Intuition this problem is easy. so, the day of arrive and leave are limited in the same year! so, simplely find out the leave day of the year and the arrive day of the year for both Alice and Bob. And then find the latest arrive day and the earliest leave day. If the leave day is bigger than or equal to the arrive day, then just calculate difference between leave day and arrive day. that's the a..

class Solution: def minimumRounds(self, tasks: List[int]) -> int: freq = collections.Counter(tasks) cnt = 0 for num, count in freq.items(): if count = 2: if (count > 2 and count % 2 != 0) or count % 3 == 0: count -= 3 else: count -= 2 cnt += 1 return cnt

class Solution: def digitSum(self, s: str, k: int) -> str: while len(s) > k: i = 0 next = '' while i < len(s): next += str(sum([int(ch) for ch in s[i:min(i + k, len(s))]])) i += k s = next return s

class Solution: def intersection(self, nums: List[List[int]]) -> List[int]: intr = set(nums[0]) for num in nums[1:]: intr = intr & set(num) return sorted(intr)

class Solution: def divideString(self, s: str, k: int, fill: str) -> List[str]: n = len(s) res = [] i = 0 while i < n: pad = k - len(s[i:i + k]) res += s[i:i + k] + fill*pad, i += k return res

class Solution: def minSwaps(self, nums: List[int]) -> int: n = len(nums) ones = nums.count(1) zeros = nums[:ones].count(0) mn_zeros = zeros for i in range(n - 1): if nums[i] == 0: zeros -= 1 if nums[(i + ones)%n] == 0: zeros += 1 mn_zeros = min(mn_zeros, zeros) return mn_zeros