计算单词数组中恰好出现两次的单词

给定一个包含 n 个单词的数组。有些单词重复了两次，我们需要对这些单词进行计数。

例子：

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.

资料来源：亚马逊采访

下面是实现：

C++

// C++ program to count all words with count
// exactly 2.
#include 
using namespace std;
 
// Returns count of words with frequency
// exactly 2.
int countWords(string str[], int n)
{
    unordered_map 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 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 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 m = new Dictionary();
         
        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 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

 
// 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));
 

输出

1

方法2：使用内置Python函数：

• 使用Counter函数计算每个单词的频率
• 在频率字典中遍历
• 检查哪个单词的频率为 2。如果是，增加计数
• 打印计数

下面是实现：

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

输出

1