Chunky Monkey (under Basic Algorithm Scripting) question

JavaScript
1

Hi, I manage to complete this challenge but I didn’t manage to use the array.push() method. I have included my code below :

/** setup

function chunkArrayInGroups(arr, size) {
// Break it up.
var totalArr = [];

if (arr.length/size <= 2) {
firstArr= arr.splice (0, size);
secondArr= arr;
totalArr = [firstArr, secondArr];
}

else if (arr.length/size <= 3) {
firstArr= arr.splice (0, size);
secondArr= arr.splice (0,size);
thirdArr = arr;
totalArr = [firstArr, secondArr, thirdArr]; }

else if (arr.length/size <= 4) {
firstArr= arr.splice (0, size);
secondArr= arr.splice (0,size);
thirdArr = arr.splice(0,size);
fourthArr= arr;
totalArr = [firstArr, secondArr, thirdArr, arr]; }

return totalArr;

}

chunkArrayInGroups([0, 1, 2, 3, 4, 5], 2);

**/ end setup

I feel like my code is not complete because if chunkArrayInGroups ([0, 1, 2, 3, 4, 5], 1), the result is only [].

Can anyone push me to the right direction to produce a better code for this challenge?

Many thanks in advance :slight_smile:

2

Why do you want to avoid array.push()? Your method cannot scale up.

3

What is any number divided by 1 equal to? You don’t have an else case to handle that. So it’s just passing you the empty array.

But as @soarer alluded, doing it this way will only solve as many test cases as you are able to think ahead to hard code in.

4

Yea, I know that. It’s just that I can’t seem to find a way to use array.push() in my code.

5

I’m stuck on this one as well. But I think you would want totalArr = [firstArr, secondArr];
to be where you have the array.push().

6

Hi guys I wanted to share my solution. I used the splice method to split the array based on the size given and then push the spliced Array into a completely newArray. Tell me what you think.:slight_smile:

function chunkArrayInGroups(arr, size) {

var arraySize  = arr.length/size;

var newArray = [];

    for (var i = 0; i < arraySize ; i++){

        var subArray= arr.splice(0,size);

        newArray.push(subArray);

    }

return newArray;

}`
15 Likes
7

@TTomaszito That looks pretty good, another option is to remember than for loops don’t need to increment by one:

function chunkArrayInGroups(arr, size) {
// Break it up.
var chunkArr = [];
for (var i=0; i < arr.length; i += size){
chunkArr = chunkArr.concat(Array.of(arr.slice(i, i + size)));
}
return chunkArr;

}

6 Likes
8

@ChadKreutzer that’s interesting I never thought about other increments. Makes loads of sense thanks for the tip.

9

Great solution TTomaszito!

If you want to force yourself to use “.slice” instead of “.splice” here is what you can do:

chunkArr.push(arr.slice(i*size+0, size*i+size));

1 Like
10

hey i did this dont know if the correct approach though(scalable)

function chunkArrayInGroups(arr, size) {
var multiD = [];

arr.map(function(element,index){
if(index % size === 0){
multiD.push(arr.slice(index,size+index));

}

});
//console.log(multiD);

return multiD;
}

chunkArrayInGroups([“a”, “b”, “c”, “d”], 2);

1 Like
11

I can see these posts are months old. I’m just starting through algorithms and using the forum to see better solutions (there is always more than one!)

thought a bit of refactoring will help to do away with concat & Array.of (and to bring in push)… since slice returns an array, using push will get us a two dimensional…

function chunkArrayInGroups(arr, size) {
// Break it up.
var chunkArr=[];
for(var index=0;index<arr.length;index+=size)
chunkArr.push(arr.slice(index,index+size));
return chunkArr;
}

5 Likes
12

Good job getting past the challenge. :thumbsup:

Here is my code:


function chunkArrayInGroups(arr, size) {
  // An array to store the results
  results = [];
  // Divide the length of 'arr' by 'size' and 'ceil' it off to next number. 'round' is different from 'ceil'
  count = Math.ceil(arr.length/size);
  
  for (var i=0; i<count; i++){
  	// Push the sliced array to 'results'
  	results.push(arr.slice(0, size));
  	// Remove the sliced elements from the array and keep the rest
  	arr = arr.slice(size, arr.length);
  }
  return results;
}
1 Like
13

Here is my code in case you are curious, this took some trial and error but I finally got it. My first solution had a FOR loop and extra function, I wanted to clean it up so here it is:

function chunkArrayInGroups(arr, size) {
  // Break it up.
  var newArr = [];
  var len = arr.length;
  for (var i = 0; i < len / size; i++){
    newArr.push(arr.slice(0,size));
    arr.splice(0,size);
    console.log( 'Loop ' + i + ' :' +newArr );
    console.log( arr );
  }

  console.log( newArr );
  return newArr;
}

chunkArrayInGroups([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 2);
14

Thanks! I was really close to finish just like you, but I was slicing my array in wrong points. Your solution clarified some things up :slight_smile:

15

Hi everyone, just want to thank you for posting your solutions. I managed to solve it but I didn’t use “slice”, only “push”. Somehow I got stuck in a line of thought and tried to get a solution no matter what. But seeing your elegant and concise solutions made me realize I should have approached it in a different way. And that’s really useful. Here’s my code:

function chunkArrayInGroups(arr, size) {
  // Break it up.
  var arrFinal = [];
  var arrToPush = [];
  var resetArr = 0;
  
  for( var i = 0; i < arr.length; i++){
      
      if (resetArr < size) {
        arrToPush.push(arr[i]);
        resetArr++;
        
        if (resetArr === size || i + 1 == arr.length){ // se exceder o size ou for a última posiçã
          arrFinal.push(arrToPush);
          arrToPush = [];
          resetArr = 0;
        }

      }

  }
  //console.log(arrFinal);
  return arrFinal;
}

chunkArrayInGroups(["a", "b", "c", "d"], 4);
16

here’s my solution if anyone is interested. :slight_smile:

This thread was helpful in coming to my answer, so maybe it will continue to help others as they look for help

function chunkArrayInGroups(arr, size) {
          //Make segment picker beginning
          var sliceStart = 0;
          //Make segment picker end
          var sliceEnd = size;
          //Make answer array
          var outarr = [];
          ///Loop through array in increments of the size param
          for (var i = 0; i < arr.length; i = i + size) {
            //push the slices to the answer array
            outarr.push(arr.slice(sliceStart,sliceEnd));
            //Increment the segment pickers
            sliceStart += size;
            sliceEnd += size;
          }
          //Answer out
          return outarr;
        }

        chunkArrayInGroups(["a", "b", "c", "d"], 2);
1 Like
17

Here’s my solution, pretty simple. Not that the challenge is that simple, my code is.

function chunkArrayInGroups(arr, size) {
  var newArray = [];
  
  for(var i = 0; i < arr.length; i+=size){
    newArray.push(arr.slice(i,i+size));
  }
  return newArray;
}

chunkArrayInGroups(["a", "b", "c", "d"], 2);
18

I chose to use splice as well, but forgot to follow up with a sub-array! This was a huge help! Thanks!

19 
var newArray = [];
        
   while (arr.length){
    newArray.push(arr.splice(0, size));
   }
  return newArray;
21

@rmdawson71 I see your point. It definitely make sense.