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"""
Question
Top K Elements in List
Medium
Given an integer array nums and an integer k, return the k most frequent elements within the array.
The test cases are generated such that the answer is always unique.
You may return the output in any order.
Example 1:
Input: nums = [1,2,2,3,3,3], k = 2
Output: [2,3]
Example 2:
Input: nums = [7,7], k = 1
Output: [7]
Constraints:
1 <= nums.length <= 10^4.
-1000 <= nums[i] <= 1000
1 <= k <= number of distinct elements in nums.
"""
class Solution:
"""
A class to solve top_k_elements from hash
"""
def hashmap(self, nums: List[int], k: int) -> List[int]:
"""
Frequent elements from the input nums: List[int] and k: int using a hashmap approach.
Args:
nums: List[int]: integer list
k: int: frequency
Returns:
List[int]: elements list
"""
return [-1, -1]
case1 = [1, 2, 2, 3, 3, 3]
k1 = 2
case2 = [7, 7]
k2 = 1
case3 = [0, 9, 0, 6, 0, 4]
k3 = 3
solution = Solution()
print(f" hashmap case1: {solution.hashmap(case1,k1)}")
print(f" hashmap case2: {solution.hashmap(case2,k2)}")
print(f" hashmap case3: {solution.hashmap(case3,k3)}")
"""
Solution
url: https://neetcode.io/problems/top-k-elements-in-list
video: https://youtu.be/YPTqKIgVk-k
1. Bucket sort
time: O(n)
space: O(n)
code:
```python
class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
count = {}
freq = [[] for i in range(len(nums) + 1)]
for n in nums:
count[n] = 1 + count.get(n, 0)
for n, c in count.items():
freq[c].append(n)
res = []
for i in range(len(freq) - 1, 0, -1):
for n in freq[i]:
res.append(n)
if len(res) == k:
return res
```
2. Heap sort (max heap)
time: O(k*logn)
space: O(k*logn)
3. Sorting
time: O(n*logn)
space: O(n*logn)
"""
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