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# string at particular index

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Hi

I was trying below example

http://codingbat.com/prob/p121596

I was not sure how to proceed on this. I wrote some dummy code as below

How to fix and improve the my code

``````public String altPairs(String str) {
for(int i=0;i<str.length();i++)
String str2=(str.substring(i,i+1)+str.substring(i+4,i+5));
return str2;
}
``````
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Commented:
``````You have defined the str2 variable inside the for() scope. It needs to be defined outside of the scope:
``````
i got it.
``````str2+=str.substring(i,i+2<=str.length()?i+2:i+1);
``````
what is the meaning of above code

You need to go through the string in jumps of 4.

So the loop will be something like "i += 4" for the step.

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Commented:
the loop starts with i=0
after executing i += 4, i will be 4
after executing i += 4 again,  i will be 8

Commented:

Given a string, return a string made of the chars at indexes 0,1, 4,5, 8,9 ... so "kittens" yields "kien".

altPairs("kitten") → "kien"
altPairs("Chocolate") → "Chole"
altPairs("CodingHorror") → "Congrr"

But the challenge statement as above did not mention about jumps of 4 right. can you please advise

``````String str2="";
for(int i=0;i<str.length();i+=4)
{
str2+=str.substring(i,i+2<=str.length()?i+2:i+1);
}
return str2;
``````

in the above working code i see jumps of 4 but not sure what is happening in below line
str2+=str.substring(i,i+2<=str.length()?i+2:i+1);
which i can translate as below more understandable to me
str2= str2+str.substring(i,i+2<=str.length()?i+2:i+1);

not clear what is being passed to substring method above as below

i,i+2<=str.length()?i+2:i+1
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Commented:
Would it have been easier to understand if it had been
i,Math.min(str.length(),i+2)
?
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Commented:
The operator ?: is called a ternary operator. It is a shortcut to write something in one line. Without this operator you would write:
``````// str2+=str.substring(i,i+2<=str.length()?i+2:i+1); is equivalent to
if(i+2<=str.length())
str2+=str.substring(i,i+2);
else
str2+=str.substring(i,i+1);
``````

Commented:
str2+=str.substring(i,i+1);
i do not see i,i+1 in below statement
str2+=str.substring(i,i+2<=str.length()?i+2:i+1); is equivalent to
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Commented:
I think the use of a tenary operator is just confusing things here.

Since this question seems to be getting more tangled rather than less, I'm going to repost the code I posted at the very top of this question and make the one small edit:

String str2=str.substring(i,i+2);
=>
String str2=str.substring(i,end);

that was required to have it solve the problem.

It relies on knowing that Math.min here:
int end = Math.min(i+2, len) ;

puts the smaller of "i+2" and "len" into the variable end.

Is this clearer?

``````public String altPairs(String str) {
int len = str.length() ;
StringBuilder result = new StringBuilder() ;

for (int i = 0 ; i < len ; i += 4) {
int end = Math.min(i+2, len) ;
String str2=str.substring(i,end);
result.append(str2) ;
}

return result.toString() ;
}
``````

Commented:
0,1, 4,5, 8,9

Given a string, return a string made of the chars at indexes 0,1, 4,5, 8,9 ... so "kittens" yields "kien".

altPairs("kitten") → "kien"
altPairs("Chocolate") → "Chole"
altPairs("CodingHorror") → "Congrr"

in the chalenge above numbers given right. Is it is like fibonacci series?
i do not see any pattern. what comes after 9?

I wonder why we are incrementing by 4
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Commented:
It is true that there are an infinite number of different series that start 0,1, 4,5, 8,9 ...,
so the problem is technically underspecified, most people seem to be interpreting it as
0,1,
4,5,
8,9,
8+4,8+4+1,
12+4,12+4+1,
16+4,16+4+1,
20+4,20+4+1,
24+4,24+4+1,
...
(and there are still an infinite number of different ways to extend that pattern, but none of the test strings are that long, so it doesn't really matter)
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Commented:
I wonder why we are incrementing by 4

We're incrementing by 4 because that's what the pattern suggested we needed to do.  It's not the fibonacci series.  The sequence is just the pair (0,1) and then adding 4 to generate the next pair (4,5) and 4 to get the next pair (8,9) etc.

