I have a long String with binary values. And i have a hash map that has the Binary digits as a key and char as a value. I have a function that supposed to read the binary string using 2 pointers and compare with hashmap and store the corresponding char in main.decodedTxt. However, im getting string out of bound exception for this. I don't know how to solve this. I'm getting exception on "String temp =" line. I have a picture link of the console output to see better picture.
public static void bitStringToText (String binText){
String bs = binText;
int from =0;
int to = 1;
while(bs != null){
String temp = bs.substring(from, to);
if (main.newMapDecoding.containsKey(temp)){
main.decodedTxt += main.newMapDecoding.get(temp);
from =to;
to = from +1;
} else {
to = to + 1;
}
}
}
Image of console exception is here
First of all there is no need to check if bs is null because no part of your code changes the value of bs. Your current code will cross the possible index of your binText at some point. It's better to loop just binText and check if you find something within it. After all you have to traverse the complete string anyways. Change your code as follows
public static void bitStringToText (String binText){
//no need to do this if you are not modifying the contents of binText
//String bs = binText;
int from =0;
int to = 1;
int size = binText.length();
String temp = "";
while(to <= size ){
temp = binText.substring(from, to);
if (main.newMapDecoding.containsKey(temp)){
main.decodedTxt += main.newMapDecoding.get(temp);
from =to;
to = from +1;
} else {
to = to + 1;
}
}
}
Hope it helps.
First, give it a try to practice debugging. It is an easy case. Either use run in debug mode (place break point on String temp = bs.substring(from, to); line) or print values of from and to before the same line. It will help to understand what is going on.
Solution:
If bs is not null you will always have StringIndexOutOfBoundsException. Because you are not checking if to is pointing to not existed index of bs String. Easiest example of the first one will be empty String: bs == "".
One of the solution could be to replace condition in while to while (to <= bs.length()).
Related
So recently I got invited to this google foo.bar challenge and I believe the code runs the way it should be. To be precise what I need to find is the number of occurrences of "abc" in a String. When I verify my code with them, I pass 3/10 test cases. I'm starting to feel bad because I don't know what I am doing wrong. I have written the code which I will share with you guys. Also the string needs to be less than 200 characters. When I run this from their website, I pass 3 tests and fail 7. Basically 7 things need to be right.
The actual question:
Write a function called answer(s) that, given a non-empty string less
than 200 characters in length describing the sequence of M&Ms. returns the maximum number of equal parts that can be cut from the cake without leaving any leftovers.
Example : Input : (string) s = "abccbaabccba"
output : (int) 2
Input: (string) s = "abcabcabcabc"
output : (int) 4
public static int answer(String s) {
int counter = 0;
int index;
String findWord ="ABC";
if(s!=null && s.length()<200){
s = s.toUpperCase();
while (s.contains(findWord))
{
index = s.indexOf(findWord);
s = s.substring(index + findWord.length(), s.length());
counter++;
}
}
return counter;
}
I see a couple of things in your code snippet:
1.
if(s.length()<200){
Why are you checking for the length to be lesser than 200? Is that a requirement? If not, you can skip checking the length.
2.
String findWord ="abc";
...
s.contains(findWord)
Can the test program be checking for upper case alphabets? Example: "ABC"? If so, you might need to consider changing your logic for the s.contains() line.
Update:
You should also consider putting a null check for the input string. This will ensure that the test cases will not fail for null inputs.
The logic of your code is well but on the other hand i found that you didn't check for if input string is empty or null.
I belief that google foo.bar wants to see the logic and the way of coding in a proper manner.
so don't be feel bad
I would go for a simpler approach
int beforeLen = s.length ();
String after = s.replace (findWord, "");
int afterLen = after.length ();
return (beforeLen - afterLen) / findWord.length ();
String pattern = "abc";
String line="<input text here>";
int i=0;
Pattern TokenPattern=Pattern.compile(pattern);
if(line!=null){
Matcher m=TokenPattern.matcher(line);
while(m.find()){
i++;
}}
System.out.println("No of occurences : "+ " "+i);
put declaration of index out before while block, isn't never good re-declare the same variable n time.
int index;
while (s.contains(findWord))
{
index = s.indexOf(findWord);
....
