# 1 - 2133. Check if Every Row and Column Contains All Numbers

	class Solution:
	def checkValid(self, matrix: List[List[int]]) -> bool:
	n = len(matrix)

	for i in range(n):
	l = [False] * n
	for j in range(n):
	l[matrix[i][j] - 1] = True
	if not all(l): return False

	for j in range(n):
	l = [False] * n
	for i in range(n):
	l[matrix[i][j] - 1] = True
	if not all(l): return False
	return True

上面的写法不太好看。下面看一个好看的：

	class Solution:
	def checkValid(self, matrix: List[List[int]]) -> bool:
	n = len(matrix)
	for a, b in zip(matrix, zip(*matrix)):
	if len(set(a)) != n or len(set(b)) != n:
	return False
	return True

参考自 Use HashSet to check each row /column. 使用序列解包 + zip 可以非常方便地获取列，再用一次 zip 让行和列一起考虑，表示行的 a 是列表，表示列的 b 是元组。

# 2 - 2134. Minimum Swaps to Group All 1's Together II

先提供一种会 TLE 的写法：

	class Solution:
	def minSwaps(self, nums: List[int]) -> int:
	ones = sum(nums)
	n = len(nums)
	if n == 1 or ones == 0: return 0

	nums = nums + nums
	deq = deque(nums[:ones])
	ans = ones - sum(deq)
	for i in range(ones, 2 * n):
	deq.popleft()
	deq.append(nums[i])
	ans = min(ans, ones - sum(deq))
	return ans

上面是 n^2 复杂度。这种题都要看数据范围，能优化就优化。一般来说 1e9 以内应该都可以，所以 1e5 的数据就表示不能用 n^2 的算法。

	class Solution:
	def minSwaps(self, nums: List[int]) -> int:
	n = len(nums)
	ones = sum(nums)
	cur = sum(nums[:ones])
	ans = ones - cur
	nums = nums + nums

	for i in range(ones, 2 * n):
	if nums[i - ones]: cur -= 1
	if nums[i]: cur += 1
	ans = min(ans, ones - cur)
	return ans

	class Solution:
	def wordCount(self, startWords: List[str], targetWords: List[str]) -> int:
	d = defaultdict(set)
	for word in startWords:
	d[len(word)].add(tuple(sorted(word)))

	print(d[4])
	ans = 0
	for word in targetWords:
	n = len(word)
	for i in range(len(word)):
	if tuple(sorted(word[:i] + word[i+1:])) in d[n-1]:
	ans += 1
	break
	return ans

上面的还可以吧。使用 bitmask 的写法：

	class Solution:
	def wordCount(self, startWords: List[str], targetWords: List[str]) -> int:
	seen = set()
	for word in startWords:
	m = 0
	for ch in word: m ^= 1 << ord(ch)-97
	seen.add(m)

	ans = 0
	for word in targetWords:
	m = 0
	for ch in word: m ^= 1 << ord(ch) - 97
	for ch in word:
	if m ^ (1 << ord(ch)-97) in seen:
	ans += 1
	break
	return ans

	class Solution:
	def earliestFullBloom(self, plantTime: List[int], growTime: List[int]) -> int:
	n = len(plantTime)
	growTime = [(x, i) for i, x in enumerate(growTime)]
	growTime.sort(key=lambda x: (-x[0], x[1]))

	bloomTime = [0] * n
	days = 0
	for t, i in growTime:
	days += plantTime[i]
	bloomTime[i] = days + t
	return max(bloomTime)

按照 growTime 从大到小贪心种植即可。贪心及其证明 —— 灵茶山艾府

	class Solution:
	def earliestFullBloom(self, plantTime: List[int], growTime: List[int]) -> int:
	a = list(zip(plantTime, growTime))
	a.sort(key=lambda x: -x[1])
	ans, day = 0, 0
	for p in a:
	day += p[0]
	ans = max(ans, day + p[1])
	return ans

不愧是大佬，好短好快。