Find the missing letter in sequence

Question

The challenge was to find the missing letter in an array in alphabetical order.

Find the missing letter in the passed letter range and return it.

If all letters are present in the range, return undefined.

My solution was this:

function fearNotLetter(str) {
  var letters = str.split('');
  var codes = [];
  var missing = 0;
  var start = 0 , next = 0 , c = 0;
  for(var i= 0; i < letters.length; i++){
    var charcode = letters[i].toString();
    codes.push(charcode.charCodeAt());
  }
  for(var j = 1; j < codes.length; j++){
    start = codes[j] -1;
    if(codes.indexOf(start) === -1){
      missing += start;
    }

  }
  if(missing === 0){
    return undefined
  }else {
    return String.fromCharCode(missing);
  }
}

I am wondering whether I have followed some bad practices or used dry code which might cause problems in a later use. Thanks in advance.

Answer 1

In general, you don't need to split the string and then process the array. Simply you must start from the first letter and store its charcode and then process the next later and check with the incremented charcode. If there is a difference than just break.

function f(str) {
    if (str && 0 < str.length) {
        var desCharCode = str.charCodeAt(0);
        for (var i = 1; i < str.length; ++i) {
            ++desCharCode;
            if (str.charCodeAt(i) != desCharCode) {
                return String.fromCharCode(desCharCode);
            }
        }
    }

    return undefined;
}

Answer 2

\$O(n)\$ time solution which is functionally equivalent to @Ozan's original solution.

Does not require the string to be sorted, nor to contain no duplicates.

function fearNotLetter(str) {
  let codes = {};
  let min = str.charCodeAt(0),
      max = min;

  for (let i = 0, l = str.length; i < l; i++) {
    const c = str.charCodeAt(i);

    codes[c] = true;

    if (c < min) {
      min = c;
    } else if (c > max) {
      max = c;
    }
  }

  for (let i = min+1; i < max; i++) {
    if (codes[i] === undefined) {
      return String.fromCharCode(i);
    }
  }
}