gudii9
asked on
string at particular index
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
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;
}
SOLUTION
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SOLUTION
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the loop starts with i=0
after executing i += 4, i will be 4
after executing i += 4 again, i will be 8
after executing i += 4, i will be 4
after executing i += 4 again, i will be 8
ASKER
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<
which i can translate as below more understandable to me
str2= str2+str.substring(i,i+2<=
not clear what is being passed to substring method above as below
i,i+2<=str.length()?i+2:i+please advise1
Would it have been easier to understand if it had been
i,Math.min(str.length(),i+?2)
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);
ASKER
str2+=str.substring(i,i+1)i do not see i,i+1 in below statement;
str2+=str.substring(i,i+2<Please advise=str.lengt h()?i+2:i+ 1); is equivalent to
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?
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() ;
}
ASKER
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
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)
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)
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,...)
ASKER
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?
please advise
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),
ASKER
unless Math.min(i+2, len) < 2this never happens right even if you take i as 0
2<2 can never happen. please advise
len < i+2 can happen
ASKER
how and when
2<=2 can happen but not 2<2 right?
please advise
2<=2 can happen but not 2<2 right?
please advise
2<=2 is always true
2<2 is never true
2<2 is never true
ASKER
correct.
i did not completely get this Math.min function use here.
Please advise
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 2Expected 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.
Please advise
Math.min returns the smaller of two values https://docs.oracle.com/javase/7/docs/api/java/lang/Math.html#min%28int,%20int%29
ASKER
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
ASKER CERTIFIED SOLUTION
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ASKER
it all make better sense now.
when i wrote as above by doing -1 i got couple of tests failied
altPairs("Chocolate") → "Chole" "Chol" X
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
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
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
ASKER
yes.
I wonder why it failed for below test case while using i < len-1
altPairs("Chocolate") → "Chole" "Chol" X
I wonder why it failed for below test case while using i < len-1
altPairs("Chocolate") → "Chole" "Chol" X
"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
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
ASKER
altPairs("y") → "y" "" X
sorry i meant to ask why it failed for above test case. please advise
ASKER
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?
"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 methodcorrect
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.substrin g(i,i+2)) ;
}
if( i==len-1 ){
result.append(str.substrin g(i)) ;
}
return result.toString() ;
}
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.substrin
}
if( i==len-1 ){
result.append(str.substrin
}
return result.toString() ;
}
ASKER
I see above solution more clear.
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?
altPairs("") → "" "" OKwill 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?
ASKER
altPairs("") → "" "" OK
how it works for above case
i<len-1(here len-1 becomes -1 for i=0 right)
Please advjes
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
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
ASKER
correct
How above solution is working then for len==0 and len==2. please advise
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
ASKER
when len==1 i can visualize it going into if loop and eppending as below
if( i==len-1 ){
result.append(str.substrin g(i)) ;
}
but if len==0 how solution passed the test case(for string like""). please advise
if( i==len-1 ){
result.append(str.substrin
}
but if len==0 how solution passed the test case(for string like""). please advise
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() ;
}
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() ;
}
ASKER
if len==0, the solution of http#a40454432 reduces toi 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?
ASKER
ozo2014-11-19 at 23:49:00ID: 40454432
i see without a
ASKER
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() ;
}
should not be as below? please advisepublic 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() ;
}
ASKER
I see if the len==0 then it directly goes to
return result.toString() ; without going to for and if loop.
Please advise if my understanding is correct?
return result.toString() ; without going to for and if loop.
Please advise if my understanding is correct?
if the len==0 then it directly goes toThat's correct
return result.toString() ;
http#a40454432
should have been
http:#a40454432
ASKER
// 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?
please advise
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
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
ASKER
altPairs("kitten") → "kien" "kien" OK
for above example str.length is 6 rightfor 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 2so 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
"kitten".substring(0,2) is "ki", which we want
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
ASKER
"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
ASKER
Oh well i see substring end is excluded right unlike start index.
Ok i got it now
Ok i got it now
challenge says at indexes 0,1
challenge does not say substring(0,1)
challenge does not say substring(0,1)
ASKER
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<=please advise
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
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
ASKER
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.
Please advise if my understanding is correct?
i+1 scenario also comes for "Chocolate"
ASKER
i+1 scenario also comes for "Chocolate"oh. When does it come for Chocolate. Please advise
When i==8
ASKER
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?
please advise
"Chocolate".length() ==9
8<9 is true
8<9 is true
ASKER
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 outputvalue 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.
Please advise where i am missing
8+2<="Chocolate".length() is not true
ASKER
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.
is my understanding above is correct now right? please advise
"Chocolate".length() is 9, not 8
ASKER
i got outputmy above understanding is correct right? please advise
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.
ASKER
Please advise as i corrected my above post as highlighted place
yes
ASKER
thank you
ASKER
Open in new window
i got it.Open in new window
what is the meaning of above codeplease elaborate on this