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LeetCode 347. Top K Frequent Elements

Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order.

Example 1:

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

Example 2:

Input: nums = [1], k = 1
Output: [1]

Option 1 - Map to count + Sort to get top k - O(nlgn)

  1. We need to count what is the freq or how many times an item appeared.
  2. The obvious thing is for each item put it in a hashmap where
    1. Key is the item
    2. Value is the number of times it appeared
  3. Now the only thing that is left is to sort this hashtable.
  4. sorting is O(nlgn), placing everyting and counting is O(n) so this makes it overall O(nlgn)

Option 2 - Improvement over option 1 - Instead of sorting use a Heap

  1. This is a common thing when you need to sort but the core of the problem is not sorting but as in this example only get top k then we could utilize a heap.
    1. We still build a map as in option 1 which takes O(n) this is the counting map.
    2. Building a heap heapify takes O(n)
    3. Remove each item from the heap is O(lg(n)) until we removed k items - as each pop from a heap is lg(n)
    4. Overall this makes it klg(n) + n which can be better than nlgn if k < n

Option 3 - O(n) - Use buckets for finding the top k

  1. Start as usual build a hashtable for counting how many times each item appears this would be O(n)
  2. Now create an array [] of size of number of items we have here index i would be a list of the number occuring i times
    1. [].get[7] --> [the items that appeared 7 times]
  3. And as you see to get the top k we just need to traverse the top k elements of this list
  4. Overall this means O(n) for creating the hashtable and O(n) for traversing this buckets list.

Here is the actual implementation of the buckets solution:

class Solution(object):
    def topKFrequent(self, nums, k):
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        # step 1 - put it in the map 
        #         [1,1,3,4] --> {
        #                             {1 --> 2}
        #                           , {3 --> 4}
        #                           , {4 --> 1}
        #                        }
        # step 2 prepare buckets [0, ... nums.length ] -> a number can appear at most nums.
        #        Bonus - buckets already sorted by index last bucket is one with highest number if count.
        # Step 3 Read buckets in reverse order.
        appearances = {} # map {number -> # times appears }
        for x in nums:
            appearances[x] = 1 + appearances.get(x, 0)
        buckets = [[] for i in range(len(nums) + 1)] # [] - An array of arrays, we are going to keep all the x that are in each bucket appeared number of times as index of bucket.
        for (x, num_appearances) in appearances.items():
        res = []    
        for i in range(len(buckets) - 1, 0, -1):
            for x in buckets[i]:
                if (len(res) == k):
                    return res;


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