347 Top K Frequent Elements

There are two approaches that simultaneously come to my mind, but both of them have the same time complexity. The fundamental differences between those approaches is that one uses a heap and the other uses sorting. (psst I like that heap one more ;p )

Heap approach: We use a heap to maintain top K elements. From looking at some of my other solutions you would come to know that heap is a pretty good idea for problems that are in the format “Top K elements …..”. Well, we do the similar thing here. First build a frequency table of the numbers in the list (this is linear time), then insert them into the heap and maintain K elements. While inserting if length of heap goes beyond K, use heappop() .

Note: If using Python, insert a tuple in the heap while doing heappush(). Python recognises the first value of the tuple as a comparator key.

Sorting approach: Similar to the heap approach, build a frequency table first. Then sort the dictionary with the dictionary value as the key. Depending if you reverse the sorting order or not, return the first K or last K keys.

Code

Python3 (heap solution)

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
 
        freq = defaultdict(int)
 
        for n in nums:
            if n in freq.keys():
                freq[n] += 1
            else: freq[n] = 1
 
        elements = []
        heapify(elements)                           # O(N)
 
        for n in freq.keys():                       # O(N)
            key = n; val = freq[n]
 
            heappush(elements, (freq[key], key))    # O(log N)
 
            if len(elements) > k:
                heappop(elements)                   # O(log N)
 
        return [t[1] for t in elements]             # O (N)

Python3 (sorting solution)

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        freq = defaultdict(int)
 
        for x in nums:
            if x in freq.keys():
                freq[x] += 1
            else:
                freq[x] = 1
 
        freq = dict(sorted(freq.items(), key=lambda x: x[1], reverse=True)) # O (n log n)
 
        tmp = list(freq.keys())
 
				# sort the dictionary based on the values
        ret = []
        for i in range(k):
            ret.append(tmp[i])
 
 
        return ret

Big O Analysis

• Runtime The runtime complexity here is $O(nlog(n))$ for both solutions.
• Memory The memory usage is $O(N)$ for both solutions since we use a dictionary and a heap.

— A

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