K 个出现次数最多的字符串

给定一个由N 个字符串的数组arr[]和一个整数K ，任务是打印在arr[] 中出现次数最多的K 个字符串。如果两个或多个字符串具有相同的频率，则打印字典序最小的字符串。
注意： K的值总是小于或等于数组中不同元素的数量。

例子：

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.

方法：

1. 计算数组中每个字符串的频率并将其存储在 HashMap 中，其中字符串为键，频率为值。
2. 现在，按照频率升序对这些键进行排序，这样做是为了将频率最低的键保持在顶部。
3. 具有相同频率的字符串按字母顺序排列优先级，即字母顺序较大的字符串具有更高的优先级。
4. 从 HashMap 中删除前N – K 个键值对。通过这样做，容器以相反的顺序留下具有最高频率的K 个键。
5. 打印存储在 HashMap 中的字符串。

下面是上述方法的实现：

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 + " ");
        }
    }
}

输出：

car 
bus

时间复杂度： O(N*log ₂ N)
辅助空间： O(N)

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