Sorting the array of string based on the custom order

Question

I have the array of strings that I want to sort in a custom order My array looks like this

["E", "D", "CC", "C", "B", "BB", "BBA", "BBD", "BBE", "BBB", "BBBC", "A", "AA"]

In my sorted array I want to show the words alphabetically but the words with large number of continuous letter should appear first.

If more number of words have same length of continuous letter then need to consider next letter alphabetically.

My expected result should look like this

["AA", "A","BBBC", "BBB","BBA", "BBD", "BBE", "BB", "B", "CC", "C", "D", "E"]

I tried to sort using the default sort() function but it didn't give me expected results. So please give your suggestions

.sort() can have a callback function as argument to be used as a comparer of 2 elements. You can write your sorting rule there — Cid
– Cid, Commented Jan 18, 2022 at 5:53
@JMP I have edited the question based on your question. Hope it helps to understand my query — Prince Solomon
– Prince Solomon, Commented Jan 18, 2022 at 6:10
How can you sort BBA, BBD, BBE, BBB to be BBB, BBA, BBD, BBE? Surely it should be BBA, BBB, BBD, BBE? Or does it depend on how the elements are stored in the original array? Or are elements with all the same character somehow worth more? — Jim Jimson
– Jim Jimson, Commented Jan 18, 2022 at 6:31

Jim Jimson · Accepted Answer · 2022-01-18 09:22:35Z

2

Ok, this should do what you want. It takes into consideration the length of the string, and whether all characters are the same and finally sorts by alpha after that:

let rawArray = ["E", "D", "CC", "C", "B", "BB", "BBA", "BBD", "BBE", "BBB", "BBBC", "A", "AA"];

rawArray.sort((a,b) => {
  
  if (a.substring(0, 1) === b.substring(0, 1) && allEqual(a) && !allEqual(b)) {
    return -1
  } 
  else if (a.length > b.length){
    return -1
  }
  else if (a < b && !allEqual(a + b)){
    return -1
  }
    
  else {
    return 1
  }
    
});


function allEqual(input) {
  
    return input.split('').every(char => char === input[0]);
}

console.log(rawArray); //["AA", "A", "BBBC", "BBB", "BBA", "BBD", "BBE", "BB", "B", "C", "CC", "D", "E"]

I made use of another answer to get the allEqual function. You can find that here.

You could also shorten the sort function, but it makes it less readable:

rawArray.sort((a,b) =>   
  (a.substring(0, 1) === b.substring(0, 1) && allEqual(a) && !allEqual(b))
              ||
              (a.length > b.length)
              ||
              (a < b && !allEqual(a + b))
              ? -1 : 1    
);

edited Jan 18, 2022 at 9:22

answered Jan 18, 2022 at 6:05

Jim Jimson

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2 Comments

Prince Solomon Over a year ago

I am getting output like [ "AA", "A", "BBBC", "BBB", "BBA", "BBD", "BBE", "BB", "B", "C", "CC", "D", "E" ] but expected one is [ "AA", "A", "BBBC", "BBB", "BBA", "BBD", "BBE", "BB", "B", "CC", "C", "D", "E" ] Note the letter C.

Jim Jimson Over a year ago

Ok, that should be fixed now.

Cid · Accepted Answer · 2022-01-18 06:28:52Z

2

I'd first write a function that compares 2 elements with the expected rules and returns 1 if the first element should be first, else -1, check the length of the 2 elements and see if the longest starts with the smallest (like in "AAA" and "A")

function myComparer(a, b)
{
  if (a.length < b.length)
  {
    if (b.startsWith(a))
    {
      return 1; // a should be after b
    }
  }
  else if (a.length > b.length)
  {
    if (a.startsWith(b))
    {
      return -1; // b should be after a
    }
  }

  if (a < b)
  {
    return -1;
  }
  return 1;
}

const CompareTester = (a, b) => {
  if (myComparer(a, b) == -1)
  {
    console.log(`"${a}" then "${b}"`);
  }
  else
  {
    console.log(`"${b}" then "${a}"`);
  }
}

const testArr = [
  { a: "AAA", b: "A" },
  { a: "A", b: "AAA" },
  { a: "BBA", b: "BB" },
  { a: "A", b: "B" },
  { a: "ABC", b: "DEF" },
  { a: "ABC", b: "ABCDEF" }
];

for (elem of testArr)
{
  CompareTester(elem.a, elem.b);
}

Now, you can use this function as a callback for the .sort() method :

const arr = ["E", "D", "CC", "C", "B", "BB", "BBA", "BBD", "BBE", "BBB", "BBBC", "A", "AA"];

function myComparer(a, b)
{
  if (a.length < b.length)
  {
    if (b.startsWith(a))
    {
      return 1; // a should be after b
    }
  }
  else if (a.length > b.length)
  {
    if (a.startsWith(b))
    {
      return -1; // b should be after a
    }
  }

  if (a < b)
  {
    return -1;
  }
  return 1;
}

arr.sort(myComparer);

console.log(arr);

answered Jan 18, 2022 at 6:28

Cid

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4 Comments

Jim Jimson Over a year ago

This one returns: ["AA", "A", "BBA", "BBBC", "BBB", "BBD", "BBE", "BB", "B", "CC", "C", "D", "E"] instead of ["AA", "A","BBBC", "BBB","BBA", "BBD", "BBE", "BB", "B", "CC", "C", "D", "E"] though.