}
I hope this help
Update:
try to compact your code
public static int answer(String s) {
int counter = 0;
int index;
String findWord = "ABC";
if (s != null && s.length() < 200) {
s = s.toUpperCase();
while ((index = s.indexOf(findWord)) > -1) {
s = s.substring(index + findWord.length(), s.length());
counter++;
}
}
return counter;
}
Update:
The logic seems good to me, I'm still try to improve the performance, if you can try this
while ((index = s.indexOf(findWord, index)) > -1) {
//s = s.substring(index + findWord.length(), s.length());
index+=findWord.length();
counter++;
}
Hopefully this will be an easy answer and i'm just overlooking something minor.
Goal: take an array list (which currently contains lines of text from a text file) and set a String variable equal a specified array list position.
At the moment, each line is raw text taken from an Encyclopedia file. I need to be able to remove non alpha's via the .replaceAll function. However, my current program returns a null pointer exception and I'm having some trouble understanding why. I'm fairly new to Java so full answers and explanations are much appreciated.
my code: (My teacher told us to use EasyReader class to make our lives, well easier...)
EasyReader fileIn = new EasyReader("Encyclopedia.txt");
public void createList()
{
String x=fileIn.readLine();
ArrayList<String> list = new ArrayList<String>();
while((!fileIn.eof()))
{
String y=fileIn.readLine();
list.add(y);
}
int count=0;
while(count<list.size())
{
String temp=list.get(count);
temp.replaceAll("[^a-zA-z ]", ""); //null pointer points to this line
temp.toLowerCase(); //and this line
list.set(count, temp);
count++;
}
count=0;
while(count<list.size());
{
System.out.println(list.get(count));
count++;
}
System.out.println(list.size());
while(count<list.size())
{
fileOut.println(list.get(count));
count++;
}
fileOut.close();
}
thanks in advance for the help :)
I think I found your error!
Your while loop should go only to list.size() - 1 not the list.size() you have there. See below:
while(count<list.size()- 1)
{
String temp=list.get(count);
temp.replaceAll("[^a-zA-z ]", ""); //null pointer points to this line
temp.toLowerCase(); //and this line
list.set(count, temp);
count++;
}
Try replacing the two lines that you commented with this instead:
temp = temp.replaceAll("[^a-zA-z ]", "");
temp = temp.toLowerCase();
This is because the replaceAll() method does not change the original string itself, but rather returns a new string with the characters replaced. Same thing with toLowerCase().
You made two mistakes:
1) the replaceAll regex should be temp.replaceAll("[^a-zA-z]", "") (no space after z)
2) you can combine two lines into one method temp.replaceAll("[^a-zA-z]", "").toLowerCase()
3) you need to save your newly returned String to the original variable temp:
temp = temp.replaceAll("[^a-zA-z]", "").toLowerCase();
as #Suitangi mentioned, replaceAll() method does not change the original string itself, but rather returns a new string with the characters replaced. Same thing with toLowerCase().
Hope that helps.
EDIT:
In the last iteration String temp = null;
Add if condition to check it, before regex:
if (temp!=null){
temp = temp.replaceAll("[^a-zA-z]", "").toLowerCase();
list.set(count, temp);
count++;
}
I need to get the values after "Swap:".
I've already developed a method to get the output from a shell command so I have a string that contains everything you see in the picture but now from the string I want to get ONLY the value after the Swap: How can i do this? These value are variable and can be even all three 0.
Let's say you have the text stored in a String called textContent. Assuming the Swap-line is the last part of your String, then you could do something like this:
int index = textContent.indexOf("Swap:");
index += "Swap:".length();
textContent.subString(index);
Try this:
String[] stringParts = text.substring(text.indexOf("Swap:") + 5).trim().split("( )+");
int[] parts = new int[stringParts.length];
for (int i = 0; i < stringParts.length; i++)
parts[i] = Integer.parseInt(stringParts[i]);
It will fill an integer array will the values after the "Swap" part.
Since you have already stored the output of the shell command, you simply need to do some string manipulation to search and extract the relevant information. The following particular string manipulation methods might be of interest to you: trim(), indexOf(), and substring().
