Related Articles

# Most frequent word in an array of strings

• Difficulty Level : Hard
• Last Updated : 15 Jun, 2021

Given an array of words find the most occurring word in it
Examples:

```Input : arr[] = {"geeks", "for", "geeks", "a",
"portal", "to", "learn", "can",
"be", "computer", "science",
"zoom", "yup", "fire", "in",
"be", "data", "geeks"}
Output : Geeks
"geeks" is the most frequent word as it
occurs 3 times```

A simple solution is to run two loops and count occurrences of every word. Time complexity of this solution is O(n * n * MAX_WORD_LEN).
An efficient solution is to use Trie data structure. The idea is simple first we will insert in trie. In trie, we keep counts of words ending at a node. We do preorder traversal and compare count present at each node and find the maximum occurring word

## CPP

 `// CPP code to find most frequent word in``// an array of strings``#include ``using` `namespace` `std;` `/*structing the trie*/``struct` `Trie {``    ``string key;``    ``int` `cnt;``    ``unordered_map<``char``, Trie*> map;``};` `/* Function to return a new Trie node */``Trie* getNewTrieNode()``{``    ``Trie* node = ``new` `Trie;``    ``node->cnt = 0;``    ``return` `node;``}` `/* function to insert a string */``void` `insert(Trie*& root, string& str)``{``    ``// start from root node``    ``Trie* temp = root;` `    ``for` `(``int` `i = 0; i < str.length(); i++) {` `        ``char` `x = str[i];` `        ``/*a new node if path doesn't exists*/``        ``if` `(temp->map.find(x) == temp->map.end())``            ``temp->map[x] = getNewTrieNode();` `        ``// go to next node``        ``temp = temp->map[x];``    ``}` `    ``// store key and its count in leaf nodes``    ``temp->key = str;``    ``temp->cnt += 1;``}` `/* function for preorder traversal */``bool` `preorder(Trie* temp, ``int``& maxcnt, string& key)``{``    ``if` `(temp == NULL)``        ``return` `false``;` `    ``for` `(``auto` `it : temp->map) {` `        ``/*leaf node will have non-zero count*/``        ``if` `(maxcnt < it.second->cnt) {``            ``key = it.second->key;``            ``maxcnt = it.second->cnt;``        ``}` `        ``// recurse for current node children``        ``preorder(it.second, maxcnt, key);``    ``}``}` `void` `mostFrequentWord(string arr[], ``int` `n)``{``    ``// Insert all words in a Trie``    ``Trie* root = getNewTrieNode();``    ``for` `(``int` `i = 0; i < n; i++)``        ``insert(root, arr[i]);` `    ``// Do preorder traversal to find the``    ``// most frequent word``    ``string key;``    ``int` `cnt = 0;``    ``preorder(root, cnt, key);` `    ``cout << ``"The word that occurs most is : "``         ``<< key << endl;``    ``cout << ``"No of times: "` `<< cnt << endl;``}` `// Driver code``int` `main()``{``    ``// given set of keys``    ``string arr[] = { ``"geeks"``, ``"for"``, ``"geeks"``, ``"a"``,``                     ``"portal"``, ``"to"``, ``"learn"``, ``"can"``, ``"be"``,``                     ``"computer"``, ``"science"``, ``"zoom"``, ``"yup"``,``                     ``"fire"``, ``"in"``, ``"be"``, ``"data"``, ``"geeks"` `};``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``mostFrequentWord(arr, n);` `    ``return` `0;``}`

Output:

```The word that occurs most is : geeks
No of times: 3```

Time Complexity: O(n * MAX_WORD_LEN)
Another efficient solution is to use hashing. Please refer Find winner of an election where votes are represented as candidate names for details.
More simple solution is to use HashMap.
Approach:
Using HashMap, one can keep track of word and it’s frequency. Next step includes iterate over it and find out the word with maximum frequency.
Below is the implementation of the above approach.

## Java

 `// Java implementation``import` `java.util.*;` `class` `GKG {` `    ``// Function returns word with highest frequency``    ``static` `String findWord(String[] arr)``    ``{` `        ``// Create HashMap to store word and it's frequency``        ``HashMap hs = ``new` `HashMap();` `        ``// Iterate through array of words``        ``for` `(``int` `i = ``0``; i < arr.length; i++) {``            ``// If word already exist in HashMap then increase it's count by 1``            ``if` `(hs.containsKey(arr[i])) {``                ``hs.put(arr[i], hs.get(arr[i]) + ``1``);``            ``}``            ``// Otherwise add word to HashMap``            ``else` `{``                ``hs.put(arr[i], ``1``);``            ``}``        ``}` `        ``// Create set to iterate over HashMap``        ``Set > set = hs.entrySet();``        ``String key = ``""``;``        ``int` `value = ``0``;` `        ``for` `(Map.Entry me : set) {``            ``// Check for word having highest frequency``            ``if` `(me.getValue() > value) {``                ``value = me.getValue();``                ``key = me.getKey();``            ``}``        ``}` `        ``// Return word having highest frequency``        ``return` `key;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String arr[] = { ``"geeks"``, ``"for"``, ``"geeks"``, ``"a"``,``                         ``"portal"``, ``"to"``, ``"learn"``, ``"can"``, ``"be"``,``                         ``"computer"``, ``"science"``, ``"zoom"``, ``"yup"``,``                         ``"fire"``, ``"in"``, ``"be"``, ``"data"``, ``"geeks"` `};``        ``String sol = findWord(arr);` `        ``// Print word having highest frequency``        ``System.out.println(sol);``    ``}``}` `// This code is contributed by Divyank Sheth`

## C#

 `// C# implementation``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `    ``// Function returns word with highest frequency``    ``static` `String findWord(String[] arr)``    ``{` `        ``// Create Dictionary to store word``        ``// and it's frequency``        ``Dictionary hs =``            ``new` `Dictionary();` `        ``// Iterate through array of words``        ``for` `(``int` `i = 0; i < arr.Length; i++)``        ``{``            ``// If word already exist in Dictionary``            ``// then increase it's count by 1``            ``if` `(hs.ContainsKey(arr[i]))``            ``{``                ``hs[arr[i]] = hs[arr[i]] + 1;``            ``}``            ` `            ``// Otherwise add word to Dictionary``            ``else``            ``{``                ``hs.Add(arr[i], 1);``            ``}``        ``}` `        ``// Create set to iterate over Dictionary``        ``String key = ``""``;``        ``int` `value = 0;` `        ``foreach``(KeyValuePair me ``in` `hs)``        ``{``            ``// Check for word having highest frequency``            ``if` `(me.Value > value)``            ``{``                ``value = me.Value;``                ``key = me.Key;``            ``}``        ``}` `        ``// Return word having highest frequency``        ``return` `key;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``String []arr = { ``"geeks"``, ``"for"``, ``"geeks"``, ``"a"``,``                        ``"portal"``, ``"to"``, ``"learn"``, ``"can"``, ``"be"``,``                        ``"computer"``, ``"science"``, ``"zoom"``, ``"yup"``,``                        ``"fire"``, ``"in"``, ``"be"``, ``"data"``, ``"geeks"` `};``        ``String sol = findWord(arr);` `        ``// Print word having highest frequency``        ``Console.WriteLine(sol);``    ``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output:

`geeks`

This article is contributed by Pranav. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.