# countHi challenge

Hi,

I am working on below challenge
http://codingbat.com/prob/p184029

Given a string, compute recursively (no loops) the number of times lowercase "hi" appears in the string.

countHi("xxhixx") → 1
countHi("xhixhix") → 2
countHi("hi") → 1
i was not clear on how to find the number of hi using recursion? please advise
LVL 7
###### Who is Participating?

Commented:
This challenge is different from the most of the recursion challenges that you posted.
But, it the same exact programming  problem as
https://www.experts-exchange.com/questions/28968900/countX.html
They use a couple of methods of the String class.  Do you know which ones ?
0

Commented:
This question and all the previous ones you have littered EE with could be solved by you if you take the time to read up on recursion. Warning....if you found if/else statements difficult, this will blow your mind!
1

Author Commented:
Substring,consist Of
0

Commented:
I used substring and indexOf.
0

Commented:
You could also use substring and startsWith.
0

Author Commented:
psedi code:
1. if length is zero return 0
2. if length is 1 return 0
3. else if str starts with h and substring with next character is equals to hi then rerun same recursive function by giving length minus one
4. return hi count.
``````public int countHi(String str) {
if(str.length()==0) return 0;
if(str.length()==1) return 1;
if(str.startsWith("h")&&str.substring(i,i+1).equals("hi")){
return  1+countHi(str.length()-1);
}
}

//substring and startsWith.
``````
i tried as above
Compile problems:

Error:      if(str.startsWith("h")&&str.substring(i,i+1).equals("hi")){
^
i cannot be resolved

see Example Code to help with compile problem
getting above error.
i do not want to use i and for loop for this but only recursion approach? please advise
0

Commented:
I think it would be easier to do this like I solved
https://www.experts-exchange.com/questions/28968900/countX.html
0

Author Commented:
``````public int countHi(String str) {
if (str.length() <2){
return 0;
}
if (str.substring(0,2).equals( "hi")) {
return 1 + countHi(str.substring(1));
}
else {
return countHi(str.substring(1));

}

}

//substring and startsWith.
``````

above passed all tests. any alternate approaches or improvements?
0

Author Commented:
``````public int countHi(String str) {
if (str.length() <2){
return 0;
}
if (str.substring(0,2) == "hi") {
return 1 + countHi(str.substring(1));
}
else {
return countHi(str.substring(1));

}

}

//substring and startsWith.
``````

why with == it failse below tests?
Expected      Run
countHi("xxhixx") → 1      0      X
countHi("xhixhix") → 2      0      X
countHi("hi") → 1      1      OK
countHi("hihih") → 2      0      X
countHi("h") → 0      0      OK
countHi("") → 0      0      OK
countHi("ihihihihih") → 4      0      X
countHi("ihihihihihi") → 5      0      X
countHi("hiAAhi12hi") → 3      0      X
countHi("xhixhxihihhhih") → 3      0      X
countHi("ship") → 1      0      X
other tests
OK
0

Commented:
why with == it failse below tests?
Don't use ==  to compare Strings. It checks  if one object is equal to the other.  In this case, they are different objects. But they do have same contents. Always use the equals method to compare String content.
0

Commented:
Here is my solution.
``````public int countHi(String str) {
int index = str.indexOf("hi");
if(index == -1) return 0;
else return 1 + countHi(str.substring(index + 2));
}
``````
0

Software engineerCommented:
Replace this code
``````    if (str.substring(0,2) == "hi") {
return 1 + countHi(str.substring(1));
}
``````

by this:
``````    if ("hi".equals(str.substring(0,2)) {
return 1 + countHi(str.substring(2));
}
``````

That are two improvements:
1) using equals instead of == (never do that for strings)
2) If you know that str.substring(0,2) equals "hi", you can start searching for antoher "hi" at index 2.
0

Author Commented:
if (str.substring(0,2).equals( "hi")) {

how above different from below
``````  if ("hi".equals(str.substring(0,2)) {
``````

to me both seems same?
0

Author Commented:
``````public int countHi(String str) {
if (str.length() <2){
return 0;
}
if (str.substring(0,2).equals( "hi")) {
return 1 + countHi(str.substring(1));
}
else {
return countHi(str.substring(1));

}

}

//substring and startsWith.
``````

is recursion bit inefficient approach as we are checking inside if and also again inside else block each character?
0

Author Commented:
2) If you know that str.substring(0,2) equals "hi", you can start searching for antoher "hi" at index 2.

i was not clear on above point? can you please elaborate on how at index 2 another "hi"?( i thought we search next character then again next character..till end)
0

Commented:
Although it can be done with either the indexOf or startsWith function, I'm going to use startsWith as I think it's a little easier to visualize.
Think about what is happening. Assume str = "xhixhix"
since the length >= 2, the program goes to the next if statement
Since the str does not start with "hi", it goes to the next if
which is return countHi(str.substring(1)) which is now the string "hixhix"
now its does start with "hi" so we add 1 to the substring from index 1 would now be ixhix but since it found "hi" during the last method call, you know that the character at the next index has to be "i" so the next substring should start at index 2 making it "xhix" which doesn't start with "hi" so get the next substring from index 1 which is hix that does start with "hi" so we add 1 to the following substring at index 2 is "x" which has a length less than 2 so 0 is returned and we are done.
0

Commented:
I think using indexOf  makes more sense. If the input was
"xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxhi"
then it would be much more efficient.
0

Commented:
rrz, I totally agree that using indexOf would likely be more efficient. I was just trying to get gudii9 to see what's happening to the strings with each recursion and thought startsWith might make it a little easier to understand.
0

Author Commented:
why with == it failse below tests?
Don't use ==  to compare Strings. It checks  if one object is equal to the other.  In this case, they are different objects. But they do have same contents

how in this case they are different objects?

That are two improvements:
1) using equals instead of == (never do that for strings)
2) If you know that str.substring(0,2) equals "hi", you can start searching for antoher "hi" at index 2.

i think i got above points

``````public int countHi(String str) {
if (str.length() <2){
return 0;
}
if (str.substring(0,2).equals( "hi")) {//does it make difference reverse if ("hi".substring(0,2).equals(str))??
return 1 + countHi(str.substring(2));
}
else {
return countHi(str.substring(1));

}

}

//substring and startsWith.
``````
0

Author Commented:
what is psedo code for startsWith?
1. check given string startsWith "hi"
2. if yes then 1+countHi(2)
3. else countHi(2)
4. then return count?
0

Author Commented:
``````public int countHi(String str) {
if (str.length() <2){
return 0;
}
if (str.startsWith( "hi")) {
return 1 + countHi(str.substring(2));
}
else {
return countHi(str.substring(1));

}

}

//substring and startsWith.
``````

like above. Above passes all tests. Any improvement in above approach?
0

Software engineerCommented:
``````str.startsWith( "hi")
``````
or
``````str.substring(0,2).equals( "hi")
``````
or
``````"hi".equals(str.substring(0,2))
``````

All do the same: checking if the first two characters of str are "hi".

It's safer to write
``````"hi".equals(str.substring(0,2))
``````
than
``````str.substring(0,2).equals( "hi")
``````
because if str = null the first will lead to a NullPointerException while the second will just work as expected.
0

Commented:
@zzynx,   I think you got that last sentence backwards.
0

Software engineerCommented:
@rrz: Right you are of course. :-)

Let me correct that:
...because if str = null
``````str.substring(0,2).equals( "hi")
``````
will lead to a NullPointerException while
``````"hi".equals(str.substring(0,2))
``````
will just work as expected.
0

Author Commented:
i got it. Thank you for reinforcing the knowledge
0
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.