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Calculate the frequency of each word in the given string
  • Difficulty Level : Easy
  • Last Updated : 20 Oct, 2020

Given a string str, the task is to find the frequency of each word in a string.

Examples: 

Input: str = “Geeks For Geeks” 
Output: 
For 1 
Geeks 2 
Explanation: 
For occurs 1 time and Geeks occurs 2 times in the given string str.

Input: str = “learning to code is learning to create and innovate” 
Output: 
and 1 
code 1 
create 1 
innovate 1 
is 1 
learning 2 
to 2 
Explanation: 
The words and, code, create, innovate, is occurs 1 time; and learning, to occurs 2 times in the given string str. 

Approach: To solve the problem mentioned above we have to follow the steps given below: 

  • Use a Map data structure to store the occurrence of each word in the string.
  • Traverse the entire string and check whether the current word is present in map or not. If it is present, then update the frequency of the current word else insert the word with frequency 1.
  • Traverse in the map and print the frequency of each word.

Below is the implementation of the above approach: 

C++




// C++ program to calculate the frequency
// of each word in the given string
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to print frquency of each word
void printFrequency(string str)
{
    map<string, int> M;
 
    // String for storing the words
    string word = "";
 
    for (int i = 0; i < str.size(); i++) {
 
        // Check if current character
        // is blank space then it
        // means we have got one word
        if (str[i] == ' ') {
 
            // If the current word
            // is not found then insert
            // current word with frequency 1
            if (M.find(word) == M.end()) {
                M.insert(make_pair(word, 1));
                word = "";
            }
 
            // update the frequency
            else {
                M[word]++;
                word = "";
            }
        }
 
        else
            word += str[i];
    }
 
    // Storing the last word of the string
    if (M.find(word) == M.end())
        M.insert(make_pair(word, 1));
 
    // Update the frequency
    else
        M[word]++;
 
    // Traverse the map
    // to print the  frequency
    for (auto& it : M) {
        cout << it.first << " - "
             << it.second
             << endl;
    }
}
 
// Driver Code
int main()
{
    string str = "Geeks For Geeks";
 
    printFrequency(str);
    return 0;
}

Java




// Java implementation of the above
// approach
 
import java.util.Map;
import java.util.TreeMap;
public class Frequency_Of_String_Words {
    
    // Function to count frequency of
    // words in the given string
    static void count_freq(String str)
    {
        Map<String,Integer> mp=new TreeMap<>();
 
        // Splitting to find the word
        String arr[]=str.split(" ");
 
        // Loop to iterate over the words
        for(int i=0;i<arr.length;i++)
        {
            // Condition to check if the
            // array element is present
            // the hash-map
            if(mp.containsKey(arr[i]))
            {
                mp.put(arr[i], mp.get(arr[i])+1);
            }
            else
            {
                mp.put(arr[i],1);
            }
        }
        
        // Loop to iterate over the
        // elements of the map
        for(Map.Entry<String,Integer> entry:
                    mp.entrySet())
        {
            System.out.println(entry.getKey()+
                    " - "+entry.getValue());
        }
    }
 
    // Driver Code
    public static void main(String[] args) {
        String str = "Geeks For Geeks";
 
        // Function Call
        count_freq(str);
    }
}

Python3




# Python3 program to calculate the frequency
# of each word in the given string
 
# Function to prfrquency of each word
def printFrequency(strr):
    M = {}
     
    # string for storing the words
    word = ""
     
    for i in range(len(strr)):
         
        # Check if current character
        # is blank space then it
        # means we have got one word
        if (strr[i] == ' '):
             
            # If the current word    
            # is not found then insert
            # current word with frequency 1
            if (word not in M):
                M[word] = 1
                word = ""
             
            # update the frequency
            else:
                M[word] += 1
                word = ""
         
        else:
            word += strr[i]
     
    # Storing the last word of the string
    if (word not in M):
        M[word] = 1
     
    # Update the frequency
    else:
        M[word] += 1
         
    # Traverse the map
    # to prthe frequency
    for it in M:
        print(it, "-", M[it])
     
# Driver Code
strr = "Geeks For Geeks"
printFrequency(strr)
 
# This code is contributed by shubhamsingh10

C#




// C# implementation of the above
// approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to count frequency of
// words in the given string
static void count_freq(String str)
{
    SortedDictionary<String,
                     int> mp = new SortedDictionary<String,
                                                    int>();
 
    // Splitting to find the word
    String []arr = str.Split(' ');
 
    // Loop to iterate over the words
    for(int i = 0; i < arr.Length; i++)
    {
         
        // Condition to check if the
        // array element is present
        // the hash-map
        if (mp.ContainsKey(arr[i]))
        {
            mp[arr[i]] = mp[arr[i]] + 1;
        }
        else
        {
            mp.Add(arr[i], 1);
        }
    }
     
    // Loop to iterate over the
    // elements of the map
    foreach(KeyValuePair<String, int> entry in mp)
    {
        Console.WriteLine(entry.Key + " - " +
                          entry.Value);
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    String str = "Geeks For Geeks";
 
    // Function call
    count_freq(str);
}
}
 
// This code is contributed by Rajput-Ji
Output: 
For - 1
Geeks - 2




 




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