Count words that appear exactly two times in an array of words

Given an array of n words. Some words are repeated twice, we need count such words.
Examples:

Input : s[] = {"hate", "love", "peace", "love", 
               "peace", "hate", "love", "peace", 
               "love", "peace"};
Output : 1
There is only one word "hate" that appears twice

Input : s[] = {"Om", "Om", "Shankar", "Tripathi", 
                "Tom", "Jerry", "Jerry"};
Output : 2
There are two words "Om" and "Jerry" that appear
twice.

Source : Amazon Interview

  1. Traverse the given array, store counts of words in a hash table
  2. Traverse hash table and count all words with count 2.

Below is the implementation.

C++

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// C++ program to count all words with count
// exavtly 2.
#include <bits/stdc++.h>
using namespace std;
  
// Returns count of words with frequency
// exactly 2.
int countWords(string str[], int n)
{
    unordered_map<string, int> m;
    for (int i = 0; i < n; i++)
        m[str[i]] += 1;
  
    int res = 0;
    for (auto it = m.begin(); it != m.end(); it++)
        if ((it->second == 2))
            res++;
  
    return res;
}
  
// Driver code
int main()
{
    string s[] = { "hate", "love", "peace", "love",
                   "peace", "hate", "love", "peace",
                   "love", "peace" };
    int n = sizeof(s) / sizeof(s[0]);
    cout << countWords(s, n);
    return 0;
}

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Java














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// Java program to count all words with count 


// exactly 2. 


import java.util.HashMap; 


import java.util.Map; 


public class GFG { 


       


    // Returns count of words with frequency 


    // exactly 2. 


    static int countWords(String str[], int n) 


    { 


        // map to store count of each word 


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


          


        for (int i = 0; i < n; i++){ 


            if(m.containsKey(str[i])){ 


                int get = m.get(str[i]); 


                m.put(str[i], get + 1); 


            } 


            else{ 


                m.put(str[i], 1); 


            } 


        } 


              


        int res = 0; 


        for (Map.Entry<String, Integer> it: m.entrySet()){ 


            if(it.getValue() == 2) 


                res++; 


        } 


                  


        return res; 


    } 


       


    // Driver code 


    public static void main(String args[]) 


    { 


        String s[] = { "hate", "love", "peace", "love", 


                       "peace", "hate", "love", "peace", 


                       "love", "peace" }; 


        int n = s.length; 


        System.out.println( countWords(s, n)); 


    } 


} 


// This code is contributed by Sumit Ghosh 



















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Python3














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# Python program to count all 


# words with count 


# exactly 2. 


   


# Returns count of words with frequency 


# exactly 2. 


def countWords(stri, n): 


    m = dict() 


    for i in range(n): 


        m[stri[i]] = m.get(stri[i],0) + 1


   


    res = 0


    for i in m.values(): 


        if i == 2: 


            res += 1


   


    return res 


   


# Driver code 


s = [ "hate", "love", "peace", "love", 


      "peace", "hate", "love", "peace", 


                "love", "peace" ] 


n = len(s) 


print(countWords(s, n)) 


  


# This code is contributed 


# by Shubham Rana 



















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C#














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// C# program to count all words with count 


// exactly 2. 


using System; 


using System.Collections.Generic; 


  


class GFG  


{ 


      


    // Returns count of words with frequency 


    // exactly 2. 


    static int countWords(String []str, int n) 


    { 


        // map to store count of each word 


        Dictionary<String,int> m = new Dictionary<String,int>(); 


          


        for (int i = 0; i < n; i++) 


        { 


            if(m.ContainsKey(str[i])) 


            { 


                int get = m[str[i]]; 


                m.Remove(str[i]); 


                m.Add(str[i], get + 1); 


            } 


            else


            { 


                m.Add(str[i], 1); 


            } 


        } 


              


        int res = 0; 


        foreach(KeyValuePair<String, int> it in m) 


        { 


            if(it.Value == 2) 


                res++; 


        } 


                  


        return res; 


    } 


      


    // Driver code 


    public static void Main(String []args) 


    { 


        String []a = { "hate", "love", "peace", "love", 


                    "peace", "hate", "love", "peace", 


                    "love", "peace" }; 


        int n = a.Length; 


        Console.WriteLine( countWords(a, n)); 


    } 


} 


  


// This code is contributed by Rajput-Ji 



















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Output:


1



This article is contributed by Saumya Tiwari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.


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