for(let i = 0; i < length; i++) {
    array.push(0);
  }
  
  //console.log(array) 
  for(let i = 0; i < sum; i++){
     array[i] ++
  }
  
  
  return array;
}

console.log(arra(0, 7));
console.log(arra(3, 7));
console.log(arra(7, 7));
console.log(arra(10, 7));

Here first, we create an array with a length, then we push 0 to each element of it. so the array will be : [0, 0, 0, 0 ,0 ,0 ,0]

Then based on the value of sum I want to add 1 from the last element of the array to the beginning. The problem is I want to come back to the first element and keep adding 1 again if there is enough sum.

So if I execute this: arra(3, 7); I would have : [0, 0, 0, 0 ,1 ,1, 1]

if I execute this: arra(7, 7) I would have: [1, 1, 1, 1 ,1 ,1 ,1]

But the problem is when I execute this: arra(10, 7) I get : [1, 1, 1, 1, 1, 1, 1, NaN, NaN, NaN] that is unexpected . I want to have : [1, 1, 1, 1 ,2 ,2 ,2]

Another problem is I can't add to array from the last element to the first...

Edit: How can fix it guys?

Upvotes: 3

Views: 1445

Answers (4)

Nina Scholz
Nina Scholz

Reputation: 386680

You could calculate the min and max values for dispersing and get the length of the right part as well.

function getArray(sum, length) {
    var max = Math.ceil(sum / length),
        min = Math.floor(sum / length),
        l = sum - min * length;

    return Array.from({ length }, (_, i) => length - i > l ? min : max);
}

console.log(...getArray(0, 7));
console.log(...getArray(3, 7));
console.log(...getArray(7, 7));
console.log(...getArray(10, 7));
console.log(...getArray(20, 7));

Upvotes: 2

Sandeep Ranjan
Sandeep Ranjan

Reputation: 834

You can just start from the end and reset back to end when you reach 1 the first element.

function arra(sum, length){
  var array = [];
  for(let i = 0; i < length; i++) {
    array.push(0);
  }
 
  let x = length -1;

  while(sum > 0) {
    if(x < 0) { // go back to the end
      x = length -1;
    }
    array[x]++;
    x--;
    sum--;
  }
  
  return array;
}

console.log(arra(0, 7));
console.log(arra(3, 7));
console.log(arra(7, 7));
console.log(arra(10, 7));

Upvotes: 1

F.bernal
F.bernal

Reputation: 2694

You get [1, 1, 1, 1, 1, 1, 1, NaN, NaN, NaN] this due to you initialize the array with the value of length, in this case you are calling arra(10,7) 

for(let i = 0; i < length; i++) {
   array.push(0);
}

Here you are getting an array of 7 elements initialized to 0

[0,0,0,0,0,0,0]

Next you execute:

 for(let i = 0; i < sum; i++){
     array[i] ++
  }

You get 1 in the first 7 elements you initialized to 0, but the next 3 you get NaN

Why?

Due to array[i] is undefined and; undefined with the operator ++ coerces to NaN.

If you want to create a circular array:

let array = null;

function circularArray(sum, length) {
  array = Array(length).fill(0);

  for (let count = 0; count < sum; count++) {
    array[count % array.length]++;
  }
}

circularArray(10, 7);

console.log(array.reverse());

Upvotes: 4

Mark
Mark

Reputation: 92460

If you want to "wrap around" you can index the array with the modulus of the length: array[i%length]. If you want to start at the other end, you can subtract from the array[length - 1 - i % length] — it wasn't clear to me which order you wanted, but the same idea works either way:

function arra(sum, length){
  var array = Array(length).fill(0)   
  
  //console.log(array) 
  for(let i = 0; i < sum; i++){
     array[length - 1 - i % length] ++
  }
  
  
  return array;
}

console.log(arra(0, 7));
console.log(arra(3, 7));
console.log(arra(7, 7));
console.log(arra(10, 7));

Upvotes: 6

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