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
- 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; } |
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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 |
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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 |
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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 |
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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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