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Array and file creation basic Java Array Problem

I happen to be having a hard time with a problem. A test will be coming up soon for me and I'm just not so sure as to how to get this problem done. (Nothing too advanced) My main issue is that the reading of and creating the final file. the problem is right below. Any help is appreciated, thanks!
I've been told to add a summary due to the difficulty to understand what im asking. So essentially you have a file, text.txt and the program takes said file and adds all the numbers in it, spaces make the numbers distinguished and if a zero is at the front it needs to get taken away. All numbers are separated (whatever number of spaces are between them) is all separated with a space + space and a space = at the end followed by space the result of the addition. I'm looking for help as how to:
a) Make the new file
b) Make the +s and =s
c) Transferring from the original file
d) How to do the addition (dealing with large space gaps and doing the summing)
IM STILL VERY NEW TO JAVA SO ANY LONGER EXPLANATIONS WOULD BE GREATLY APPRECIATED
The approach you are to implement is to store each integer in an array of digits, with one digit per array element. We will be using arrays of length 25, so we will be able to store integers up to 25 digits long. We have to be careful in how we store these digits. Consider, for example, storing the numbers 38423 and 27. If we store these at the “front” of the array with the leading digit of each number in index 0 of the array, then when we go to add these numbers together, we’re likely to add them like this:
38423
27
Thus, we would be adding 3 and 2 in the first column and 8 and 7 in the second column. Obviously this won’t give the right answer. We know from elementary school arithmetic that we have to shift the second number to the right to make sure that we have the appropriate columns lined up:
38423
27
To simulate this right-shifting of values, we will store each value as a sequence of exactly 25 digits, but we’ll allow the number to have leading 0’s. For example, the problem above is converted into:
0000000000000000000038423
0000000000000000000000027
Now the columns line up properly and we have plenty of space at the front in case we have even longer numbers to add to these.
The data for your program will be stored in a file called sum.txt. Each line of the input file will have a different addition problem for you to solve. Each line will have one or more integers to be added together. Take a look at the input file at the end of this write-up and the output you are supposed to produce. Notice that you produce a line of output for each input line showing the addition problem you are solving and its answer. Your output should also indicate at the end how many lines of input were processed. You must exactly reproduce this output.
You should use the techniques described in chapter 6 to open a file, to read it line by line, and to process the contents of each line. In reading these numbers, you won’t be able to read them as ints or longs because many of them are too large to be stored in an int or long. So you’ll have to read them as String values using calls on the method next(). Your first task, then, will be to convert a String of digits into an array of 25 digits. As described above, you’ll want to shift the number to the right and include leading 0’s in front. Handout #15 provides an example of how to process an array of digits. Notice in particular the use of the String method charAt and the method Character.getNumericValue. These will be helpful for solving this part of the problem.
You are to add up each line of numbers, which means that you’ll have to write some code that allows you to add together two of these numbers or to add one of them to another. This is something you learned in Elementary School to add starting from the right, keeping track of whether there is a digit to carry from one column to the next. Your challenge here is to take a process that you are familiar with and to write code that performs the corresponding task.
