Leetcode: Top K Frequent Elements

Top K Frequent Elements

Similar Problems:

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

For example,
Given [1,1,1,2,2,3] and k = 2, return [1,2].

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.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## Blog link: https://code.dennyzhang.com/top-k-frequent-elements
## Basic Ideas: priority queue: heapq
## Complexity:
import collections, heapq
class Solution(object):
    def topKFrequent(self, nums, k):
        :type nums: List[int]
        :type k: int
        :rtype: List[int]
        q = []
        m = collections.defaultdict(lambda: 0)
        for num in nums: m[num] += 1
        # python heapq doesn't support max heap by default
        for num in m: heapq.heappush(q, (-m[num], num))

        res = []
        for i in xrange(k):
            (count, num) = heapq.heappop(q)
        return res

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