Count words that appear exactly two times in an array of words (original) (raw)

Last Updated : 23 Mar, 2023

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

Below is the implementation:

Try It Yourself

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

Time Complexity : O(N)
Auxiliary Space: O(N)

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:

C++ `

// C++ program for the above approach

#include #include #include using namespace std;

int countWords(vector s, int n) { // Calculating frequency using unordered_map unordered_map<string, int> freq; for (int i = 0; i < n; i++) { freq[s[i]]++; } int count = 0; // Traversing in freq dictionary for (auto& it : freq) { if (it.second == 2) { count++; } } return count; }

// Driver code int main() { vector s = { "hate", "love", "peace", "love", "peace", "hate", "love", "peace", "love", "peace" }; int n = s.size(); cout << countWords(s, n) << endl; return 0; }

// This code is contributed by prince

Java

import java.util.HashMap; import java.util.Map; import java.util.List; import java.util.ArrayList;

public class Main { public static int countWords(List s, int n) { // Calculating frequency using HashMap Map<String, Integer> freq = new HashMap<String, Integer>(); for (int i = 0; i < n; i++) { String key = s.get(i); if (freq.containsKey(key)) { freq.put(key, freq.get(key) + 1); } else { freq.put(key, 1); } } int count = 0; // Traversing in freq dictionary for (Map.Entry<String, Integer> entry : freq.entrySet()) { if (entry.getValue() == 2) { count++; } } return count; }

// Driver code
public static void main(String[] args) {
    List<String> s = new ArrayList<String>();
    s.add("hate");
    s.add("love");
    s.add("peace");
    s.add("love");
    s.add("peace");
    s.add("hate");
    s.add("love");
    s.add("peace");
    s.add("love");
    s.add("peace");
    int n = s.size();
    System.out.println(countWords(s, n));
}

}

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

C#

using System; using System.Collections.Generic; using System.Linq;

class Program { static int CountWords(List s, int n) { // Calculating frequency using Dictionary Dictionary<string, int> freq = new Dictionary<string, int>(); foreach(string word in s) { if (freq.ContainsKey(word)) freq[word]++; else freq.Add(word, 1); } int count = 0; // Traversing in freq dictionary foreach(KeyValuePair<string, int> pair in freq) { if (pair.Value == 2) count++; } return count; }

static void Main(string[] args)
{
    List<string> s = new List<string>{
        "hate", "love", "peace", "love", "peace",
        "hate", "love", "peace", "love", "peace"
    };
    int n = s.Count();
    Console.WriteLine(CountWords(s, n));
}

}

JavaScript

// Returns count of words with frequency exactly 2. function countWords(strList) { // Calculating frequency using Counter const m = {}; for (let i = 0; i < strList.length; i++) { const word = strList[i]; m[word] = m[word] ? m[word] + 1 : 1; }

let count = 0; // Traversing in freq dictionary for (const i in m) { if (m[i] === 2) { count += 1; } }

return count; }

// Driver code const s = ["hate", "love", "peace", "love", "peace", "hate", "love", "peace", "love", "peace"]; console.log(countWords(s));

// This code is contributed by adityashatmfh

Time Complexity : O(n)
Auxiliary Space: O(n)