Medium
Longest Semi-Repeating Subarray — Python
Full explanation · Time O(n) · Space O(1)
# Time: O(n)
# Space: O(n)
import collections
# freq table, two pointers
class Solution(object):
def longestSubarray(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
cnt = collections.defaultdict(int)
result = left = repeat = 0
for right in xrange(len(nums)):
cnt[nums[right]] += 1
if cnt[nums[right]] == 2:
repeat += 1
if repeat > k:
if cnt[nums[left]] == 2:
repeat -= 1
cnt[nums[left]] -= 1
if not cnt[nums[left]]:
del cnt[nums[left]]
left += 1
return len(nums)-left
# Time: O(n)
# Space: O(n)
import collections
# freq table, two pointers
class Solution2(object):
def longestSubarray(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
cnt = collections.defaultdict(int)
result = left = repeat = 0
for right in xrange(len(nums)):
cnt[nums[right]] += 1
if cnt[nums[right]] == 2:
repeat += 1
while repeat > k:
if cnt[nums[left]] == 2:
repeat -= 1
cnt[nums[left]] -= 1
if not cnt[nums[left]]:
del cnt[nums[left]]
left += 1
result = max(result, right-left+1)
return result