So overall we get (0,1,4,5,8,9,...)

Commented:
The sequence is just the pair (0,1) and then adding 4 to generate the next pair (4,5) and 4 to get the next pair (8,9) etc.

now i understand the pattern

for (int i = 0 ; i < len ; i += 4) {
int end = Math.min(i+2, len) ;
String str2=str.substring(i,end);

what are we doing in above 4 lines after incrementing by 4 to get the next set of 2 numbers? why we are getting substring i till end(excluding end ofcourse) all the way?
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Commented:
when i==0, we want str.substring(0,2), unless Math.min(i+2, len)  < 2, in which case we want  str.substring(0,len),

Commented:
unless Math.min(i+2, len)  < 2
this never happens right even if you take i as 0
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Commented:
len < i+2 can happen

Commented:
how and when
2<=2 can happen but not 2<2 right?
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Commented:
2<=2 is always true
2<2  is never true

Commented:
correct.

``````public String altPairs(String str) {
int len = str.length() ;
StringBuilder result = new StringBuilder() ;

for (int i = 0 ; i < len ; i += 4) {
// int end = Math.min(i+2, len) ;
int end=len;
String str2=str.substring(i,end);
result.append(str2) ;
}

return result.toString() ;
}
``````
when i commented above Math.min function line i am failing in all test cases where length is more than 2
``````Expected	Run
altPairs("kitten") → "kien"	"kittenen"	X
altPairs("Chocolate") → "Chole"	"Chocolateolatee"	X
altPairs("CodingHorror") → "Congrr"	"CodingHorrorngHorrorrror"	X
altPairs("yak") → "ya"	"yak"	X
altPairs("ya") → "ya"	"ya"	OK
altPairs("y") → "y"	"y"	OK
altPairs("") → ""	""	OK
altPairs("ThisThatTheOther") → "ThThThth"	"ThisThatTheOtherThatTheOtherTheOtherther"	X
``````

i did not completely get this Math.min function use here.
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Commented:

Commented:
0 length string("") take 2 as end..so appends 0,1 index characters which are"" and ""
1 length string("y") take 2 as end..so appends 0,1 index characters which are"y" and ""
2 length string("ya" take 2 as end..so appends 0,1 index characters which are"y" and "a"
3 length string("yak") take 2 as end..so appends 0,1 index characters which are"y" and "a" in first iteration(when i=0) and appends k in second iteration(when i=1)
6 length string("kitten") 2 as end....so appends 0,1 index characters which are"k" and "i" in first iteration(when i=0) and appends t,t in second iteration(when i=1) and  appends e,n in second iteration(when i=2)

but we are moving i by 4 after each iteration right?? please advise
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Commented:
it all make better sense now.
``````public String altPairs(String str) {
int len = str.length() ;
StringBuilder result = new StringBuilder() ;

for (int i = 0 ; i < len-1 ; i += 4) {
int end = Math.min(i+2, len) ;
//int end=len;
String str2=str.substring(i,end);
result.append(str2) ;
}

return result.toString() ;
}
``````

when i wrote as above by doing -1 i got couple of tests failied
``````Expected	Run
altPairs("kitten") → "kien"	"kien"	OK
altPairs("Chocolate") → "Chole"	"Chol"	X
altPairs("CodingHorror") → "Congrr"	"Congrr"	OK
altPairs("yak") → "ya"	"ya"	OK
altPairs("ya") → "ya"	"ya"	OK
altPairs("y") → "y"	""	X
altPairs("") → ""	""	OK
altPairs("ThisThatTheOther") → "ThThThth"	"ThThThth"	OK
``````
i wonder why they failed esp.
altPairs("Chocolate") → "Chole"      "Chol"      X
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Commented:
for (int i = 0 ; i < len-1 ; i += 4)

Is causing the loop to stop too soon (the -1 makes it stop 1 before the last character).

