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Need to generate a new string in javascript

I'm trying to solve a question to create a new compressed string. Each character has a count next to it and I was supposed to come with up a new string in alphabetical order followed by the total count of that character within the string.

For eg:

Input: "a10b1a5c1" Output: "a15b1c1"

Input: "b10a1b5c1" Output: "a1b15c1"

How I approached this question?

  1. Have an object with key as char and value as count
  2. Convert the object to string

function stringCompression(s) {
const object={}

    for(var i=0; i<s.length-1; i++) {
      if(s[i].match(/[0-9]/)!==null && s[i+1].match(/[a-zA-Z]/) === null) {
        if(object.hasOwnProperty("s[i]")){
          // Add previous and current value
          } else {
          object[s[i-1]]=parseInt(s[i]+s[i+1]) 
          }
    }
  }
return object
}


// output should be a15b1c1
const output= stringCompression("a10b1a5c1");
const final = Object.keys(output).map(key => `${key}${output[key]}`).join("");
console.log(final)

Question:

  1. Can't think of a way to sort alphabetically
  2. The function I wrote doesn't handle if the character has a single digit count
  3. Is there any other way I can optimise the solution?
about 4 years ago · Juan Pablo Isaza
2 Respuestas
Responde la pregunta

0

You could reduce() everything to a single "counter" object, and then use Object.entries() to convert that object to an array and perform sorting, flattening and joining on that array:

const compress = (s) => Object.entries(
  [...s.matchAll(/([A-Za-z])(\d+)/g)].reduce((a, [_, char, number]) => ({
    ...a,
    [char]: (a[char] || 0) + +number
  }), {})
).sort(([a], [b]) => a.localeCompare(b)).flat().join('');

console.log(compress('b10a1b5c1'));

about 4 years ago · Juan Pablo Isaza Denunciar

0

const STRING = 'a10c8b3a4'


const isDigit = (charCode) => {
    return charCode >= 48 && charCode <=57
}

const isAlphabet = (charCode) => {
    return charCode >= 97 && charCode <= 122
}

const getSortedObject = (obj) => {
return Object.keys(obj).sort().reduce(
  (object, key) => { 
    object[key] = obj[key]; 
    return object;
  }, {}
);
}

const getCompressedString = (string) => {
    let obj = {}
  let alphabet = ''
  let digit = ''
  let i = 0
  
  while(i<string.length) {
    const char = string[i]
    const charCode = string.charCodeAt(i)
    
    if(isAlphabet(charCode)) {
        alphabet = char
      if(!obj[alphabet]) {
        obj[alphabet] = '0'
      }
      i++
    } else if(isDigit(charCode)) {
        digit = '0'
        while(isDigit(string.charCodeAt(i)) && i<string.length) {
        digit += string[i]
        i++
      }
      obj[alphabet] = parseInt(obj[alphabet]) + parseInt(digit)
    }
  }
  
  let compressedString = ''
    Object.keys(getSortedObject(obj)).forEach(key => {compressedString += key + obj[key]})
  
  return compressedString
}

console.log(getCompressedString(STRING))

about 4 years ago · Juan Pablo Isaza Denunciar
Responde la pregunta
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