Calculate the frequency of each word in the given string

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 frequency 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 print frequency 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 print the 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

Javascript

//Javascript code for above

// Function to print frequency of each word
function printFrequency(str) {
    // Create a new Map to store the frequency counts
    let M = new Map();

    // String for storing the words
    let word = "";

    // Loop through each character in the string
    for (let i = 0; i < str.length; i++) {

      // Check if current character is a blank space
      if (str[i] === " ") {

        // If the current word is not found then insert it
        // into the Map with frequency 1
        if (!M.has(word)) {
          M.set(word, 1);
          word = "";
        }
        // If the current word is already in the Map, update its frequency count
        else {
          M.set(word, M.get(word) + 1);
          word = "";
        }
      }
      // If the current character is not a blank space,
      // add it to the current word
      else {
        word += str[i];
      }
    }

    // Check if the last word in the string is already in the Map,
    // and update its frequency count if it is
    if (!M.has(word)) {
      M.set(word, 1);
    } else {
      M.set(word, M.get(word) + 1);
    }

    // sorting map key in increasing order
    M = new Map([...M.entries()].sort());

    // Loop through each key-value pair in the Map and
    // print the frequency count of each word
      for (let [key, value] of M) {
        console.log(`${key} - ${value}`);
      }
}

// Driver Code
let str = "Geeks For Geeks";
printFrequency(str);

Output:

For - 1
Geeks - 2

Time Complexity: O(L * log (M)) , Where L is the length of the string and M is the number of words present in the string.

Auxiliary Space: O(M), where M is the number of words present in the string.