Leetcode Problems

3 Steps Method Using HashMap, Priority Queue (Max Heap) and Lambda Expression in Java

1. Top K Frequent Elements

Given a non-empty array of integers, return the k most frequent elements.

Example 1:

Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]

Example 2:

Input: nums = [1], k = 1
Output: [1]

Note:

• You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
• Your algorithm’s time complexity must be better than O(n log n), where n is the array’s size.
• It’s guaranteed that the answer is unique, in other words, the set of the top k frequent elements is unique.
• You can return the answer in any order.

Algorithm:

fun topKFrequent(int[] nums, int k)

1. Store the frequency of each element in the HashMap -> O(N)
2. Sort the frequency using Priority Queue (Max Heap) -> O(log N)
3. Store the top K frequent element in an Array from the Max Heap and return -> O(K)

Complexity:

Time: O(N) + O(log N) + O(K) = O(N)

Space: O(N)

Code:

2. Top K Frequent Words

Given a non-empty list of words, return the k most frequent elements.

Your answer should be sorted by frequency from highest to lowest. If two words have the same frequency, then the word with the lower alphabetical order comes first.

Example 1:

Input: ["i", "love", "leetcode", "i", "love", "coding"], k = 2
Output: ["i", "love"]
Explanation: "i" and "love" are the two most frequent words.
    Note that "i" comes before "love" due to a lower alphabetical order.

Example 2:

Input: ["the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"], k = 4
Output: ["the", "is", "sunny", "day"]
Explanation: "the", "is", "sunny" and "day" are the four most frequent words,
    with the number of occurrence being 4, 3, 2 and 1 respectively.

Note:

1. You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
2. Input words contain only lowercase letters.

Follow up:

1. Try to solve it in O(n log k) time and O(n) extra space.

Algorithm:

fun topKFrequent(int[] nums, int k)

1. Store the frequency of each element in the HashMap -> O(N)
2. Sort the frequency using Priority Queue (Max Heap) -> O(log N)
3. Store the top K frequent element in an Array from the Max Heap and return -> O(K)

Complexity:

Time: O(N) + O(log N) + O(K) = O(N)

Space: O(N)

Code:

Happy Coding: )