Merge pull request #2199 from changhyumm/main

[changhyumm] WEEK 06 solutions

Author: changhyumm

best-time-to-buy-and-sell-stock/changhyumm.py

@@ -0,0 +1,9 @@
+class Solution:
+    def maxProfit(self, prices: List[int]) -> int:
+        min_price = prices[0]
+        max_profit = 0
+        for i in range(0, len(prices) - 1):
+            min_price = min(prices[i], min_price)
+            profit = prices[i+1] - min_price
+            max_profit = max(profit, max_profit)
+        return max_profit

container-with-most-water/changhyumm.py

@@ -0,0 +1,23 @@
+class Solution:
+    def maxArea(self, height: List[int]) -> int:
+        # 시간복잡도 O(n2)으로 타임아웃
+        # max_amount = 0
+        # for i in range(len(height) - 1):
+        #     for j in range(i + 1, len(height)):
+        #         amount = (j-i) * min(height[i], height[j])
+        #         max_amount = max(amount, max_amount)
+        # return max_amount
+        
+        # 투포인터 활용
+        # 시간복잡도 O(n)
+        max_amount = 0
+        left = 0
+        right = len(height) - 1
+        while left < right:
+            amount = (right - left) * min(height[left], height[right])
+            max_amount = max(amount, max_amount)
+            if height[left] < height[right]:
+                left += 1
+            else:
+                right -= 1
+        return max_amount

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

@@ -0,0 +1,34 @@
+# prefix 기반으로 탐색할 수 있는 트라이노드 자료구조 활용
+class TrieNode:
+    def __init__(self):
+        self.children = {}
+        self.is_ending = False
+
+class WordDictionary:
+
+    def __init__(self):
+        self.root = TrieNode()
+
+    def addWord(self, word: str) -> None:
+        current_node = self.root
+        for char in word:
+            if char not in current_node.children:
+                current_node.children[char] = TrieNode()
+            current_node = current_node.children[char]
+        current_node.is_ending = True
+
+    def search(self, word):
+        def dfs(node, index):
+            # 단어자리수만큼 왔고, 끝단어인경우 찾기 성공
+            if index == len(word):
+                return node.is_ending
+            # 와일드카드인 경우 자식들 전부 탐색
+            if word[index] == ".":
+                for child in node.children.values():
+                    if dfs(child, index+1):
+                        return True
+            # 일반적인 경우 다음 자식 탐색
+            if word[index] in node.children:
+                return dfs(node.children[word[index]], index+1)
+            return False
+        return dfs(self.root, 0)

longest-increasing-subsequence/changhyumm.py

@@ -0,0 +1,14 @@
+class Solution:
+    def lengthOfLIS(self, nums: List[int]) -> int:
+        # dp[i] = nums[0]부터 nums[i]까지의 LIS 길이
+        # 처음은 1
+        n = len(nums)
+        dp = [1] * n
+
+        for i in range(1, n):
+            for j in range(i):
+                if nums[j] < nums[i]:
+                    # nums[i]가 더 큰 경우 길이 + 1
+                    dp[i] = max(dp[i], dp[j] + 1)
+        
+        return max(dp) # 전체 중 최대

valid-parentheses/changhyumm.py

@@ -0,0 +1,20 @@
+class Solution:
+    def isValid(self, s: str) -> bool:
+        brackets = {"(": ")", "{": "}", "[": "]"}
+        stack = []
+        # 괄호의 특성상 stack 자료구조 활용
+        # 시간 복잡도 O(n)
+        # 공간 복잡도 O(n)
+        for string in s:
+            if string not in brackets:
+                if len(stack) == 0:
+                    return False
+                else:
+                    target = stack.pop()
+                    if string != target:
+                        return False
+            else:
+                stack.append(brackets[string])
+        if stack:
+            return False
+        return True