[HodaeSsi] week 06

container-with-most-water/HodaeSsi.py

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+# 시간 복잡도 : O(n)
+# 공간 복잡도 : O(1)
+# 문제 유형 : Two Pointer
+class Solution:
+	def maxArea(self, height: List[int]) -> int:
+		left, right = 0, len(height) - 1
+		max_area = 0
+
+		while left < right:
+			max_area = max(max_area, (right - left) * min(height[left], height[right]))
+			if height[left] < height[right]:
+				left += 1
+			else:
+				right -= 1
+
+		return max_area
+

design-add-and-search-words-data-structure/HodaeSsi.py

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+# 문제유형 : Trie
+class WordDictionary:
+	def __init__(self):
+		self.trie = {}
+
+	# 시간 복잡도 : O(N)
+	# 공간 복잡도 : O(N)
+	def addWord(self, word: str) -> None:
+		node = self.trie
+		for char in word:
+			if char not in node:
+				node[char] = {}
+			node = node[char]
+		node['$'] = True
+
+	# 시간 복잡도 : O(m^N) (m : 철자의 종류, N : 문자열의 길이)
+	def search(self, word: str) -> bool:
+		return self.dfs(0, self.trie, word)
+
+	def dfs(self, i, node, word):
+		if i == len(word):
+			return '$' in node
+		if word[i] == '.':
+			for child in node.values():
+				if isinstance(child, dict) and self.dfs(i + 1, child, word):
+					return True
+		if word[i] in node:
+			return self.dfs(i + 1, node[word[i]], word)
+		return False
+

longest-increasing-subsequence/HodaeSsi.py

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+# 시간 복잡도 : O(nlogn)
+# 공간 복잡도 : O(n)
+# 문제 유형 : DP, Binary Search
+class Solution:
+	def lengthOfLIS(self, nums: List[int]) -> int:
+		dp = []
+		for num in nums:
+			if not dp or num > dp[-1]:
+				dp.append(num)
+			else:
+				left, right = 0, len(dp) - 1
+				while left < right:
+					mid = left + (right - left) // 2
+					if dp[mid] < num:
+						left = mid + 1
+					else:
+						right = mid
+				dp[right] = num
+		return len(dp)
+

spiral-matrix/HodaeSsi.py

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+# 시간 복잡도 : O(m * n)
+# 공간 복잡도 : O(1)
+# 문제 유형 : 구현
+class Solution:
+	def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
+		row, col = len(matrix), len(matrix[0])
+		up, down, left, right = 0, row - 1, 0, col - 1
+		result = []
+		while True:
+			for i in range(left, right + 1):
+				result.append(matrix[up][i])
+			up += 1
+			if up > down:
+				break
+			for i in range(up, down + 1):
+				result.append(matrix[i][right])
+			right -= 1
+			if right < left:
+				break
+			for i in range(right, left - 1, -1):
+				result.append(matrix[down][i])
+			down -= 1
+			if down < up:
+				break
+			for i in range(down, up - 1, -1):
+				result.append(matrix[i][left])
+			left += 1
+			if left > right:
+				break
+		return result
+

valid-parentheses/HodaeSsi.py

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+# 시간 복잡도 : O(n)
+# 공간 복잡도 : O(n)
+# 문제 유형 : Stack
+class Solution:
+	def isValid(self, s: str) -> bool:
+		stack = []
+
+		for char in s:
+			if char in (')', '}', ']'):
+				if not stack or stack.pop() != {')': '(', '}': '{', ']': '['}[char]:
+					return False
+			else:
+				stack.append(char)
+
+		return not stack
+