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K most occurring strings
• Last Updated : 13 Jul, 2020

Given an array arr[] of N strings and an integer K, the task is to print K strings which occurred the most number of times in the arr[]. If two or more strings have the same frequency then print the lexicographically smallest string.
Note: The value of K is always less than or equal to the number of distinct elements in the array.

Examples:

Input: str[] = {“geeks”, “geeksforgeeks”, “geeks”, “article”}, K = 1
Output: “geeks”
Explanation:
“geeks” –> 2
“geeksforgeeks” –> 1
“article” –> 1
Hence, the most occurring string is “geeks”

Input: str[] = {“car”, “bus”, “car”, “bike”, “car”, “bus”, “bike”, “cycle”}, K = 2
Output : “car”, “bus”
Explanation:
“car” –> 3
“bus” –> 2
“bike” –> 2
“cycle” –> 1
string “car” has highest frequency, string “bus” and “bike” both have second highest frequency, but string “bus” is lexicographically small because it has shorter length.

Approach:

1. Count the frequency of each string in the array and store it in a HashMap where the string is the key and frequency as the value.
2. Now, sort these keys according to their frequencies in ascending order, this is done to keep keys with the least frequencies at the top.
3. The strings with equal frequencies are prioritized alphabetically, i.e., the string which is alphabetically greater has a higher priority.
4. Delete the top N – K key-value pairs from the HashMap. By doing this the container is left with K keys with the highest frequencies, in reverse order.
5. Print the strings stored in the HashMap.

Below is the implementation of the above approach:

## Java

 `// Java program for the above approach``import` `java.util.*;`` ` `class` `FrequentWords {`` ` `    ``// Function that returns list of K``    ``// most frequent strings``    ``public` `static` `ArrayList``    ``frequentWords(ArrayList arr, ``int` `K)``    ``{`` ` `        ``// Hash map to store the frequency``        ``// of each string``        ``HashMap Freq``            ``= ``new` `HashMap<>();`` ` `        ``// Set the default frequency of``        ``// each string 0``        ``for` `(String word : arr) {``            ``Freq.put(word,``                     ``Freq.getOrDefault(word, ``0``)``                         ``+ ``1``);``        ``}`` ` `        ``// Using a priority queue to store``        ``// the strings in accordance of the``        ``// frequency and alphabetical order``        ``// (if frequency is equal)`` ` `        ``// Lambda expression is used set the``        ``// priority, if frequencies are not``        ``// equal than the string with greater``        ``// frequency is returned else the``        ``// string which is lexically smaller``        ``// is returned``        ``PriorityQueue Order``            ``= ``new` `PriorityQueue<>(``                ``(a, b)``                    ``-> (Freq.get(a) != Freq.get(b))``                           ``? Freq.get(a) - Freq.get(b)``                           ``: b.compareTo(a));`` ` `        ``// Traverse the HashMap``        ``for` `(String word : Freq.keySet()) {``            ``Order.offer(word);`` ` `            ``// Remove top (N - K) elements``            ``if` `(Order.size() > K) {``                ``Order.poll();``            ``}``        ``}`` ` `        ``// Order queue has K most frequent``        ``// strings in a reverse order``        ``ArrayList ans``            ``= ``new` `ArrayList<>();`` ` `        ``while` `(!Order.isEmpty()) {``            ``ans.add(Order.poll());``        ``}`` ` `        ``// Reverse the ArrayList so as``        ``// to get in the desired order``        ``Collections.reverse(ans);`` ` `        ``return` `ans;``    ``}`` ` `    ``// Driver Code``    ``public` `static` `void``        ``main(String[] args)``    ``{``        ``int` `N = ``3``, K = ``2``;`` ` `        ``// Given array``        ``ArrayList arr``            ``= ``new` `ArrayList();``        ``arr.add(``"car"``);``        ``arr.add(``"bus"``);``        ``arr.add(``"car"``);`` ` `        ``// Function Call``        ``ArrayList ans``            ``= frequentWords(arr, K);`` ` `        ``// Print the K most occurring strings``        ``for` `(String word : ans) {``            ``System.out.println(word + ``" "``);``        ``}``    ``}``}`
Output:
```car
bus
```

Time Complexity: O(N*log2N)
Auxiliary Space: O(N)

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