Java > String-1 > hasBad (CodingBat Solution)

Problem:

Given a string, return true if "bad" appears starting at index 0 or 1 in the string, such as with "badxxx" or "xbadxx" but not "xxbadxx". The string may be any length, including 0. Note: use .equals() to compare 2 strings.

hasBad("badxx") → true
hasBad("xbadxx") → true
hasBad("xxbadxx") → false


Solution:

public boolean hasBad(String str) {
  if(str.length() < 3)
    return false;
  else if ((str.substring(0,3)).equals("bad"))
    return true;
  else if (str.length() > 3){
    if ((str.substring(1,4)).equals("bad"))
      return true;
  }
    return false;
}

7 comments:

  1. AnonymousMay 4, 2019 at 6:13 AM

    public boolean hasBad(String str) {
    String bad = "bad";

    if(str.length()<=3 && !str.contains(bad)){
    return false;
    }
    if(str.startsWith(bad) || str.substring(1, 4).startsWith(bad)){
    return true;
    }
    return false;
    }

    ReplyDelete
  2. UnknownJuly 21, 2019 at 7:08 AM

    String badStr = "bad";

    if (str.length() < 3) {
    return false;
    }

    if (str.substring(0, 3).equals(badStr)) {
    return true;
    }

    if ( str.length() > 3 && str.substring(1, 4).equals(badStr)) {
    return true;
    } else {
    return false;
    }
    }

    ReplyDelete
  3. UnknownMay 9, 2020 at 12:37 AM

    public boolean hasBad(String str) {
    return str.length()>=3?str.substring(0,3).equals("bad")
    ||
    str.length()>3&&str.substring(1,4).equals("bad"):false;
    }

    ReplyDelete
  4. AnarAghaAugust 12, 2020 at 7:55 PM

    public boolean hasBad(String str) {
    if(str.length()<3) {
    return false;
    }
    if (str.substring(0).startsWith("bad") || str.substring(1).startsWith("bad")){
    return true;
    }
    return false;
    }

    ReplyDelete
  5. AnonymousSeptember 13, 2020 at 5:58 AM

    public final boolean hasBad(String str) {
    return !str.isEmpty() && (str.startsWith("bad")
    || str.substring(1).startsWith("bad"));
    }

    ReplyDelete
  6. UnknownDecember 20, 2021 at 6:22 PM

    public boolean hasBad(String str) {
    return ((str.length() >= 3 && str.startsWith("bad"))
    || (str.length() >= 4 && str.substring(1, 4).equals("bad")));
    }

    ReplyDelete