You can either have:
for (int i = 0 ; i < len ; i += 4)
or
for (int i = 0 ; i <= len-1 ; i += 4)

Both are equivalent.  So if you prefer to think about "len-1" you need to use "<=".

The reason most people write:
for (int i = 0 ; i < len ; i += 4)
is because the loop starts from 0, not 1.

So
for (int i = 0 ; i < 10 ; i++)
means to loop 10 times (0,1,2,3,4,5,6,7,8,9).  So it's easier to read when the "10" is the limit or "len" is the limit in this loop.

For that to work, we use i < 10 or i < len.

Does that make sense?

Doug

Commented:
yes.

I wonder why it failed for below test case while using i < len-1
altPairs("Chocolate") → "Chole"      "Chol"      X
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Commented:
"Chocolate" is a string of length 9.  So len - 1 is 8.

This means the loop becomes:
for (i = 0 ; i < 8 ; i += 4)
which means we get (0,4) as the values for i instead of (0,4,8) if the test was i < len.

So we miss the last group of characters - the trailing "e" in this case.

Doug

Commented:
``````altPairs("y") → "y"	""	X
``````

Commented:
``````int end = Math.min(i+2, len) ;
``````

so basically we are getting end to feed to substring method to get next set of characters to be appeneded to string builder right?
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Commented:
"y" is a string of length 1, so len - 1 is 0, so altPairs("y") failed for the same reason as altPairs("Chocolate")

we are getting end to feed to substring method
correct
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Commented:
It can actually be useful to loop only for  i<len-1, because that guarantees that within the loop, str.substring(i,i+2) will not be out of bounds, which makes it unnecessary to repeatedly guard for that condition inside the loop body.
After the loop is done, we can then do a single check to see if we need to append one more character in the case where we stopped to soon to get the final length 1 piece of str.substring(i,len):

public String altPairs(String str) {
int len = str.length() ;
StringBuilder result = new StringBuilder() ;
int i;
for ( i = 0 ; i < len-1 ; i += 4) {
result.append(str.substring(i,i+2)) ;
}
if( i==len-1 ){
result.append(str.substring(i)) ;
}
return result.toString() ;
}

Commented:
I see above solution more clear.
altPairs("") → ""      ""      OK
will not go to if loop of above solution right.

only when len is one like "y" string it goes to aboe if loop for rest of scenarios just go to for loop and appends the string using substring method and returns as result.toString right?

Commented:
``````altPairs("") → ""	""	OK
``````

how it works for above case
i<len-1(here len-1 becomes -1 for i=0 right)
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Commented:
for ( i = 0 ; i < len-1 ; i += 4) does not go into the loop body when len==0 or len==1
for ( i = 0 ; i < len ; i += 4) does not go into the loop body when len==0 but does go into the loop body when len==1

Commented:
correct
public String altPairs(String str) {
int len = str.length() ;
StringBuilder result = new StringBuilder() ;
int i;
for ( i = 0 ; i < len-1 ; i += 4) {
result.append(str.substring(i,i+2)) ;
}
if( i==len-1 ){
result.append(str.substring(i)) ;
}
return result.toString() ;
}

for ( i = 0 ; i < len-1 ; i += 4) does not go into the loop body when len==0 or len==1

How above solution is working then for len==0 and len==2. please advise

Commented:
when len==1 i can visualize it going into if loop and eppending as below
if( i==len-1 ){
result.append(str.substring(i)) ;
}

but if len==0 how solution passed the test case(for string like""). please advise
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Commented:
if len==0, the solution of http#a40454432 reduces to

public String altPairs(String str) {
int len = str.length() ;
StringBuilder result = new StringBuilder() ;
int i;
// don't execute code in for loop body, don't execute code in if body
return result.toString() ;
}

Commented:
if len==0, the solution of http#a40454432 reduces to
i could not see above mentioned url--a40454432

https://www.experts-exchange.com/Programming/Languages/Java/Q_40454432.html

i tied this also
https://www.experts-exchange.com/Programming/Languages/Java/Q_a40454432.html
is it is spearate question or separate comment on this question?

