Finding repeated words on a string and counting the repetitions

Question

I need to find repeated words on a string, and then count how many times they were repeated. So basically, if the input string is this:

String s = \"House, House         


        

Answer 1

You've got the hard work done. Now you can just use a Map to count the occurrences:

Map<String, Integer> occurrences = new HashMap<String, Integer>();

for ( String word : splitWords ) {
   Integer oldCount = occurrences.get(word);
   if ( oldCount == null ) {
      oldCount = 0;
   }
   occurrences.put(word, oldCount + 1);
}

Using map.get(word) will tell you many times a word occurred. You can construct a new list by iterating through map.keySet():

for ( String word : occurrences.keySet() ) {
  //do something with word
}

Note that the order of what you get out of keySet is arbitrary. If you need the words to be sorted by when they first appear in your input String, you should use a LinkedHashMap instead.

Answer 2

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;

public class DuplicateWord {

    public static void main(String[] args) {
        String para = "this is what it is this is what it can be";
        List < String > paraList = new ArrayList < String > ();
        paraList = Arrays.asList(para.split(" "));
        System.out.println(paraList);
        int size = paraList.size();

        int i = 0;
        Map < String, Integer > duplicatCountMap = new HashMap < String, Integer > ();
        for (int j = 0; size > j; j++) {
            int count = 0;
            for (i = 0; size > i; i++) {
                if (paraList.get(j).equals(paraList.get(i))) {
                    count++;
                    duplicatCountMap.put(paraList.get(j), count);
                }

            }

        }
        System.out.println(duplicatCountMap);
        List < Integer > myCountList = new ArrayList < > ();
        Set < String > myValueSet = new HashSet < > ();
        for (Map.Entry < String, Integer > entry: duplicatCountMap.entrySet()) {
            myCountList.add(entry.getValue());
            myValueSet.add(entry.getKey());
        }
        System.out.println(myCountList);
        System.out.println(myValueSet);
    }

}

Input: this is what it is this is what it can be

Output:

[this, is, what, it, is, this, is, what, it, can, be]

{can=1, what=2, be=1, this=2, is=3, it=2}

[1, 2, 1, 2, 3, 2]

[can, what, be, this, is, it]

Answer 3

If you pass a String argument it will count the repetition of each word

/**
 * @param string
 * @return map which contain the word and value as the no of repatation
 */
public Map findDuplicateString(String str) {
    String[] stringArrays = str.split(" ");
    Map<String, Integer> map = new HashMap<String, Integer>();
    Set<String> words = new HashSet<String>(Arrays.asList(stringArrays));
    int count = 0;
    for (String word : words) {
        for (String temp : stringArrays) {
            if (word.equals(temp)) {
                ++count;
            }
        }
        map.put(word, count);
        count = 0;
    }

    return map;

}

output:

 Word1=2, word2=4, word2=1,. . .

Answer 4

public class Counter {

private static final int COMMA_AND_SPACE_PLACE = 2;

private String mTextToCount;
private ArrayList<String> mSeparateWordsList;

public Counter(String mTextToCount) {
    this.mTextToCount = mTextToCount;

    mSeparateWordsList = cutStringIntoSeparateWords(mTextToCount);
}

private ArrayList<String> cutStringIntoSeparateWords(String text)
{
    ArrayList<String> returnedArrayList = new ArrayList<>();


    if(text.indexOf(',') == -1)
    {
        returnedArrayList.add(text);
        return returnedArrayList;
    }

    int position1 = 0;
    int position2 = 0;

    while(position2 < text.length())
    {
        char c = ',';
        if(text.toCharArray()[position2] == c)
        {
            String tmp = text.substring(position1, position2);
            position1 += tmp.length() + COMMA_AND_SPACE_PLACE;
            returnedArrayList.add(tmp);
        }
        position2++;
    }

    if(position1 < position2)
    {
        returnedArrayList.add(text.substring(position1, position2));
    }

    return returnedArrayList;
}

public int[] countWords()
{
    if(mSeparateWordsList == null) return null;


    HashMap<String, Integer> wordsMap = new HashMap<>();

    for(String s: mSeparateWordsList)
    {
        int cnt;

        if(wordsMap.containsKey(s))
        {
            cnt = wordsMap.get(s);
            cnt++;
        } else {
            cnt = 1;
        }
        wordsMap.put(s, cnt);
    }                
    return printCounterResults(wordsMap);
}

private int[] printCounterResults(HashMap<String, Integer> m)
{        
    int index = 0;
    int[] returnedIntArray = new int[m.size()];

    for(int i: m.values())
    {
        returnedIntArray[index] = i;
        index++;
    }

    return returnedIntArray;

}

}

Answer 5

You can use Prefix tree (trie) data structure to store words and keep track of count of words within Prefix Tree Node.

  #define  ALPHABET_SIZE 26
  // Structure of each node of prefix tree
  struct prefix_tree_node {
    prefix_tree_node() : count(0) {}
    int count;
    prefix_tree_node *child[ALPHABET_SIZE];
  };
  void insert_string_in_prefix_tree(string word)
  {
    prefix_tree_node *current = root;
    for(unsigned int i=0;i<word.size();++i){
      // Assuming it has only alphabetic lowercase characters
            // Note ::::: Change this check or convert into lower case
    const unsigned int letter = static_cast<int>(word[i] - 'a');

      // Invalid alphabetic character, then continue
      // Note :::: Change this condition depending on the scenario
      if(letter > 26)
        throw runtime_error("Invalid alphabetic character");

      if(current->child[letter] == NULL)
        current->child[letter] = new prefix_tree_node();

      current = current->child[letter];
    }
  current->count++;
  // Insert this string into Max Heap and sort them by counts
}

    // Data structure for storing in Heap will be something like this
    struct MaxHeapNode {
       int count;
       string word;
    };

After inserting all words, you have to print word and count by iterating Maxheap.

Answer 6

package string;

import java.util.HashMap;
import java.util.Map;
import java.util.Set;

public class DublicatewordinanArray {
public static void main(String[] args) {
String str = "This is Dileep Dileep Kumar Verma Verma";
DuplicateString(str);
    }
public static void DuplicateString(String str) {
String word[] = str.split(" ");
Map < String, Integer > map = new HashMap < String, Integer > ();
for (String w: word)
if (!map.containsKey(w)) {
map.put(w, 1);
    }
else {
map.put(w, map.get(w) + 1);
        }
Set < Map.Entry < String, Integer >> entrySet = map.entrySet();
 for (Map.Entry < String, Integer > entry: entrySet)
if (entry.getValue() > 1) {
 System.out.printf("%s : %d %n", entry.getKey(), entry.getValue());
}
 }
}