Changing the value of an index inside an array

Question

I'm stuck with this problem where I need to eliminate a letter from each of the array values and change the first letter if it is in index 0. My current approach is this:

function capital(arr, char){
    let str = "";
    let result = "";
for (let i = 0; i < arr.length; i++){
    str = arr[i] + (i < arr.length - 1 ? ",": "");;

    for (let j = 0; j < str.length; j++){
        if (str[j] === char){
            result += "";
             if (str[j] === char){
            result += (j === 0? "A": "");
        }
        
        else {
            result += str[j];
        }
    }
}
    console.log(result);
}
capital(["doritos","sandking","bandana", "demand"], "d");

The program should eliminate each occurrence of letter d found in the strings and change the index 0 to letter A if the d is in index 0.

The current result is

Aoritos,sanking,banana,Aeman

but it should look like

Aritos,sanking,banana,Aman

The requirement does not allow any use of built in functions, and the actual program requires the letters to be case insensitive, but I can work on it by substituting codes and adding couple if elses, I just need help with the changing the index 0 condition. Any help would be appreciated, thank you!

your example seems off to me, please correct it.

Harsh Saini
– Harsh Saini

2021-08-06 05:58:49 +00:00
Commented Aug 6, 2021 at 5:58 — Harsh Saini
– Harsh Saini, Commented Aug 6, 2021 at 5:58

Yogen Darji · Accepted Answer · 2021-08-06 06:56:07Z

1

You can check if your input char is at first index str.indexOf(char) then simply add prefix "A" to your string without first char i.e str.substring(1)

function capital(arr, char) {
  let str = "";
  let result = "";
  for (let i = 0; i < arr.length; i++) {
    str = arr[i];
    if(str.indexOf(char) === 0) {
      result = 'A' + str.substring(1);
    }else {
      result = str;
    }

    console.log(result);
  }
}
capital(["doritos", "sandking", "bandana", "demand"], "d");

edited Aug 6, 2021 at 6:56

answered Aug 6, 2021 at 6:01

Yogen Darji

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Comments

Azad · Accepted Answer · 2021-08-06 06:23:18Z

0

this logic may help you

function capital(arr, char) {

  return arr.map(e => {

    let isFirst = false;
    if (e[0] == "d") isFirst = true;
    e = e.replace(new RegExp(char, 'ig'), '');
    if (isFirst)
      e = e.replace(e.charAt(0), "A");
    return e;
  });

}
console.log(capital(["doritos", "sandking", "bandana", "demand"], 'd'))

answered Aug 6, 2021 at 6:23

Azad

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Comments

Hassan Imam · Accepted Answer · 2021-08-06 07:14:26Z

You can iterate through each letter of your word and update the first letter if it is matched with passed letter and at 0 index, for other index if it doesn't match with passed letter add to your result.

const capitalize = ch => {
 const letter = ch.charCodeAt(0);
 if(letter >= 97 && letter <= 122) {
    return String.fromCharCode(letter - 32);
 }
 return ch;
}

const capital = (words, ch) => {
  let result = '';
  for(let i = 0; i < words.length; i++) {
    let newWord = '';
    for(let j = 0; j < words[i].length; j++) {
      if(capitalize(words[i][j]) === capitalize(ch) && j === 0) {
        newWord = 'A';
      }
      if(capitalize(words[i][j]) !== capitalize(ch)) {
        newWord += words[i][j];
      }
    }
    result += newWord;
    if(i < words.length - 1 ) {
      result += ',';
    }
  }
  return result;
}

const result = capital(["doritos","sandking","bandana", "demand", "sandDking"], "d");
console.log(result);

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