Java > String-1 > endsLy (CodingBat Solution)

Problem:

Given a string, return true if it ends in "ly".

endsLy("oddly") → true
endsLy("y") → false
endsLy("oddy") → false


Solution:

public boolean endsLy(String str) {
  int len = str.length();
  String ly = "ly";
 
  if (len < 2)
    return false;
  else if (ly.equals(str.substring(len-2,len)))
    return true;
  else
    return false;
}

14 comments:

  1. AnonymousJuly 19, 2016 at 3:12 PM

    This below code will also work.

    public boolean endsLy(String str) {
    if(str.length() <2 ){
    return false;
    }
    if(str.substring(str.length()-2).equals("ly"))
    return true;
    else
    return false;
    }

    ReplyDelete
  2. AnonymousJune 13, 2017 at 6:46 AM

    public boolean endsLy(String str) {
    if(str.length() < 2){
    return false;
    }
    return(str.substring(str.length()-2).equals("ly"));
    }

    ReplyDelete
  3. AnonymousNovember 15, 2019 at 5:32 PM

    public boolean endsLy(String str) {
    return str.endsWith("ly");
    }

    ReplyDelete
  4. ShuaibJuly 24, 2020 at 9:37 AM

    public boolean endsLy(String str) {
    int len = str.length();

    return (len>=2) && str.substring(len-2, len).equals("ly");

    }

    ReplyDelete
  5. Anmol ParikhJanuary 10, 2021 at 11:02 AM

    public boolean endsLy(String str) {
    if(str.length()<2){
    return false;
    }else{
    String a=str.substring(str.length()-2);
    String ly="ly";
    if(ly.equals(a)){
    return true;
    }return false;
    }
    }

    ReplyDelete
  6. Anmol ParikhJanuary 10, 2021 at 11:04 AM

    public boolean endsLy(String str) {
    if(str.length()<2){
    return false;
    }else{
    String a=str.substring(str.length()-2);
    String ly="ly";
    if(ly.equals(a)){
    return true;
    }return false;
    }
    }

    ReplyDelete
  7. GabeFebruary 3, 2021 at 4:37 PM

    if(str.length() < 2) return false;

    return (str.substring(str.length()-2).equals("ly"));

    ReplyDelete
  8. AnonymousFebruary 8, 2021 at 12:59 PM

    public boolean endsLy(String str) {
    return (str.length()>1&&str.substring(str.length()-2,str.length()).equals("ly"));
    }

    ReplyDelete
  9. UnknownMarch 9, 2021 at 6:57 AM

    public boolean endsLy(String str) { if(str.length()>=2) { if(str.substring(str.length()-2,str.length())==("ly")) { return true; } else { return false; } } else { return false; } }

    ReplyDelete
  10. UnknownMarch 9, 2021 at 6:57 AM

    what is the error..!explain in detail

    ReplyDelete
  11. UnknownOctober 3, 2021 at 9:55 PM

    public boolean endsLy(String str) {

    int length=str.length();
    if(length<2)
    return false;
    if(str.charAt(length-1)=='y' && str.charAt(length-2)=='l')
    return true;
    else
    return false;
    }

    ReplyDelete
  12. UnknownFebruary 12, 2022 at 11:34 AM

    public boolean endsLy(String str) {
    if(str.endsWith("ly")){
    return true;
    }else return false;
    }

    ReplyDelete
  13. sam456August 31, 2022 at 12:13 PM

    public boolean endsLy(String str) {
    int len=str.length()-2;
    if(str.length()<2) return false;
    String s=str.substring(len);
    if(s.equals("ly")){
    return true;
    }
    return false;
    }

    ReplyDelete
  14. Shre_Jeni❤️August 24, 2024 at 9:25 AM

    return str.endsWith("ly") ? true:false;

    ReplyDelete