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 code to find most frequent word in // an array of strings #include <bits/stdc++.h> 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[0]); 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<String, Integer> hs = new HashMap<String, Integer>(); // 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<Map.Entry<String, Integer> > set = hs.entrySet(); String key = ""; int value = 0; for (Map.Entry<String, Integer> 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<String, int> hs = new Dictionary<String, int>(); // 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<String, int> 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 |
Output:
geeks
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