Cid Over a year ago

@JimJimson can you reduce your example so I can see what you mean?

Jim Jimson Over a year ago

According to the OP, BBB needs to come first, then BBA, not BBA first and then BBB after. It's a very, very weird requirement! I gave you an upvote for effort though haha

Cid Over a year ago

I see, could be a typo in OP example, because "AA" and "AAA" are just supposed to be string samples

danh · Accepted Answer · 2022-01-18 07:33:04Z

2

The challenge was understanding the sort rules implicit in the question, but I think I can restate the OP to suggest that the strings should be sorted first by the lexical order of the first letter, then by the run length of the first set of repeating characters, then by the whole string length, and last by the lexical order of the remaining chars.

To expose that logic clearly in the sort, This snippet preprocesses the input strings into objects like:

{ input: "BBBC", firstChar: "B", runLength: 3, fallback: "C", length: 4 }

If all that's right, then the sort is simple...

function sortInfo(original) {
  const firstLetter = original[0];
  const length = original.length;
  let runLength = 0;
  for (let letter of original) {
    if (letter === firstLetter) runLength++;
    else break;
  }
  const fallback = original.slice(runLength)
  return {  original, firstLetter, runLength, length, fallback }
}

let input = ["E", "D", "CC", "C", "B", "BB", "BBA", "BBD", "BBE", "BBB", "BBBC", "A", "AA"];
let sortable = input.map(sortInfo)

sortable.sort((a,b) => {
  if (a.firstLetter !== b.firstLetter) return a.firstLetter.localeCompare(b.firstLetter);
  if (a.runLength !== b.runLength) return b.runLength - a.runLength;
  if (a.length !== b.length) return b.length - a.length;
  return a.fallback.localeCompare(b.fallback);
});

console.log(sortable.map(o => o.original))

edited Jan 18, 2022 at 7:33

answered Jan 18, 2022 at 7:17

danh

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4 Comments

ceving Over a year ago

I think the prefix rule must be done recursively. He did not answer to the question if AABB comes before AABA, but I think this must be so.

Jim Jimson Over a year ago

AABA comes before AABB. Length first (BBBC), then all same letter followed by alpha sort (BBB, BBA, BBC, BBD)

danh Over a year ago

Yeah, @JimJimson, dunno. I'm still not sure about the OP intent. My aim was to expose a useful way of articulating the problem. Cheers.

danh Over a year ago

@ceving, I see, so it would be first letter, then length of the first repeating string, then recursively on the remainder. That's interesting. New website idea: "overflowstack"... people post answers, and other people try to work out what the question is. :-)

Nimantha · Accepted Answer · 2022-01-19 12:29:56Z

2

You need your own custom sorting algorithm for this kind of sorting.

var yourArray = ["E", "D", "CC", "C", "B", "BB", "BBB", "A", "AA"];  // Unsorted Array  

let customerSort = (arr) => 
{
    // This loop is for iterating array
    for (var i = 0; i < arr.length; i++) 
    {   
        // This loop is for comparing each item of the array with all items in array
        for (var j = 0; j < arr.length; j++)
        {
            // This condition is for sorting array in alphabetical order
            if (arr[i] < arr[j]) 
            {
                var x = arr[i];
                arr[i] = arr[j];
                arr[j] = x; 
            }

            // This condition is for sorting array in the custom order needed
            if (arr[i].charAt(0) == arr[j].charAt(0) && arr[i].length > arr[j].length) 
            {
                var x = arr[i];
                arr[i] = arr[j];
                arr[j] = x; 
            }
        }
    }
    // Return Sorted Array
    return arr;
}

console.log(customerSort(yourArray));

edited Jan 19, 2022 at 12:29

Nimantha

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answered Jan 18, 2022 at 6:29

Coderrr47

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3 Comments

Jim Jimson Over a year ago

Close. But this also produces the incorrect result: ["AA", "A", "BBBC", "BBA", "BBB", "BBD", "BBE", "BB", "B", "CC", "C", "D", "E"] vs ["AA", "A", "BBBC", "BBB", "BBA", "BBD", "BBE", "BB", "B", "CC", "C", "D", "E"]

Coderrr47 Over a year ago

Is the length of the string a consideration while sorting

Coderrr47 Over a year ago

Is BBB the only exception

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