# 347-top-k-frequent-elements


Try it on <a href='https://leetcode.com/problems/347-top-k-frequent-elements'>leetcode</a>

## Description
<div class="description">
<div><p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <em>the</em> <code>k</code> <em>most frequent elements</em>. You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [1,1,1,2,2,3], k = 2
<strong>Output:</strong> [1,2]
</pre><p><strong>Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [1], k = 1
<strong>Output:</strong> [1]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>k</code> is in the range <code>[1, the number of unique elements in the array]</code>.</li>
	<li>It is <strong>guaranteed</strong> that the answer is <strong>unique</strong>.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Your algorithm's time complexity must be better than <code>O(n log n)</code>, where n is the array's size.</p>
</div>
</div>

## Solution(Python)
```Python
class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        return self.prioirtyQueue(nums, k)

    # Time Complexity: O(nlogn)
    # Space Complexity: O(n)
    def naive(self, nums: List[int], k: int) -> List[int]:
        freqmap = Counter(nums)
        freqmap = sorted(freqmap, key=freqmap.get, reverse=True)
        return freqmap[:k]

    # Time Complexity: O(nlogk)
    # Space Complexity: O(n)
    def prioirtyQueue(self, nums: List[int], k: int) -> List[int]:
        if k == len(nums):
            return nums

        count = Counter(nums)

        return heapq.nlargest(k, count.keys(), key=count.get)

```