Printing Individual Words Without Repeating?

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I want to list every unique word in a text file and how many times every word is found in it. I tried using an if cycle but I'm not sure how to eliminate the already listed words after they are being counted.

for (int i = 0; i < words.size(); i++) {
    count = 1;
    //Count each word in the file and store it in variable count 
    for (int j = i + 1; j < words.size(); j++) {
        if (words.get(i).equals(words.get(j))) {

    System.out.println("The word " + words.get(i) + " can be 
    found " + count + " times in the file.");

The contents of the text file is "Hello world. Hello world.", and the program will print the following: The word Hello can be found 2 times in the file. The word world can be found 2 times in the file. The word Hello can be found 1 times in the file. The word world can be found 1 times in the file.

You need to use an ArrayList to store the already found words, and after that, you need to check every word in the file, whether it is present within the ArrayList or not. If the word is present inside the ArrayList, you need to ignore that word. Otherwise, add that word to the ArrayList. A sample code for you:

ArrayList<String> found_words=new ArrayList<String>();
public static void main(String arguments[])
    String data=""; //data from your file
    String[] words=data.split("\\s"); //split the string into individual words
    for(int i=0;i<words.length;i++)
        String current_word=words[i];
            int count=1;
            for(int j=i+1;j<words.length;j++)
            System.out.println("The word "+current_word+" can be found "+count+" times in the file.");
static boolean is_present(String word)
    for(int i=0;i<found_words.size();i++)
            return true;
    return false;

Print the Non-repeating words | Practice, Print all the non-repeating words out of two given sentences. 1 st come Statement 1 String words those are not Repeated in statement 2 and After same as� How to Create and Print Labels for a Single Item or Address in Microsoft Word. This tutorial will teach you how to create and print labels for a single item or address in Microsoft Word. Click on "Mailings" tab and then select "Labels".

I would suggest to leverage a HashMap to solve this problem. simply put, HashMap is a key value pair that hashes the keys and has a search complexity of O(1).

Iterate the list of words only once and keep on storing the encountered word in a HashMap. when you encounter a word, check if it already exists in the HashMap. If it does not exist, add it to the map with key as the word itself and value as 1. if The word alrady exists, Increase the value by 1.

After completing the iteration, the HashMap would contain key value pairs of unique words vs their count !!

just in case if you are not aware of maps in java -

Given a string, find its first non-repeating character, Given a string, find the first non-repeating character in it. Else break the loop and print the current character as the answer. Method 2: HashMap and single string traversal. We can augment the count array by storing not just counts but also the index of the first time you encountered the character e.g. (3� Printing elements of an array without repeating an element. Ask Question Asked 6 years, 11 months ago. I want someone help me to print each element (String) one

You could do this :

public void printWordOccurence(String filePath) throws FileNotFoundException {
        if (filePath.isEmpty())

        File file = new File(filePath);
        Scanner input = new Scanner(file);
        HashMap<String, Integer> wordOccurence = new HashMap<>();

        while (input.hasNext()) {
            wordOccurence.merge(, 1, Integer::sum);

        for (String word : wordOccurence.keySet()) {
            System.out.println(word + " appears " + wordOccurence.get(word) + " times");

Program to Find the Duplicate Words in a String, In this program, we need to find out the duplicate words present in the string words using built-in function; words = string.split(" ");; print("Duplicate words in a� I know a set of six 4-letter words without repeating a letter (i.e. using 24 different letters). Here is an example: gasp, verb, jinx, flow, duck, myth. Are there sets of words with no repeated letters and having. four 6-letter words? five 5-letter words? three 8-letter words? Bonus: Find a set of words (non-repeating letters) of sizes 1,2,3,4,5,6.

Repeat a string, If there is a simpler/more efficient way to repeat a single “character” (i.e. creating a string std::string repeat( const std::string &word, int times ) { Here are some ideas to help you get the labels you want the first time, without a lot of fuss and bother. Note: This article is also available as a PDF download . 1: Print a single label

Programming with Python: Repeating Actions with Loops, We want to create plots for all of our data sets with a single statement. To do that, we'll word[0]. One way to print each character is to use four print statements: ' Char' is not a keyword in Python that pulls the characters from words or strings. Kangaroo word or marsupial This refers to a word carrying another word within it (without transposing any letters). Example: encourage contains courage , cog, cur , urge, core, cure, nag, rag, age

Python - Filter Duplicate Words, Then we apply set() function which creates an unordered collection of unique elements. The result has unique words which are not ordered. import nltk word_data� The printer driver tells Word which way the envelope should be loaded into the printer, and this information is displayed in the Printing Options tab of the Envelope Options dialog box. 1. The feed method determines the position of the envelope (right, middle, left) and whether the long or short edge is being fed into the printer.

  • whats the type of words ?
  • Possible duplicate of How to count the number of occurrences of each word?
  • instead of writing if-else in your while loop you could use merge method of Map interface introduced in Java 8.
  • now you can use Integer::sum method reference instead of (val1, val2) -> val1 + val2 :)