47 lines
1.5 KiB
Python
47 lines
1.5 KiB
Python
|
class Solution:
|
||
|
|
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
|
||
|
|
'''
|
||
|
|
Approach using bucket sort
|
||
|
|
'''
|
||
|
|
|
||
|
|
c = Counter(nums)
|
||
|
|
|
||
|
|
# Idea here is that the list indices will represent
|
||
|
|
# the actual counts and we store elements matching thise
|
||
|
|
# counts under the index. If size n list is provided, max
|
||
|
|
# count possible is n but lists are 0-indexed so we set
|
||
|
|
# num_bucket to have n+1 size.
|
||
|
|
num_bucket = [[] for _ in range(len(nums) + 1)]
|
||
|
|
|
||
|
|
# Later down, we will iterate in reverse through `num_bucket` and
|
||
|
|
# instead of iterating through the entire list, we can take note
|
||
|
|
# of the max count we observe and start from there instead.
|
||
|
|
max_count_seen = -1
|
||
|
|
|
||
|
|
for num, count in c.items():
|
||
|
|
num_bucket[count].append(num)
|
||
|
|
|
||
|
|
max_count_seen = max(max_count_seen, count)
|
||
|
|
|
||
|
|
result = []
|
||
|
|
j = max_count_seen
|
||
|
|
|
||
|
|
# Iterate reverse through num_bucket (note the max_count_seen
|
||
|
|
# optimization) and fill the result array until k elements are
|
||
|
|
# added
|
||
|
|
while j >= 0:
|
||
|
|
for num in num_bucket[j]:
|
||
|
|
result.append(num)
|
||
|
|
|
||
|
|
# Possible that we filled k elements already in this inner
|
||
|
|
# loop, so get out
|
||
|
|
if len(result) == k:
|
||
|
|
break
|
||
|
|
|
||
|
|
if len(result) == k:
|
||
|
|
break
|
||
|
|
|
||
|
|
j -= 1
|
||
|
|
|
||
|
|
return result
|