# Calculate the frequency of each word in the given string

Last Updated : 02 Mar, 2023

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 ` `using` `namespace` `std;`   `// Function to print frequency of each word` `void` `printFrequency(string str)` `{` `    ``map 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 mp=``new` `TreeMap<>();`   `        ``// Splitting to find the word` `        ``String arr[]=str.split(``" "``);`   `        ``// Loop to iterate over the words` `        ``for``(``int` `i=``0``;i 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 mp = ``new` `SortedDictionary();`   `    ``// 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 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.