Your program also must write out these numbers. In doing so, it should not print any leading 0’s. Even though it is convenient to store the number internally with leading 0’s, a person reading your output would rather see these numbers without any leading 0’s.
You can assume that the input file has numbers that have 25 or fewer digits and that the answer is always 25 digits or fewer. Notice, however, that you have to deal with the possibility that an individual number might be 0 or the answer might be 0. There will be no negative integers in the input file.
You should solve this problem using arrays that are exactly 25 digits long. Certain bugs can be solved by stretching the array to something like 26 digits, but it shouldn’t be necessary to do that and you would lose style points if your arrays require more than 25 digits.
The choice of 25 for the number of digits is arbitrary (a magic number), so you should introduce a class constant that you use throughout that would make it easy to modify your code to operate with a different number of digits.
Consider the input file as an example of the kind of problems your program must solve. We might use a more complex input file for actual grading.
The Java class libraries include classes called BigInteger and BigDecimal that use a strategy similar to what we are asking you to implement in this program. You are not allowed to solve this problem using BigInteger or BigDecimal. You must solve it using arrays of digits.
The sample program of handout #18 should be particularly helpful to study to prepare you for this programming problem. It has some significant differences from the task you are asked to solve, but it also has some similarities that you will find helpful to study. The textbook has a discussion of the program starting in section 7.6.
You may assume that the input file has no errors. In particular, you may assume that each line of input begins with at least one number and that each number and each answer will be 25 digits or fewer. There will be whitespace separating the various numbers, although there is no guarantee about how much whitespace there will be between numbers.
You will again be expected to use good style throughout your program and to comment each method and the class itself. A major portion of the style points will be awarded based on how you break this program down into static methods. As with the sample program in handout #18, try to think in terms of logical subtasks of the overall task and create different methods for different subtasks. You should have at least four static methods other than main and you are welcome to introduce more than four if you find it helpful.
Your program should be stored in a file called Sum.java. You will need to include the files Scanner.java and sum.txt from the class web page (under the “assignments” link) in the same folder as your program. For those using DrJava, you will either have to use a full path name for the file sum.txt (see section 6.2.2 of the book) or you will have to put the file in the same directory as the DrJava program.
Input file sum.txt
82384
204 435
22 31 12
999 483
28350 28345 39823 95689 234856 3482 55328 934803
7849323789 22398496 8940 32489 859320
729348690234239 542890432323 534322343298
3948692348692348693486235 5834938349234856234863423
999999999999999999999999 432432 58903 34
82934 49802390432 8554389 4789432789 0 48372934287
0
0 0 0
7482343 0 4879023 0 8943242
3333333333 4723 3333333333 6642 3333333333
Output that should be produced
82384 = 82384
204 + 435 = 639
22 + 31 + 12 = 65
999 + 483 = 1482
28350 + 28345 + 39823 + 95689 + 234856 + 3482 + 55328 + 934803 = 1420676
7849323789 + 22398496 + 8940 + 32489 + 859320 = 7872623034
729348690234239 + 542890432323 + 534322343298 = 730425903009860
3948692348692348693486235 + 5834938349234856234863423 = 9783630697927204928349658
999999999999999999999999 + 432432 + 58903 + 34 = 1000000000000000000491368
82934 + 49802390432 + 8554389 + 4789432789 + 0 + 48372934287 = 102973394831
0 = 0
0 + 0 + 0 = 0
7482343 + 0 + 4879023 + 0 + 8943242 = 21304608
3333333333 + 4723 + 3333333333 + 6642 + 3333333333 = 10000011364
Total lines = 14

