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

Difficulty Level : Easy
Last Updated : 07 Aug, 2021

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

Examples:

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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

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
// exactly 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

C#

// 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

Javascript

<script>
 
// Javascript program to count all words with count
// exactly 2.
 
// Returns count of words with frequency
// exactly 2.
function countWords(str, n)
{
    var m = new Map();
    for (var i = 0; i < n; i++)
    {
        if(m.has(str[i]))
            m.set(str[i], m.get(str[i])+1)
        else
            m.set(str[i], 1)
    }
 
    var res = 0;
 
    m.forEach((value, key) => {
         
        if ((value == 2))
            res++;
    });
    return res;
}
 
// Driver code
var s = ["hate", "love", "peace", "love",
               "peace", "hate", "love", "peace",
               "love", "peace" ];
var n = s.length;
document.write( countWords(s, n));
 
</script>

Output

Method 2: Using Built-in Python functions:

Count the frequencies of every word using the Counter function
Traverse in frequency dictionary
Check which word has frequency 2. If so, increase the count
Print Count

Below is the implementation:

Python

# importing Counter from collections
from collections import Counter
 
# Python program to count all words with count exactly 2.
# Returns count of words with frequency exactly 2.
def countWords(stri, n):
   
    # Calculating frequency using Counter
    m = Counter(stri)
 
    count = 0
    # Traversing in freq dictionary
    for i in m:
        if m[i] == 2:
            count += 1
 
    return count
 
# 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 vikkycirus

Output

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