I wanna make my code better but don't know how at this point. I'd like someone else's opnion to tell me how can I improve [closed]

Question

So a friend of mine asked if I could do a function that compresses strings. That function compresses a string if one character repeats itself more than four times, like: "abbbbbbcccdddd" will turn into "a#6bccc#4d". I'm noob and I wanna improve. Thank u for who is answering me :).

let text = '92HSSSS9U27888YSGGGGGG28SLKHFKLSPPPPSKWWW'

function compressText(text){
    return searchIqualLetters(text)
}

function searchIqualLetters(originalText){
    let modifyedText = ''

    for(let i = 0; i < originalText.length - 1; i++){

        for(let j = i + 1; j < originalText.length; j++){
            
            modifyedText = repeatFourTimes(i, j, originalText)
            if(modifyedText !== undefined && modifyedText !== ''){
                originalText = modifyedText
            }
        }
    }
    return originalText
}

function checkTextModifyed(){

}

function repeatFourTimes(i, j, text){
    if(text[i] == text[j] && text[i] == text[j + 1] && text[i] == text[j + 2]){
        return getLastIqualLetter(i, j, text)
    }
}

function getLastIqualLetter(i, j, text){
    let lastIqualLetterPosition = 0

    for(let k = 0; text[i] == text[j + k]; k++){
        lastIqualLetterPosition = j + k
    }

    return makeNewText(i, lastIqualLetterPosition, text)
}

function makeNewText(i, lastLetter, text){
    let cutedPart = text.slice(i, lastLetter + 1)

    let startPart = text.slice(0, i)
    let centerPart = '#' + cutedPart.length + cutedPart[0]
    let endPart = text.slice(lastLetter + 1, text.length)
    
    return startPart + centerPart + endPart 
}

console.log(compressText(text)) // Result: 92H#4S9U27888YS#6G28SLKHFKLS#4PSKWWW

Answer 1

Here's my take on the problem:

function condense(text) {
    let count = 0;
    let out = "";
    text.split("").forEach((el, i, arr) => { // el = current letter, i = index of letter, arr = the input text as an array
        if (arr[i + 1] != el) { // if the next letter is not the same as the current one
            if (count < 3) {
                out += el.repeat(count + 1) // if the sequence is less than 3 letters, then just do it normally
            } else {
                out += "#" + (count + 1) + el; // else, condense it
            }
            count = 0; // reset sequence length
        } else {
            count += 1; // increase count of repeated letter sequence length
        }
    })
    return out;
}
console.log(condense("aaabbbbcdddddee")); // Output: "aaab#4cd#5ee"

console.log(condense('92HSSSS9U27888YSGGGGGG28SLKHFKLSPPPPSKWWW'));
console.log("92H#4S9U27888YS#6G28SLKHFKLS#4PSKWWW (reference)");

Answer 2

Here is my take on it:

const text = '92HSSSS9U27888YSGGGGGG28SLKHFKLSPPPPSKWWW';

function compr(txt){
  let  cnt=1,arr=[];
  txt.split("").reduce((p,c,i)=>{
    if(c==p) cnt++;
    if(c!=p || i==txt.length-1) {
     arr.push(cnt>3?`#${cnt}${p}`:p.repeat(cnt));
     cnt=1;
    }
    return c;
   });
   return arr.join("");
}

console.log(compr(text));
console.log("92H#4S9U27888YS#6G28SLKHFKLS#4PSKWWW (reference)");

The core of the action happens in the .reduce() loop. Here I compare the current character c with the previous one(p). If it is the same I simply increase the counter cnt otherwise - or in case I have reached the end of the string (i==txt.length-1) - I add the latest group of character in its suitable form to arr and reset cnt to 1. The arr array is eventually .join()ed and returned as the resulting string.