String array - instances of unique values

Question

I wrote a program that loops through a string array and prints the unique words and their occurrences, and then prints them to a file. This part works perfectly, but I'm having trouble figuring out how to get the total count of unique words. I have a feeling I need to create an int array for a counter, but I'm not exactly grasping how to do this. Unfortunately I am NOT allowed to use Hashmap, Sets, or Lists; have to stick with Arrays.

boolean [] done = new boolean[textfile.length];
for(int i = 0; i<textfile.length; i++){
    if(done[i])
        continue;
    int nb = 0;
    for(int j = i; j < textfile.length; j++){
        if(done[j])
            continue;
        if(textfile[i].equals(textfile[j])){
            done[j] = true;
            nb++;

        }
    }

    pw.println(textfile[i] + "occurs " + nb + " times");
}

are you allowed to use set?

codeMan
– codeMan

2013-10-09 10:26:10 +00:00
Commented Oct 9, 2013 at 10:26 — codeMan
– codeMan, Commented Oct 9, 2013 at 10:26
what is the variable nb??

codeMan
– codeMan

2013-10-09 10:27:47 +00:00
Commented Oct 9, 2013 at 10:27 — codeMan
– codeMan, Commented Oct 9, 2013 at 10:27
counts the instances of each unique word

user2791187
– user2791187

2013-10-09 10:28:26 +00:00
Commented Oct 9, 2013 at 10:28 — user2791187
– user2791187, Commented Oct 9, 2013 at 10:28

Prasad · Accepted Answer · 2013-10-09 10:50:38Z

1

I will suggest you to do as follows-

-Add all the values in an array and sort this array.(so that it will be easy to get unique entry)

-Compare all the elements of sorted array (one by one) with the word to be compared.

-Now while comparing a word in a text file with a word present in an array, maintain a global variable i.e. counter which will be incremented on every occurrence of a unique character and will persist the value for future use.

answered Oct 9, 2013 at 10:50

Prasad

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Comments

Ron Teller · Accepted Answer · 2013-10-09 10:36:58Z

1

Right now you are comparing each word with all the words after it in the file, which takes O(N^2) times, you can instead create an array containing all the words, sort it by lexicographic order which takes O(NlogN) time, then iterate through the array and count the occurrences of each unique word (if 2 adjacent words are equal, keep counting, otherwise print and reset the counter) which takes O(N) times.

edited Oct 9, 2013 at 10:36

answered Oct 9, 2013 at 10:31

Ron Teller

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1 Comment

user2791187 Over a year ago

Thanks, this is just the info I needed. I'll see if I can get this to work.

alex · Accepted Answer · 2013-10-09 10:27:59Z

0

You could set uniqe_counter outside of first for and increment that every time when nb == 1 before printing

answered Oct 9, 2013 at 10:27

alex

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Comments

codeMan · Accepted Answer · 2013-10-09 10:29:38Z

0

You are reinitializing the variable nb in every iteration of the outer loop. thus u end up loosing the count of unique words processed in the last iteration.

U have to place the int nb = 0; out side the first for loop.

answered Oct 9, 2013 at 10:29

codeMan

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Comments

Milan Baran · Accepted Answer · 2013-10-09 10:34:16Z

0

The easiest way is:

Sort the array
Go trough the sorted array and remember last unique entry
If lastUniqueEntry is different as next entry. UniqueCount + 1
Do everything else, like prints the unique words and their occurrences

answered Oct 9, 2013 at 10:34

Milan Baran

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