Commented:
ozo2014-11-19 at 23:49:00ID: 40454432

i see without a

Commented:
if len==0, the solution of http#a40454432 reduces to

so the code supposed to be as below
``````public String altPairs(String str) {
int len = str.length() ;
StringBuilder result = new StringBuilder() ;
int i;
// don't execute code in for loop body, don't execute code in if body
return result.toString() ;
}
``````
``````public String altPairs(String str) {
int len = str.length() ;
StringBuilder result = new StringBuilder() ;
int i;
for ( i = 0 ; i < len-1 ; i += 4) {
result.append(str.substring(i,i+2)) ;
}
if( i==len-1 ){
result.append(str.substring(i)) ;
}
return result.toString() ;
}
``````

Commented:
I see if the len==0 then it directly goes to

return result.toString() ; without going to for and if loop.

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Commented:
if the len==0 then it directly goes to

return result.toString() ;
That's correct

http#a40454432
should have been
http:#a40454432

Commented:
``````// str2+=str.substring(i,i+2<=str.length()?i+2:i+1); is equivalent to
if(i+2<=str.length())
str2+=str.substring(i,i+2);
else
str2+=str.substring(i,i+1)
``````

i am still trying to understand how these two are  equivalent.
substring second argument is
i+2<=str.length()?i+2:i+1
which means
if i+2<=str.length() is true take i+2 //when this case happen?
if i+2<=str.length() is false take i+1 as value.//when this case happen?
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Commented:
when i+2<=str.length() is true,  i+2<=str.length()?i+2:i+1 is i+2
when i+2<=str.length() is false, i+2<=str.length()?i+2:i+1 is i+1

https://docs.oracle.com/javase/tutorial/java/nutsandbolts/op2.html

Commented:
``````altPairs("kitten") → "kien"	"kien"	OK
``````
for above example str.length is 6 right
for the first iteration i=0

so
``````str2+=str.substring(i,i+2<=str.length()?i+2:i+1);
``````
i+2<=str.length()?i+2:i+1 becomes 2<=6 since true takes i+2 which is 2
so it should print 0,2 indexed characters as first set right before incrementing by 4 for next set.

But we do not want 0,2 but instead we want 0,1.

Please correct where i am thinking wrong. The test cases are passing so my understanding is wrong above some where. please advise
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Commented:
"kitten".substring(0,2) is "ki", which we want
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Commented:
One way to visualize it could be to think of the index numbers as being between the characters in the string, and the substring character being between the selected index numbers
`````` _ _
k i t t e n
| ^ | ^ ^ ^ ^
0 1 2 3 4 5 6
``````

Commented:
"kitten".substring(0,2) is "ki", which we want

Given a string, return a string made of the chars at indexes 0,1, 4,5, 8,9 ... so "kittens" yields "kien".

altPairs("kitten") → "kien"
altPairs("Chocolate") → "Chole"
altPairs("CodingHorror") → "Congrr"

challenge says 0,1 then 4,5 then 8,9 right

Commented:
Oh well i see substring end is excluded right unlike start index.
Ok i got it now
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Commented:
challenge says at indexes 0,1
challenge does not say substring(0,1)