Having problems formatting Strings to a specific precision

I am trying to format a string and write it into a file. See my code below
StringBuilder sb=new StringBuilder();
sb.insert(0, String.format("%-30s", recDataWithWarId.getRiaFirmName() != null ? recDataWithWarId.getRiaFirmName() : " "));
The conditions I would like to achieve are (I already have a handle on condition 1 mentioned below):
If recDataWithWarId.getRiaFirmName() is not thirty characters in length or null I want to fill up rest of spaces with blank spaces.
I would also like to limit recDataWithWarId.getRiaFirmName() to 30 if the length is greater than 30 and which I had hoped this code would do automatically.
What is the most efficient way of achieving these two conditions without having to write lengthy code one for substring and one to format? I have to do it on multiple place as this is a fairly large text file.

You need to specify both a precision and a width in your format string. You want "%-30.30s"
The first -30 specifies that the value will be padded on the right to 30 characters. The 30 after the decimal point specifies the maximum number of characters.

Suffix array nlogn creation

I have been learning suffix arrays creation, & i understand that We first sort all suffixes according to first character, then according to first 2 characters, then first 4 characters and so on while the number of characters to be considered is smaller than 2n.
But my doubt is why don't we choose the first 3 characters, then 9... and so on. Why only 2 characters are taken into account since the strings are a part of same strings and not different random strings?

I haven't analyzed the suffix array construction algorithm thoroughly, but still would like to share my thoughts.
In my humble opinion, your question is similar to the following ones:
Why do computers use binary encoding of information instead of ternary?
Why does binary search bisect the range instead of trisecting it?
Why are there two sexes rather than three?
The reason is that the number 2 is special - it is the smallest plural number. The difference between 1 and 2 is qualitative, whereas the difference between 2 and 3 (as well as any other positive integer) is quantitative and therefore not as drastic.
As a result, binary formulation of many algorithms and data structures turns out to be the simplest one, though some of them may be generalized, with various degrees of added complexity, for an arbitrary base.

Answer is given from the post you linked. And as #Leon answered, the algorithm work because it use a dichotomous approach to solve the sorting problem. if you correctly read the answer, the main purpose is to divide word be small 2 character fragments. So that 4 characters can be easily sort base on the arrangement of the 2 pair of characters, 6 characters with 4-2 or 2-4 or 2-2-2 and so one. Thus have a word of 3 letters in the table is non-sense since word of 3 characters may be seen has 2 characters + the position in the alphabet of the last character.

I think you are considering only the speed of 2^x versus 3^x where you obviously would prefer the latter.
But you have to consider the effort you need for each step.
Since 3^x needs about 1.58 less steps than 2^x you would need to be able to compute a single step for the 3^x growth in less than 1.58 times what you need for a single step in the 2^x growth to perform better.
Generally the problems will get much more complex when you have to handle three elements in each step instead of two.
Also if you could expand it to 3^x you could also do it for a bigger n^x and then with big n your algorithm is suddenly not exponential but effectively linear.

Unique ID Code in Java

Hi I want you guys suggestion.
I want to create a 15 length key code form current time stamp. Code should contain small and capital characters with digits. Anyone can suggest how I can create a 15 length unique code from current time stamp? Using Java in servlet side.

Given your constraints, I'd probably create a GUID, append or prepend a timestamp, and transform it into the 15-letters format. GUIDs are 32 hexadecimal digits and so have 32^16 (1.20892582x1024) possible values (although not all of them are used). 15 characters with digits or upper or lower case English letters (so, 62 possible values per digit) gives you 15^62 (8.272905461x1072) — plenty of room. If you were okay adding + and / to your list of possible characters, you could use Base64 encoding rather than doing it yourself.

Parsing an arithmetic expression and building a tree from it in Java

I needed some help with creating custom trees given an arithmetic expression. Say, for example, you input this arithmetic expression:
(5+2)*7
The result tree should look like:
*
/ \
+ 7
/ \
5 2
I have some custom classes to represent the different types of nodes, i.e. PlusOp, LeafInt, etc. I don't need to evaluate the expression, just create the tree, so I can perform other functions on it later.
Additionally, the negative operator '-' can only have one child, and to represent '5-2', you must input it as 5 + (-2).
Some validation on the expression would be required to ensure each type of operator has the correct the no. of arguments/children, each opening bracket is accompanied by a closing bracket.
Also, I should probably mention my friend has already written code which converts the input string into a stack of tokens, if that's going to be helpful for this.
I'd appreciate any help at all. Thanks :)
(I read that you can write a grammar and use antlr/JavaCC, etc. to create the parse tree, but I'm not familiar with these tools or with writing grammars, so if that's your solution, I'd be grateful if you could provide some helpful tutorials/links for them.)

