From 6544db9a9a41550f5a2c8c3cad02c249f35b002d Mon Sep 17 00:00:00 2001
From: siroo <fordevelop@pm.me>
Date: Sun, 5 Jan 2025 21:55:05 +0900
Subject: [PATCH 1/3] #221 & #236 solution

---
 best-time-to-buy-and-sell-stock/jungsiroo.py | 23 ++++++++++
 group-anagrams/jungsiroo.py                  | 44 ++++++++++++++++++++
 2 files changed, 67 insertions(+)
 create mode 100644 best-time-to-buy-and-sell-stock/jungsiroo.py
 create mode 100644 group-anagrams/jungsiroo.py

diff --git a/best-time-to-buy-and-sell-stock/jungsiroo.py b/best-time-to-buy-and-sell-stock/jungsiroo.py
new file mode 100644
index 000000000..351db1877
--- /dev/null
+++ b/best-time-to-buy-and-sell-stock/jungsiroo.py
@@ -0,0 +1,23 @@
+class Solution:
+    def maxProfit(self, prices: List[int]) -> int:
+        """
+        가장 수익을 많이 얻을 수 있도록 저점에 매수, 고점에 매도
+        매수와 매도는 서로 다른 날
+   		
+		min_price를 빼가면서 price 업데이트
+
+		Time Complexity : O(n)
+		Space Complexity : O(1)
+        """
+
+        min_price = max(prices)
+        days = len(prices)
+    
+        for day in range(days):
+			min_price = min(prices[day], min_price)
+            prices[day] -= min_price
+    
+        return max(prices)
+    
+
+        
diff --git a/group-anagrams/jungsiroo.py b/group-anagrams/jungsiroo.py
new file mode 100644
index 000000000..a70024778
--- /dev/null
+++ b/group-anagrams/jungsiroo.py
@@ -0,0 +1,44 @@
+from collections import defaultdict
+
+class Solution:
+    def groupAnagrams(self, strs: List[str]) -> List[List[str]]:
+        # Naive Solution : Sort string
+
+        # 각 단어마다 모두 정렬을 한 뒤 해당 값을 hash로 사용하여 딕셔너리에 추가하는 방법
+        # strs[i].length = k
+        # Time Complexity : O(n*klog(k))
+        # Space Complexity : O(n*k)
+
+        """
+        
+        n = len(strs)
+        word_dict = defaultdict(list)
+
+        for word in strs:
+            key = hash(''.join(sorted(word)))
+            word_dict[key].append(word)
+
+        ret = []
+        for value in word_dict.values():
+            ret.append(value)
+        return ret
+        """
+
+        # Better Solution : Counting 
+        
+        # anagram 의 특성 중 알파벳 카운트 갯수가 같다는 것을 이용
+        # 카운트 갯수를 활용하여 key 값으로 처리
+        # Time Complexity : O(n*k)
+        # Space Complexity : O(n*k)
+        word_dict = defaultdict(list)
+        
+        for word in strs:
+            freq = [0]*26
+            for char in word:
+                freq[ord(char) - ord('a')] += 1
+            word_dict[tuple(freq)].append(word)
+
+        return list(word_dict.values())
+        
+
+

From fd2cd7a4334ca86d607ae7bf17e1b81aaf817e4f Mon Sep 17 00:00:00 2001
From: siroo <fordevelop@pm.me>
Date: Sun, 5 Jan 2025 23:33:28 +0900
Subject: [PATCH 2/3] word-break solution

---
 word-break/jungsiroo.py | 27 +++++++++++++++++++++++++++
 1 file changed, 27 insertions(+)
 create mode 100644 word-break/jungsiroo.py

diff --git a/word-break/jungsiroo.py b/word-break/jungsiroo.py
new file mode 100644
index 000000000..5024e3cae
--- /dev/null
+++ b/word-break/jungsiroo.py
@@ -0,0 +1,27 @@
+class Solution:
+    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
+        # DP를 활용한 문제
+
+        """
+        KMP나 Rabin Karp를 이용하여 푸는 문제인 줄 알았으나 전혀 아닌 문제
+        neetcode의 도움을 받았음
+        
+        계속 순회를 하면서 if dp[j]를 통해 길이만큼 끊어주고 word_set에 있는지를 확인함
+        Time Complexity : O(n^2)
+        Space Complexity : O(n)
+        """
+        
+        word_set = set(wordDict)
+        n = len(s)
+
+        dp = [False] * (n + 1)
+        dp[0] = True  
+
+        for i in range(1, n + 1):
+            for j in range(i):
+                if dp[j] and s[j:i] in word_set:
+                    dp[i] = True
+                    break
+
+        return dp[n]
+

From 3e2c6d92f56a745875afc2cf41a8be209f5c437b Mon Sep 17 00:00:00 2001
From: siroo <fordevelop@pm.me>
Date: Mon, 6 Jan 2025 23:59:34 +0900
Subject: [PATCH 3/3] implement trie solution

---
 implement-trie-prefix-tree/jungsiroo.py | 47 +++++++++++++++++++++++++
 1 file changed, 47 insertions(+)
 create mode 100644 implement-trie-prefix-tree/jungsiroo.py

diff --git a/implement-trie-prefix-tree/jungsiroo.py b/implement-trie-prefix-tree/jungsiroo.py
new file mode 100644
index 000000000..fd33a563e
--- /dev/null
+++ b/implement-trie-prefix-tree/jungsiroo.py
@@ -0,0 +1,47 @@
+"""
+Recursive vs Iterative
+
+재귀 방식의 경우, 만약 문자열의 길이가 굉장히 길다면 그만큼의 콜이 일어나고 이는 성능적으로 느려질 수 있음.
+두 가지 모두 시간 복잡도 면에서는 O(m) 임 (m = len(string))
+
+Node 클래스를 따로 두어 처리하면 가독성 높게 처리할 수 있다.
+"""
+
+class Node:
+    def __init__(self, key=None):
+        self.key = key
+        self.children = {}
+        self.is_end = False
+
+class Trie:
+    def __init__(self):
+        self.head = Node()
+
+    def insert(self, word: str) -> None:
+        curr = self.head
+
+        for ch in word:
+            if ch not in curr.children:
+                curr.children[ch] = Node(ch)
+            curr = curr.children[ch]
+        curr.is_end = True
+    
+    def search(self, word: str) -> bool:
+        curr = self.head
+        
+        for ch in word:
+            if ch not in curr.children:
+                return False
+            curr = curr.children[ch]
+        
+        return curr.is_end
+        
+    def startsWith(self, prefix: str) -> bool:
+        curr = self.head
+
+        for ch in prefix:
+            if ch not in curr.children:
+                return False
+            curr = curr.children[ch]
+        return True
+