Commented:
altPairs("y") → "y"      "y"      OK
altPairs("") → ""      ""      OK

these are only two scenarios where false returns
``````i+2<=str.length()?i+2:i+1
``````

so  i+1 is returned.
so the substring becomes

altPairs("y") → "y"      "y"      OK        ===>substring(0,1)  --->so give back y
altPairs("") → ""               ====>substring(0,1) so gives back ""

But to begin with how we know to compare with i+2 as below(not i or i+1)
i+2<=
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Commented:
altPairs("") makes len==0
so whether we loop for ( i = 0 ; i < len-1 ; i += 4)
or for ( i = 0 ; i < len ; i += 4)
the loop condition starts out false, so we don't even enter the loop body

altPairs("y") makes str.length()==1
so i+2<str.length() is false, which means i+2<=str.length()?i+2:i+1 is i+1

Commented:
``````altPairs("CodingHorror") → "Congrr"	"Congrr"	OK
``````

how we got Congrr??

substring second argument is
i+2<=str.length()?i+2:i+1
which means
if i+2<=str.length() is true take i+2

so we get substring(0,2) which is "Co" in the first iteration

then i incremented by 4 so 0 becomes 4
6<=12 true so take i+2
so we get substring(4,6) which is "ng" in the second iteration

then i incremented by 4 so 4 becomes 8
8<=12 true so take i+2
so we get substring(8,10) which is "rr" in the second iteration

so i+1 scenario only comes for "y"(1 character strings)
if empty string wil not even go inside the loop.

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Commented:
i+1 scenario also comes for "Chocolate"

Commented:
i+1 scenario also comes for "Chocolate"
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Commented:
When i==8

Commented:
for(int i=0;i<str.length();i+=4)
the loop condition starts out false, so we don't even enter the loop body when i=8 right?
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Commented:
"Chocolate".length() ==9
8<9 is true

Commented:
``````public class Test33 {

/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
//altPairs("chocolate");
System.out.println("value is-->"+altPairs("chocolate"));
}

public static String altPairs(String str) {
String str2="";
for(int i=0;i<str.length();i+=4)
{
str2+=str.substring(i,i+2<=str.length()?i+2:i+1);
}
return str2;
}
}
``````
i got output
value is-->chole

how we got chole??

substring second argument is
i+2<=str.length()?i+2:i+1
which means
if i+2<=str.length() is true take i+2

so we get substring(0,2) which is "ch" in the first iteration

then i incremented by 4 so 0 becomes 4
6<=8 true so take i+2
so we get substring(4,6) which is "ol" in the second iteration

//chocolate
then i incremented by 4 so 4 becomes 8
8<=8 true so take i+2
so we get substring(8,10) which is "e?" in the third iteration

so i+1 is not coming right.

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Commented:
8+2<="Chocolate".length() is not true

Commented:
i got output
value is-->chole

how we got chole??

substring second argument is
i+2<=str.length()?i+2:i+1
which means
if i+2<=str.length() is true take i+2 (since2<=8)

so we get substring(0,2) which is "ch" in the first iteration

then i incremented by 4 so 0 becomes 4
6<=8 true so take i+2
so we get substring(4,6) which is "ol" in the second iteration

//chocolate
then i incremented by 4 so 4 becomes 8
10<=8 false so take i+1
so we get substring(8,9) which is "e" in the third iteration

so i+1 is  coming right.

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Commented:
"Chocolate".length() is 9, not 8

Commented:
i got output
value is-->chole

how we got chole??

substring second argument is
i+2<=str.length()?i+2:i+1
which means
if i+2<=str.length() is true take i+2 (since2<=9)

so we get substring(0,2) which is "ch" in the first iteration

then i incremented by 4 so 0 becomes 4
6<=9 true so take i+2
so we get substring(4,6) which is "ol" in the second iteration

//chocolate
then i incremented by 4 so 4 becomes 8
10<=9 false so take i+1
so we get substring(8,9) which is "e" in the third iteration
so i+1 is  coming right.

Commented:
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Commented:
yes

Commented:
thank you

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