Assuming this is some kind of homework and you want to do it yourself..
I did this once, you need a stack
So what you do for the example is:
parse what to do? Stack looks like
( push it onto the stack (
5 push 5 (, 5
+ push + (, 5, +
2 push 2 (, 5, +, 2
) evaluate until ( 7
* push * 7, *
7 push 7 +7, *, 7
eof evaluate until top 49
The symbols like "5" or "+" can just be stored as strings or simple objects, or you could store the + as a +() object without setting the values and set them when you are evaluating.
I assume this also requires an order of precedence, so I'll describe how that works.
in the case of: 5 + 2 * 7
you have to push 5 push + push 2 next op is higher precedence so you push it as well, then push 7. When you encounter either a ) or the end of file or an operator with lower or equal precedence you start calculating the stack to the previous ( or the beginning of the file.
Because your stack now contains 5 + 2 * 7, when you evaluate it you pop the 2 * 7 first and push the resulting *(2,7) node onto the stack, then once more you evaluate the top three things on the stack (5 + *node) so the tree comes out correct.
If it was ordered the other way: 5 * 2 + 7, you would push until you got to a stack with "5 * 2" then you would hit the lower precedence + which means evaluate what you've got now. You'd evaluate the 5 * 2 into a *node and push it, then you'd continue by pushing the + and 3 so you had *node + 7, at which point you'd evaluate that.
This means you have a "highest current precedence" variable that is storing a 1 when you push a +/-, a 2 when you push a * or / and a 3 for "^". This way you can just test the variable to see if your next operator's precedence is < = your current precedence.
if ")" is considered priority 4 you can treat it as other operators except that it removes the matching "(", a lower priority would not.

I wanted to respond to Bill K.'s answer, but I lack the reputation to add a comment there (that's really where this answer belongs). You can think of this as a addendum to Bill K.'s answer, because his was a little incomplete. The missing consideration is operator associativity; namely, how to parse expressions like:
49 / 7 / 7
Depending on whether division is left or right associative, the answer is:
49 / (7 / 7) => 49 / 1 => 49
or
(49 / 7) / 7 => 7 / 7 => 1
Typically, division and subtraction are considered to be left associative (i.e. case two, above), while exponentiation is right associative. Thus, when you run into a series of operators with equal precedence, you want to parse them in order if they are left associative or in reverse order if right associative. This just determines whether you are pushing or popping to the stack, so it doesn't overcomplicate the given algorithm, it just adds cases for when successive operators are of equal precedence (i.e. evaluate stack if left associative, push onto stack if right associative).

The "Five minute introduction to ANTLR" includes an arithmetic grammar example. It's worth checking out, especially since antlr is open source (BSD license).

Several options for you:
Re-use an existing expression parser. That would work if you are flexible on syntax and semantics. A good one that I recommend is the unified expression language built into Java (initially for use in JSP and JSF files).
Write your own parser from scratch. There is a well-defined way to write a parser that takes into account operator precedence, etc. Describing exactly how that's done is outside the scope of this answer. If you go this route, find yourself a good book on compiler design. Language parsing theory is going to be covered in the first few chapters. Typically, expression parsing is one of the examples.
Use JavaCC or ANTLR to generate lexer and parser. I prefer JavaCC, but to each their own. Just google "javacc samples" or "antlr samples". You will find plenty.
Between 2 and 3, I highly recommend 3 even if you have to learn new technology. There is a reason that parser generators have been created.
Also note that creating a parser that can handle malformed input (not just fail with parse exception) is significantly more complicated that writing a parser that only accepts valid input. You basically have to write a grammar that spells out the various common syntax errors.
Update: Here is an example of an expression language parser that I wrote using JavaCC. The syntax is loosely based on the unified expression language. It should give you a pretty good idea of what you are up against.
Contents of org.eclipse.sapphire/plugins/org.eclipse.sapphire.modeling/src/org/eclipse/sapphire/modeling/el/parser/internal/ExpressionLanguageParser.jj

the given expression (5+2)*7 we can take as infix
Infix : (5+2)*7
Prefix : *+527
from the above we know the preorder and inorder taversal of tree ... and we can easily construct tree from this.
Thanks,