return the first non repeating character in a string in javascript

Question

So I tried looking for this in the search but the closest I could come is a similar answer in several different languages, I would like to use Javascript to do it.

T

Answer 1

let arr = [10, 5, 3, 4, 3, 5, 6];
outer:for(let i=0;i<arr.length;i++){
    for(let j=0;j<arr.length;j++){
        if(arr[i]===arr[j+1]){
            console.log(arr[i]);
            break outer;
        }
    }
}


//or else you may try this way...
function firstDuplicate(arr) {
    let findFirst = new Set()
    for (element of arr)
        if (findFirst.has(element ))
            return element 
        else
            findFirst.add(element )
}

Answer 2

You can iterate through each character to find() the first letter that returns a single match(). This will result in the first non-repeated character in the given string:

const first_nonrepeated_character = string => [...string].find(e => string.match(new RegExp(e, 'g')).length === 1);
const string = 'aabcbd';

console.log(first_nonrepeated_character(string)); // c

Answer 3

Two further possibilities, using ECMA5 array methods. Will return undefined if none exist.

Javascript

function firstNonRepeatedCharacter(string) {
    return string.split('').filter(function (character, index, obj) {
        return obj.indexOf(character) === obj.lastIndexOf(character);
    }).shift();
}

console.log(firstNonRepeatedCharacter('aabcbd'));

On jsFiddle

Or if you want a bit better performance, especially on longer strings.

Javascript

function firstNonRepeatedCharacter(string) {
    var first;

    string.split('').some(function (character, index, obj) {
        if(obj.indexOf(character) === obj.lastIndexOf(character)) {
            first = character;
            return true;
        }

        return false;
    });

    return first;
}

console.log(firstNonRepeatedCharacter('aabcbd'));

On jsFiddle

Answer 4

Easy way to solve this algorithm, very straight forward.

function firstNonRepeatChar(str){
    let map = {};
    for(let i=0; i<str.length; i++){
        if(Object.keys(map).includes(str[i])){
            map[str[i]]++
        }
        else{
            map[str[i]] = 1;
        }
    }

    for(let j=0; j< Object.values(map).length; j++){
        if(Object.values(map)[j] == 1){
            console.log(Object.keys(map)[j]);
            return
        }
        if (j == Object.values(map).length-1 && Object.values(map)[j] != 1){
            console.log('_');
            return;
        }
        else{
            continue;
        }
    }
}

nonRepeat("aaabbcccdeeef");

Answer 5

    //To find first non repeating letter 
    //It will check for both upper and lower case
    //only use one String.indexOf()

    var mystr="ohvhvtccggt";
    var checkFirstNonRepeating=function(){
    var ele=[];
      for(var i=0;i<mystr.length;i++) {
      var key=mystr.charAt(i);
      if(!ele[key])
          ele[key]=0;
      ele[key]++;

   //Just check for second occurance of character 
   //no need to use indexOf twice

       if(mystr.indexOf(key,i+1)==-1 && ele[key]<2)  
         return mystr[i];
       }
     return "All repeating letters";
    }
    console.log(checkFirstNonRepeating());
              
        /*
        Input  : "ohvhvtoccggt"
        Output : All repeating letters
        Input  :"oohjtht"
        Output :j 
        */

Answer 6

This solution should works with array with integers and string.

function firstNoneRepeating(list, map = new Map()) {
  for (let item of list) {
    if (map.has(item)) {
      map.set(item, map.get(item) + 1);
    } else {
      map.set(item, 1);
    }
  }
  for (let [key, value] of map.entries()) {
    if (value === 1) {
      return key;
    }
  }
}
console.log(firstNoneRepeating("aabcbd"));
console.log(firstNoneRepeating([5, 2, 3, 4, 2, 6, 7, 1, 2, 3]));