Below is a simple example code on how to extract the value under the total's column using the above String methods:
public class ShellOutput {
public ShellOutput() {
final String extract = "Swap:"; // the keyword to search
String shellOutput = "Swap: 75692 29657 0"; // your shell output
int position = shellOutput.indexOf(extract); // get the position of the Swap: text
if (position != -1) {
String swapLine = shellOutput.substring(position + extract.length()); // remove everything except the swap line
String numbers = swapLine.trim(); // assuming they are spaces, otherwise do some operations to remove tabs if used
int firstSpace = numbers.indexOf(' '); // get the first space or change to a tab character if it is used
String totalNumber = numbers.substring(0, firstSpace); // remove up to the first found after the number
System.out.println("Total = " + totalNumber);
} else {
System.out.println("No '" + extract + "' segment found.");
}
}
public static void main(String[] args) {
new ShellOutput();
}
}
Output: Total = 75692
I'm using codingbat.com to get some java practice in. One of the String problems, 'withoutString' is as follows:
Given two strings, base and remove, return a version of the base string where all instances of the remove string have been removed (not case sensitive).
You may assume that the remove string is length 1 or more. Remove only non-overlapping instances, so with "xxx" removing "xx" leaves "x".
This problem can be found at: http://codingbat.com/prob/p192570
As you can see from the the dropbox-linked screenshot below, all of the runs pass except for three and a final one called "other tests." The thing is, even though they are marked as incorrect, my output matches exactly the expected output for the correct answer.
Here's a screenshot of my output:
And here's the code I'm using:
public String withoutString(String base, String remove) {
String result = "";
int i = 0;
for(; i < base.length()-remove.length();){
if(!(base.substring(i,i+remove.length()).equalsIgnoreCase(remove))){
result = result + base.substring(i,i+1);
i++;
}
else{
i = i + remove.length();
}
if(result.startsWith(" ")) result = result.substring(1);
if(result.endsWith(" ") && base.substring(i,i+1).equals(" ")) result = result.substring(0,result.length()-1);
}
if(base.length()-i <= remove.length() && !(base.substring(i).equalsIgnoreCase(remove))){
result = result + base.substring(i);
}
return result;
}
Your solution IS failing AND there is a display bug in coding bat.
The correct output should be:
withoutString("This is a FISH", "IS") -> "Th a FH"
Yours is:
withoutString("This is a FISH", "IS") -> "Th a FH"
Yours fails because it is removing spaces, but also, coding bat does not display the correct expected and run output string due to HTML removing extra spaces.
This recursive solution passes all tests:
public String withoutString(String base, String remove) {
int remIdx = base.toLowerCase().indexOf(remove.toLowerCase());
if (remIdx == -1)
return base;
return base.substring(0, remIdx ) +
withoutString(base.substring(remIdx + remove.length()) , remove);
}
Here is an example of an optimal iterative solution. It has more code than the recursive solution but is faster since far fewer function calls are made.
public String withoutString(String base, String remove) {
int remIdx = 0;
int remLen = remove.length();
remove = remove.toLowerCase();
while (true) {
remIdx = base.toLowerCase().indexOf(remove);
if (remIdx == -1)
break;
base = base.substring(0, remIdx) + base.substring(remIdx + remLen);
}
return base;
}
I just ran your code in an IDE. It compiles correctly and matches all tests shown on codingbat. There must be some bug with codingbat's test cases.
If you are curious, this problem can be solved with a single line of code:
public String withoutString(String base, String remove) {
return base.replaceAll("(?i)" + remove, ""); //String#replaceAll(String, String) with case insensitive regex.
}
Regex explaination:
The first argument taken by String#replaceAll(String, String) is what is known as a Regular Expression or "regex" for short.
Regex is a powerful tool to perform pattern matching within Strings. In this case, the regular expression being used is (assuming that remove is equal to IS):
(?i)IS
This particular expression has two parts: (?i) and IS.
IS matches the string "IS" exactly, nothing more, nothing less.
(?i) is simply a flag to tell the regex engine to ignore case.
With (?i)IS, all of: IS, Is, iS and is will be matched.
As an addition, this is (almost) equivalent to the regular expressions: (IS|Is|iS|is), (I|i)(S|s) and [Ii][Ss].
EDIT
Turns out that your output is not correct and is failing as expected. See: dansalmo's answer.
public String withoutString(String base, String remove) {
String temp = base.replaceAll(remove, "");
String temp2 = temp.replaceAll(remove.toLowerCase(), "");
return temp2.replaceAll(remove.toUpperCase(), "");
}
Please find below my solution
public String withoutString(String base, String remove) {
final int rLen=remove.length();
final int bLen=base.length();
String op="";
for(int i = 0; i < bLen;)
{
if(!(i + rLen > bLen) && base.substring(i, i + rLen).equalsIgnoreCase(remove))
{
i +=rLen;
continue;
}
op += base.substring(i, i + 1);
i++;
}
return op;
}
Something things go really weird on codingBat this is just one of them.
