2026-06-01 16:12:21 +08:00
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#+PROPERTY: STUDY_DECK_02
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2026-06-01 17:12:10 +08:00
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* TODO 0347. Top K Frequent Elements :medium:
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2026-06-01 02:33:30 +08:00
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:PROPERTIES:
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2026-06-01 17:22:07 +08:00
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:NEETCODE: [[file:../../roadmap.org::*0347. Top K Frequent Elements][0347. Top K Frequent Elements]]
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2026-06-01 02:33:30 +08:00
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:END:
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2026-06-01 17:22:07 +08:00
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Given an integer array ~nums~ and an integer ~k~, return /the/ ~k~ /most frequent elements/. You may return the answer in *any order*.
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*Example 1:*
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*Input:* nums = [1,1,1,2,2,3], k = 2
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*Output:* [1,2]
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*Example 2:*
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*Input:* nums = [1], k = 1
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*Output:* [1]
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*Example 3:*
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*Input:* nums = [1,2,1,2,1,2,3,1,3,2], k = 2
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*Output:* [1,2]
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*Constraints:*
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- ~1 <= nums.length <= 10^{5}~
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- ~-10^{4} <= nums[i] <= 10^{4}~
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- ~k~ is in the range ~[1, the number of unique elements in the array]~.
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- It is *guaranteed* that the answer is *unique*.
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*Follow up:* Your algorithm's time complexity must be better than ~O(n log n)~, where n is the array's size.
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2026-06-01 02:39:53 +08:00
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** TODO Approach
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Write your approach here.
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** TODO Python
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#+begin_src python
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2026-06-01 17:22:07 +08:00
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class Solution:
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def topKFrequent(self, nums: List[int], k: int) -> List[int]:
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2026-06-01 02:39:53 +08:00
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#+end_src
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** TODO C++
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2026-06-01 02:33:30 +08:00
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#+begin_src cpp
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2026-06-01 17:22:07 +08:00
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class Solution {
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public:
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vector<int> topKFrequent(vector<int>& nums, int k) {
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}
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};
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2026-06-01 02:33:30 +08:00
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#+end_src
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