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# countHi challenge

Posted on 2016-09-09
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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
0
Question by:gudii9
• 11
• 7
• 3
• +2

LVL 15

Expert Comment

ID: 41791977
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

LVL 27

Accepted Solution

rrz earned 250 total points
ID: 41792702
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

LVL 7

Author Comment

ID: 41793543
Substring,consist Of
0

LVL 27

Expert Comment

ID: 41793628
I used substring and indexOf.
0

LVL 31

Assisted Solution

awking00 earned 125 total points
ID: 41794636
You could also use substring and startsWith.
0

LVL 7

Author Comment

ID: 41796851
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

LVL 27

Expert Comment

ID: 41796950
I think it would be easier to do this like I solved
https://www.experts-exchange.com/questions/28968900/countX.html
0

LVL 7

Author Comment

ID: 41796958
``````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

LVL 7

Author Comment

ID: 41796960
``````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

LVL 27

Expert Comment

ID: 41797012
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

LVL 27

Expert Comment

ID: 41797015
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

LVL 37

Assisted Solution

zzynx earned 125 total points
ID: 41797359
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

LVL 7

Author Comment

ID: 41798204
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

LVL 7

Author Comment

ID: 41798217
``````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

LVL 7

Author Comment

ID: 41798225
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

LVL 31

Expert Comment

ID: 41798368
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

LVL 27

Expert Comment

ID: 41798409
I think using indexOf  makes more sense. If the input was
"xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxhi"
then it would be much more efficient.
0

LVL 31

Expert Comment

ID: 41798457
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

LVL 7

Author Comment

ID: 41798472
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

LVL 7

Author Comment

ID: 41798477
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

LVL 7

Author Comment

ID: 41798480
``````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

LVL 37

Expert Comment

ID: 41799242
``````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

LVL 27

Expert Comment

ID: 41799266
@zzynx,   I think you got that last sentence backwards.
0

LVL 37

Expert Comment

ID: 41799288
@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

LVL 7

Author Comment

ID: 41799835
i got it. Thank you for reinforcing the knowledge
0

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