I am adding to a previous solution, but using a StringBuilder for better practice. Most credit goes to Anirudh.
public String withoutString(String base, String remove) {
//create a constant integer the size of remove.length();
final int rLen=remove.length();
//create a constant integer the size of base.length();
final int bLen=base.length();
//Create an empty string;
StringBuilder op = new StringBuilder();
//Create the for loop.
for(int i = 0; i < bLen;)
{
//if the remove string lenght we are looking for is not less than the base length
// and the base substring equals the remove string.
if(!(i + rLen > bLen) && base.substring(i, i + rLen).equalsIgnoreCase(remove))
{
//Increment by the remove length, and skip adding it to the string.
i +=rLen;
continue;
}
//else, we add the character at i to the string builder.
op.append(base.charAt(i));
//and increment by one.
i++;
}
//We return the string.
return op.toString();
}
Taylor's solution is the most efficient one, however I have another solution that is a naive one and it works.
public String withoutString(String base, String remove) {
String returnString = base;
while(returnString.toLowerCase().indexOf(remove.toLowerCase())!=-1){
int start = returnString.toLowerCase().indexOf(remove.toLowerCase());
int end = remove.length();
returnString = returnString.substring(0, start) + returnString.substring(start+end);
}
return returnString;
}
#Daemon
your code works. Thanks for the regex explanation. Though dansalmo pointed out that codingbat is displaying the intended output incorrectly, I through in some extra lines to your code to unnecessarily account for the double spaces with the following:
public String withoutString(String base, String remove){
String result = base.replaceAll("(?i)" + remove, "");
for(int i = 0; i < result.length()-1;){
if(result.substring(i,i+2).equals(" ")){
result = result.replace(result.substring(i,i+2), " ");
}
else i++;
}
if(result.startsWith(" ")) result = result.substring(1);
return result;
}
public String withoutString(String base, String remove){
return base.replace(remove,"");
}
i have a string like this
106.826820,-6.303850,0
which i get from parsing the google Maps KML Document.
now i wanna parse its string to Double
this is my code :
public Double getLatitude(){
for(int posisi = GPS_NAME.indexOf(","); posisi > GPS_NAME.length(); posisi++){
TEMP_LAT = "" + GPS_NAME.indexOf(posisi);
}
GPS_LATITUDE = Double.valueOf(TEMP_LAT);
return GPS_LATITUDE;
}
public Double getLongitude(){
int posisiakhir = GPS_NAME.indexOf(",");
for(int i = 0; i < posisiakhir; i++){
TEMP_LON = "" + GPS_NAME.indexOf(i);
}
GPS_LONGITUDE = Double.valueOf(TEMP_LON);
return GPS_LONGITUDE;
}
but when i try to run it i got error like this
could somebody help me solving my problems >_<
and also confirm GPS_LATITUDE = Double.valueOf(TEMP_LAT); TEMP_LAT is not null there - as exception is NullPointerException not NumberFormatException.
You have null pointer exception so you should to it like this:
public Double getLatitude(){
for(int posisi = GPS_NAME.indexOf(","); posisi > GPS_NAME.length(); posisi++){
TEMP_LAT = "" + GPS_NAME.indexOf(posisi);
}
if (TEMP_LAT != null) {
GPS_LATITUDE = Double.parseDouble(TEMP_LAT);
}
return GPS_LATITUDE;
}
And for converting to double rather you should use Double.parseDouble() or also you can use new Double(TEMP_LAT).doubleValue() but first approach is cleaner.
Besides the probable causes for the NPE, I don't really get your logic; it looks completely dodgy to me.
Just some examples:
GPS_NAME.indexOf(",") will either return -1 or an index that is smaller than the length of the string in which is being searched. Then why have a condition that checks if it is larger than the length? posisi > GPS_NAME.length() will never be true, hence the for loops are useless...
Then inside the loops you do TEMP_LAT = "" + GPS_NAME.indexOf(posisi). From the earlier remark we know that posisi is either -1 or some other number that is smaller than the length of the string. So GPS_NAME.indexOf(posisi) will try to find a character repesented by the integer posisi (which will be a rather small number) in the string. How does that make sense?
I'd like to advise you to rethink your logic - perhaps String.split(",") is a good starting point.
Use Double.parseDouble(TEMP_LAT);