last two characters of string challenge

Do the  len>=2 on the left side of the && so it can prevent the substring(len-2) when it is false

public boolean endsLy(String str) {

int len=str.length();

if(len>=2 && str.substring(len-2).equals("ly"))
{
return true;
}
  return false;
}

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now it passed all as below

Expected	Run		
endsLy("oddly") → true	true	OK	    
endsLy("y") → false	false	OK	    
endsLy("oddy") → false	false	OK	    
endsLy("oddl") → false	false	OK	    
endsLy("olydd") → false	false	OK	    
endsLy("ly") → true	true	OK	    
endsLy("") → false	false	OK	    
endsLy("falsey") → false	false	OK	    
endsLy("evenly") → true	true	OK	    
other tests
OK

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i did not realize first come first serve(i was under impression as long as both conditions there that is sufficient but in this case order also important)

The && and || operators perform Conditional-AND and Conditional-OR operations on two boolean expressions. These operators exhibit "short-circuiting" behavior, which means that the second operand is evaluated only if needed.

https://docs.oracle.com/javase/tutorial/java/nutsandbolts/op2.html
In this case, it also means that the second operand is evaluated only if valid.

if(len>=2 && str.substring(len-2).equals("ly"))

instead of above shortcircuit And if i use simle & as below

if(len>=2 & str.substring(len-2).equals("ly"))

i would have avoided this bottle neck. please advise

No, & does not shortcut.  && is the shortcut operator

public boolean endsLy(String str) {

int len=str.length();

if(str.substring(len-2).equals("ly")& len>=2 )
{
return true;
}
  return false;
}

Correct. When write like above i am getting string index out of bound

Expected	Run		
endsLy("oddly") → true	true	OK	    
endsLy("y") → false	Exception:java.lang.StringIndexOutOfBoundsException: String index out of range: -1 (line number:5)	X	    
endsLy("oddy") → false	false	OK	    
endsLy("oddl") → false	false	OK	    
endsLy("olydd") → false	false	OK	    
endsLy("ly") → true	true	OK	    
endsLy("") → false	Exception:java.lang.StringIndexOutOfBoundsException: String index out of range: -2 (line number:5)	X	    
endsLy("falsey") → false	false	OK	    
endsLy("evenly") → true	true	OK	    
other tests
OK	

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Think of what you are asking. If the substring beginning at the index of len - 2 equals "ly" then ...
If the length of the string is 0 or 1, then it's trying to evaluate the substring beginning at index -1 or -2, which will always be out of bounds.

unlike with len>=2 && str.substring(len-2).equals("ly")
when you do str.substring(len-2).equals("ly")& len>=2, the str.substring(len-2) is executed regardless of whether the len>=2 is satisfied

Another chance to use StringBuilder -
      public static boolean endsLy(String str) {
            StringBuilder sb = new StringBuilder(str);
            return sb.reverse().substring(0, 2).equalsIgnoreCase("yl");
      }

    public static boolean endsLy(String str) {
            StringBuilder sb = new StringBuilder(str);
            return sb.reverse().substring(0, 2).equalsIgnoreCase("yl");
      }

Expected      Run      
endsLy("y") → false      Exception:java.lang.StringIndexOutOfBoundsException: String index out of range: 2 (line number:3)      X
endsLy("") → false      Exception:java.lang.StringIndexOutOfBoundsException: String index out of range: 2 (line number:3)      X

but
  return sb.reverse().toString().startsWith("yl");
would work

Good point, ozo. I didn't test with 1 character or empty string. :-(
One more refinement would be using toLowerCase to make case insensitive.
return sb.reverse().toString().toLowerCase().startsWith("yl");

public boolean endsLy(String str) {

int len=str.length();

if(len>=2 &str.substring(len-2).equals("ly") )
{
return true;
}
  return false;
}

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without shortcut && if i use only one & and wrote as above still getting string out of bound which does not make sense to me.
for endsLy("y") it should not even go and check second condition in if loop right to throw exception?
  please advise

ndsLy("oddly") → true	true	OK	    
endsLy("y") → false	Exception:java.lang.StringIndexOutOfBoundsException: String index out of range: -1 (line number:5)	X	    
endsLy("oddy") → false	false	OK	    
endsLy("oddl") → false	false	OK	    
endsLy("olydd") → false	false	OK	    
endsLy("ly") → true	true	OK	    
endsLy("") → false	Exception:java.lang.StringIndexOutOfBoundsException: String index out of range: -2 (line number:5)	X	    
endsLy("falsey") → false	false	OK	    
endsLy("evenly") → true	true	OK	    
other tests

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I like stringbuilder solution too.

http://codingbat.com/prob/p103895 does not call for case insensitivity

& does not shortcut.  && is the shortcut operator

& tries to evaluate both operands regardless of the value of the other operand.

i see now why we only have to use &&

does not call for case insensitivity

what you mean by above statement. you mean like upper and and lower character does not matter for the challenge right?

The CodingBat site says Given a string, return true if it ends in "ly", which is lower case and doesn't mention if it ends in "LY", so "LOVELY" would return false. I was interpreting the question (likely incorrectly) to return true if it ended in "LY", "Ly', "lY", or "ly".

   public  boolean endsLy(String str) {
            StringBuilder sb = new StringBuilder(str);
           // return sb.reverse().substring(0, 2).equalsIgnoreCase("yl");
           return sb.reverse().toString().startsWith("ly");
      }

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when i wrote as above it failed

Expected	Run		
endsLy("oddly") → true	false	X	    
endsLy("y") → false	false	OK	    
endsLy("oddy") → false	false	OK	    
endsLy("oddl") → false	false	OK	    
endsLy("olydd") → false	false	OK	    
endsLy("ly") → true	false	X	    
endsLy("") → false	false	OK	    
endsLy("falsey") → false	false	OK	    
endsLy("evenly") → true	false	X	    
other tests
X	  

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nor does it test any cases that end in "yl" so it might also be argued that sorting before testing could also be valid.

i think it is failing if sorting or ordering(what is the difference?") is changing as above.

It needs to be startsWith("yl") as you have reversed the string.

Then again, while it doesn't explicitly specify the the case of "ly" must be exact, neither does it explicitly specify that the order of "ly" must be exact, nor does it test any cases that end in "yl" so it might also be argued that sorting before testing could also be valid.

i have not clearly understood above statement. Can you please elaborate.

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

ozo,
That's even less complicated and the best solution yet.