diff --git a/best-time-to-buy-and-sell-stock/gcount85.py b/best-time-to-buy-and-sell-stock/gcount85.py
new file mode 100644
index 0000000000..238bf8cf5a
--- /dev/null
+++ b/best-time-to-buy-and-sell-stock/gcount85.py
@@ -0,0 +1,21 @@
+"""
+# Approach
+지금까지의 최저 가격을 갱신함과 동시에 최선의 이익도 업데이트합니다.
+
+# Complexity
+- Time complexity: O(N)
+
+- Space complexity: O(1)
+"""
+
+
+class Solution:
+    def maxProfit(self, prices: list[int]) -> int:
+        min_price = float("inf")
+        answer = 0
+
+        for price in prices:
+            min_price = min(min_price, price)
+            answer = max(answer, price - min_price)
+
+        return answer
diff --git a/group-anagrams/gcount85.py b/group-anagrams/gcount85.py
new file mode 100644
index 0000000000..65d4890ddd
--- /dev/null
+++ b/group-anagrams/gcount85.py
@@ -0,0 +1,21 @@
+"""
+# Approach
+strs 배열을 순회하며 문자열을 정규화(정렬)하고,
+정규화 값이 같은 원소들끼리 모이도록 딕셔너리에 추가하여 최종 값을 반환한다.
+
+# Complexity
+strs의 길이를 N, 문자열의 길이를 K라고 할 때,
+
+- Time complexity:  O(N*KlogK)
+- Space complexity: O(N*K)
+"""
+
+from collections import defaultdict
+
+
+class Solution:
+    def groupAnagrams(self, strs: list[str]) -> list[list[str]]:
+        anagram = defaultdict(list)  # normalized str : str list
+        for s in strs:
+            anagram["".join(sorted(s))].append(s)
+        return list(anagram.values())
diff --git a/implement-trie-prefix-tree/gcount85.py b/implement-trie-prefix-tree/gcount85.py
new file mode 100644
index 0000000000..764e78800a
--- /dev/null
+++ b/implement-trie-prefix-tree/gcount85.py
@@ -0,0 +1,41 @@
+class TrieNode:
+    def __init__(self):
+        self.children = {}
+        self.is_end = False
+
+
+class Trie:
+
+    def __init__(self):
+        self.root = TrieNode()
+
+    def insert(self, word: str) -> None:
+        node = self.root
+        for ch in word:
+            if ch not in node.children:
+                node.children[ch] = TrieNode()
+            node = node.children[ch]
+        node.is_end = True
+
+    def search(self, word: str) -> bool:
+        node = self.root
+        for ch in word:
+            if ch not in node.children:
+                return False
+            node = node.children[ch]
+        return node.is_end
+
+    def startsWith(self, prefix: str) -> bool:
+        node = self.root
+        for ch in prefix:
+            if ch not in node.children:
+                return False
+            node = node.children[ch]
+        return True
+
+
+# Your Trie object will be instantiated and called as such:
+# obj = Trie()
+# obj.insert(word)
+# param_2 = obj.search(word)
+# param_3 = obj.startsWith(prefix)
diff --git a/word-break/gcount85.py b/word-break/gcount85.py
new file mode 100644
index 0000000000..ef5e4f014b
--- /dev/null
+++ b/word-break/gcount85.py
@@ -0,0 +1,35 @@
+"""
+# Intuition
+wordDict를 word들의 길이로 분류하고,
+s의 각 위치까지의 문자열을 완성할 수 있는지 dp 배열로 확인합니다.
+
+# Complexity
+wordDict의 길이를 N, s의 길이를 K
+- Time complexity: O(N+K)
+- Space complexity: O(N+K)
+"""
+
+from collections import defaultdict
+
+
+class Solution:
+    def wordBreak(self, s: str, wordDict: list[str]) -> bool:
+        n = len(s)
+        dp = [False] * (n + 1)
+        word_dict = defaultdict(set)
+        for word in wordDict:
+            word_dict[len(word)].add(word)
+
+        dp[0] = True
+        for i in range(1, n + 1):
+            for (
+                k,
+                v,
+            ) in word_dict.items():
+                if i + k - 1 > n:
+                    continue
+                if dp[i - 1] == False:
+                    continue
+                if s[i - 1 : i + k - 1] in v:
+                    dp[i + k - 1] = True
+        return dp[n]