Java > Recursion-1 > parenBit (CodingBat Solution)

Problem:

Given a string that contains a single pair of parenthesis, compute recursively a new string made of only of the parenthesis and their contents, so "xyz(abc)123" yields "(abc)".

parenBit("xyz(abc)123") → "(abc)"
parenBit("x(hello)") → "(hello)"
parenBit("(xy)1") → "(xy)"


Solution:

public String parenBit(String str) {
  if (str.equals("")) return str;
  if (str.charAt(0) == '(') {
    if (str.charAt(str.length()-1) == ')')
      return str;
    else 
      return parenBit(str.substring(0, str.length()-1));
  } else
    return parenBit(str.substring(1));
}

11 comments:

  1. UnknownNovember 17, 2015 at 7:19 AM

    CodingBat's solution:

    public String parenBit(String str) {
      if (str.charAt(0) != '(') {
        return parenBit(str.substring(1));
      }
      if (str.charAt(str.length()-1) != ')') {
        return parenBit(str.substring(0, str.length()-1));
      }
      return str;
    }

    ReplyDelete
  2. UnknownFebruary 28, 2017 at 1:54 AM

    3 lines best solution)))

    public String parenBit(String str) {
    if(str.length() == 0) return str;
    if(str.charAt(0) == '(') return str.charAt(0) + str.substring(1, str.indexOf(')')+1);
    return parenBit(str.substring(1));
    }

    ReplyDelete
    Replies
    1. UnknownMay 2, 2017 at 2:33 AM

      I'm pretty sure you can't use indexOf() in this example. You should use recursion instead.

      Delete
    2. AnonymousDecember 13, 2019 at 7:14 AM

      return str.substring(str.indexOf("("),str.indexOf(")") + 1);
      **Without recursion**

      Delete
  3. AnonymousDecember 7, 2017 at 12:28 PM

    @ЯКОВ ШТЕФФЕН nah, it works his solution.

    ReplyDelete
  4. MikhailMarch 2, 2019 at 3:09 PM

    A little bit stupid compared to codingbat's solution but it works:

    public String parenBit(String str) {
    if (str.length() == 0) return str;

    if (str.charAt(0) == '(')
    return "(" + parenBit("*" + str.substring(1));
    if (str.length() > 1 && str.charAt(0) == '*')
    return str.charAt(1) == ')' ? ")" : str.charAt(1) + parenBit("*" + str.substring(2));
    else
    return parenBit(str.substring(1));
    }

    ReplyDelete
  5. Mohamed KarmApril 17, 2020 at 6:50 AM

    hi From Egypt:
    public String parenBit(String str) {
    if(str.length() > 0){
    return str.substring(str.indexOf('('), str.indexOf(')')+1);}
    else return "";

    }

    ReplyDelete
  6. UnknownJuly 28, 2020 at 1:50 PM

    public String parenBit(String str) {
    if (str.charAt(0) == '(' && str.charAt(str.length() - 1) == ')')
    {
    return str;
    }
    else
    {
    if (str.charAt(0) == '(')
    {
    return parenBit(str.substring(0, str.length() - 1));
    }
    else if (str.charAt(str.length() - 1) == ')')
    {
    return parenBit(str.substring(1));
    }
    else
    {
    return parenBit(str.substring(1, str.length() - 1));
    }
    }
    }

    ReplyDelete
  7. UnknownApril 8, 2021 at 12:09 PM

    My solution with cheats

    public String parenBit(String str) {
    if(str.length() < 2) return str;
    if(str.startsWith("("))
    return str.substring(str.indexOf("("), str.indexOf(")")) + ")";
    return parenBit(str.substring(1));
    }

    ReplyDelete
  8. muhammadFebruary 6, 2022 at 6:22 AM

    int s1=str.indexOf('(');
    int s2=str.indexOf(')');
    return str.substring(s1,s2+1);

    ReplyDelete
  9. UnknownMarch 22, 2022 at 6:20 AM

    return str.substring(str.indexOf("("), str.indexOf(")") +1 );

    ReplyDelete