Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order.
Example 1:
Input: nums = [1,1,1,2,2,3], k = 2 Output: [1,2]
Example 2:
Input: nums = [1], k = 1 Output: [1]
1 <= nums.length <= 10^5 -10^4 <= nums[i] <= 10^4 k is in the range [1, the number of unique elements in the array]. It is guaranteed that the answer is unique.
Follow up: Your algorithm's time complexity must be better than O(n log n), where n is the array's size.
class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
count = {}
for num in nums:
if num in count:
count[num] += 1
else:
count[num] = 1
# find ordered list
# Sort the dictionary keys based on their values in descending order
sorted_keys = sorted(count, key=lambda x: count[x], reverse=True)
return sorted_keys[:k]Pretty straightforward solution. The only tricky part was getting the sorted keys acccording to dictionary values. Took help in writing the lambda function for reverse sorting the dictionary.
Time Complexity = O(nlogn) - for sorting Space complextiy = O(nlogn) - again for sorting , the dictionary takes only O(n) space