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

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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

Below is the implementation.

C++

// 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; 
} 

Java

// 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 

Python3

# 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 

Output:

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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