Java > String-1 > frontAgain (CodingBat Solution)

Problem:

Given a string, return true if the first 2 chars in the string also appear at the end of the string, such as with "edited".

frontAgain("edited") → true
frontAgain("edit") → false
frontAgain("ed") → true


Solution:

public boolean frontAgain(String str) {
  if (str.length() < 2)
    return false;
  else if (str.substring(0,2).equals(str.substring(str.length()-2, str.length())))
    return true;
  else
    return false;
}

14 comments:

  1. UnknownOctober 17, 2016 at 4:09 AM

    Thaks for nice logic

    ReplyDelete
  2. UnknownOctober 17, 2016 at 4:23 AM

    Thaks for nice logic

    ReplyDelete
  3. AnonymousDecember 31, 2016 at 2:53 AM

    Thank you
    There is a solution

    public boolean frontAgain(String str) {
    return str.length()>=2 && str.substring(0, 2).equals(str.substring(str.length()-2, str.length()));
    }

    ReplyDelete
  4. UnknownApril 6, 2018 at 11:02 AM

    public boolean frontAgain(String str) {
    if(str.length()>1 && str.charAt(0) == str.charAt(str.length()-2)
    && str.charAt(1) == str.charAt(str.length()-1))
    return true;
    else return false;
    }

    ReplyDelete
  5. UnknownApril 6, 2018 at 11:04 AM

    public boolean frontAgain(String str) {
    if(str.length()>1 && str.substring(0,2).equals(str.substring(str.length()-2)))
    return true;
    else return false;
    }

    ReplyDelete
  6. UnknownAugust 23, 2018 at 9:13 AM

    public boolean frontAgain(String str) {
    if (str.length() < 2) return false;
    if (str.endsWith(str.substring(0, 2))) return true;
    else return false;
    }

    ReplyDelete
  7. UnknownAugust 23, 2018 at 9:14 AM

    public boolean frontAgain(String str) {
    if (str.length() < 2) return false;
    if (str.endsWith(str.substring(0, 2))) return true;
    else return false;
    }

    ReplyDelete
  8. AnonymousMay 5, 2019 at 12:01 PM

    public boolean frontAgain(String str) {
    if(str.length()<2){
    return false;
    }
    String first2 = str.substring(0,2);
    return str.startsWith(first2) && str.endsWith(first2);
    }

    ReplyDelete
  9. UnknownApril 21, 2020 at 8:02 PM

    public boolean frontAgain(String str) {
    if(str.length()>1&&str.endsWith(str.substring(0,2)))
    return true;
    return false;
    }

    ReplyDelete
  10. UnknownApril 22, 2020 at 8:15 AM

    public boolean frontAgain(String str) {
    return (str.length()>1&&str.endsWith(str.substring(0,2)));
    }

    ReplyDelete
  11. Anmol ParikhJanuary 11, 2021 at 11:01 AM

    public boolean frontAgain(String str) {
    if(str.length()>1){
    String a=str.substring(0,2);
    String b=str.substring(str.length()-2,str.length());
    if(a.equals(b)){
    return true;
    }else return false;
    }else return false;

    }

    ReplyDelete
  12. ewedaaNovember 13, 2021 at 1:03 PM

    public boolean frontAgain(String str) {
    if(str.length()<2){
    return false;
    }
    return str.substring(0,2).equals(str.substring(str.length()-2));
    }

    ReplyDelete
  13. UNKNOWNovember 23, 2021 at 5:09 AM

    public boolean frontAgain(String str) {
    if (str.length()>1 && str.endsWith(str.substring(0,2))) {
    return true;
    }
    return false;
    }

    ReplyDelete
  14. UnknownFebruary 3, 2022 at 6:38 AM

    public boolean frontAgain(String str) {
    return (str.length()>=2 && (str.substring(0,2).equals(str.substring(str.length()-2))));
    }

    ReplyDelete