Daily Challenge #206 - Pound Means Backspace

Assume "#" is like a backspace in string. This means that string "a#bc#d" would actually be "bd".

Implement a function that will process a string with "#" symbols and understand them as backspaces.

Examples

"abc#def##ghi###" ==> "abd"
"abc#d##c" ==> "ac"
"abc##d######" ==> ""
"#######" ==> ""
"" ==> ""

Tests

cleanString('abc#de##c')
cleanString('abc####dhh##c#')
cleanString('Thee# Empires# SS#tateBuildingg#')

Good Luck!

---

This challenge comes from vetalpaprotsky on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

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Comments

[Cameron Desautels]

A quick Clojure solution, leveraging a (linked) list as a stack:

(defn clean-string [s]
  (->> (reduce #(if (= \# %2) (pop %1) (conj %1 %2)) '() s)
       (reverse)
       (apply str)))

[Patrick Tingen]

Progress 4GL

FUNCTION cleanString RETURNS CHARACTER
  (pcString AS CHARACTER):

  DEFINE VARIABLE i AS INTEGER   NO-UNDO.
  DEFINE VARIABLE c AS CHARACTER NO-UNDO.

  DO i = 1 TO LENGTH(pcString):
    CASE SUBSTRING(pcString,i,1):
      WHEN '#' THEN IF c > '' THEN c = SUBSTRING(c, 1, LENGTH(c) - 1).
      OTHERWISE c = c + SUBSTRING(pcString,i,1).
    END CASE. 
  END.

  RETURN c.
END FUNCTION. 

MESSAGE cleanString("abc#def##ghi###") = "abd" 
   SKIP cleanString("abc#d##c") = "ac"
   SKIP cleanString("abc##d######") = ""
   SKIP cleanString("#######") = ""
   SKIP cleanString("") = ""
   VIEW-AS ALERT-BOX INFORMATION BUTTONS OK.

[M#3]

PHP:

function cleanString(string $text)
{
    $result = '';
    for ($i = 0; $i < strlen($text); $i++){
        if($text[$i] === '#') {
            if(strlen($result) > 0) {
                $result = substr($result, 0, -1);
            }
        } else {
            $result .= $text[$i];
        }
    }
    return $result;
}

[Sabin Pandelovitch]

Another JS solution

const cleanString = string => string.split('').reduce((a,v)=> v==="#" ? a.substr(0,a.length-1) : a+=v,'');

[Valts Liepiņš]

Haskell:

cleanString :: String -> String
cleanString = reverse . clean 0 . reverse
  where
    clean :: Int -> String -> String
    clean _ "" = ""
    clean d ('#':rest) = clean (d + 1) rest
    clean 0 (ch:rest)  = ch : clean 0 rest
    clean d (ch:rest)  = clean (d - 1) rest

[Empereol]

TypeScript

function cleanString(input: string): string {
  return [...input].reduce(
    (total, cur) => (cur === '#' ? (total = total.slice(0, -1)) : (total += cur)),
    ''
  );
}

[Vidit Sarkar]

Here is a ugly Python one-liner using lambda.
Use Python 3.8

cleanString = lambda s,countPound=0 : ''.join(filter(lambda char : not isinstance(char, int), [(countPound := countPound + 1) if c == '#' else c if countPound == 0 else (countPound := countPound - 1) for c in s[::-1]]))[::-1]

print(cleanString("Thee# Empires# SS#tateBuildingg#")) # The Empire StateBuilding
print(cleanString("abc#d##c")) # ac
print(cleanString("abc#def##ghi###")) # abd
print(cleanString("abc##d######")) # empty string
print(cleanString("#######")) # empty string
print(cleanString("abc#de##c")) # abc
print(cleanString("abc####dhh##c#")) # d
print(cleanString("abcdef")) # abcdef
print(cleanString("########c")) # c

Idea is based on

def cleanString(s):
    countPound = 0
    ans = ""
    for c in s[::-1]:
        if c == '#':
            countPound += 1
        else:
            if countPound == 0:
                ans += c
            else:
                countPound -= 1
    return ans[::-1]

[Jürgen Hermann]

Python – No doc strings etc., it's a challenge after all…

import re

def clean_text(text, _back_hash = re.compile(r'[^#]#')):
    changes = '#' in text
    while changes:
        text, changes = _back_hash.subn('', text)
    return text.replace('#', '')


def tests():
    data = (
        ("abc#def##ghi###", "abd"),
        ("abc#d##c", "ac"),
        ("abc##d######", ""),
        ("#######", ""),
        ("", ""),
    )
    for inp, out in data:
        assert clean_text(inp) == out, \
               f"{inp!r} cleaned is not {out!r} but {clean_text(inp)!r}"
    print('All tests OK!')


if __name__ == '__main__':
    for x in ('abc#de##c', 'abc####dhh##c#', 'Thee# Empires# SS#tateBuildingg#'):
        print(f"{x!r} ⇒ {clean_text(x)!r}")
    tests()

[Amin]

This is always a good idea to comment your code. Beginners trying to solve these challenges can learn a lot from veterans like you!

[Avalander]

Scala

  def cleanString (str: String): String = {
    (str.toList.foldLeft(List[Char]()) {
      case (Nil, '#')     => Nil
      case (x :: xs, '#') => xs
      case (xs, c)        => c :: xs
    }).reverse.mkString
  }

[Vidit Sarkar]

Hope this is right, C++

string cleanString(string s){
    string ans = ""; // the answer string
    int length = 0; // length of answer string
    for(char c : s){

        if(c == '#'){
            // if c == '#' and length > 0, then pop one character and decrement length
            if(length > 0){
                ans.pop_back();
                length--;
            }
        }
        // else add the character to string and increment length
        else{
            ans += c;
            length++;
        }
    }

    return ans;
}