diff --git a/contains-duplicate/Chaedie.py b/contains-duplicate/Chaedie.py
new file mode 100644
index 000000000..822394382
--- /dev/null
+++ b/contains-duplicate/Chaedie.py
@@ -0,0 +1,21 @@
+
+
+'''
+풀이:
+    중복된 요소가 있는지 찾는 문제입니다.
+
+    hash set 으로 중복을 제거하고
+    기존 nums 의 길이와 중복 제거된 nums_set 의 길이가 같은지 return 했습니다.
+
+시간 복잡도:
+    O(n) - has set 을 만드는 시간
+
+공간 복잡도:
+    O(n) - n개의 요소를 set에 담기 때문
+'''
+
+
+class Solution:
+    def containsDuplicate(self, nums: List[int]) -> bool:
+        nums_set = set(nums)
+        return len(nums_set) != len(nums)
diff --git a/house-robber/Chaedie.py b/house-robber/Chaedie.py
new file mode 100644
index 000000000..265ff1c7e
--- /dev/null
+++ b/house-robber/Chaedie.py
@@ -0,0 +1,103 @@
+'''
+Solution:
+    스스로 풀지 못해 학습만 진행했습니다.
+    다음 기회에 다시 풀어보도록 하겠습니다.
+'''
+class Solution:
+     def rob(self, nums: List[int]) -> int:
+        prev, curr = 0, 0
+        for num in nums:
+            prev, curr = curr, max(num + prev, curr)
+
+        return curr
+
+
+# class Solution:
+    # '''
+    # 4. 달레의 코드 풀이 - DP, prev, curr
+    # O(n) time
+    # O(1) space
+    # '''
+
+    # # 달레의 코드 풀이 - DP, prev, cur
+
+    # def rob(self, nums: List[int]) -> int:
+    #     prev, curr = 0, 0
+    #     for num in nums:
+    #         prev, curr = curr, max(num + prev, curr)
+
+    #     return curr
+
+    # '''
+    # 3. 달레의 코드 풀이 - DP
+    # [1,2,3,1]
+    # [1, 2, 3, 1]
+    # DP:[0, 1, 2, 4, 4]
+    # MAX(1 + DP[2], DP[1]) = MAX(1 + 2, 4) = 4
+    # '''
+    # # 달레의 코드 풀이 - DP
+    # # def rob(self, nums: List[int]) -> int:
+    # #     dp = [0] * (len(nums) + 1)
+    # #     dp[1] = nums[0]
+    # #     for n in range(2,len(nums) + 1):
+    # #         dp[n] = max(nums[n - 1] + dp[n - 2], dp[n - 1])
+    # #     return dp[-1]
+    # '''
+    # 2. 달레의 코드 풀이 - 재귀, 메모이제이션
+    # time: O(n)
+    # space: O(n)
+    # '''
+    # # 달레의 코드 풀이 - 재귀, 메모이제이션
+    # # def rob(self, nums: List[int]) -> int:
+    # #     memo = {}
+        
+    # #     def dfs(start):
+    # #         if start in memo:
+    # #             return memo[start]
+    # #         if not start < len(nums):
+    # #             memo[start] = 0
+    # #         else:
+    # #             memo[start] = max(nums[start] + dfs(start + 2), dfs(start + 1))
+    # #         return memo[start]
+    # #     return dfs(0)
+    # '''
+    # 1. 달레의 코드 풀이 - 재귀
+    # time: O(2^n)
+    # space: O(n)
+    
+    # F([1,2,3,1]) => MAX(1 + F([3,1], f([2,3,1])))
+    #     F([3,1]) => MAX(3 + F([]), F([1]))
+    #         F([]) => 0
+    #         F([1]) => MAX(1 + F([]), F([])) => MAX(1 + 0, 0) => 1
+    #             F([]) => 0
+    #             F([]) => 0
+    #     F([2,3,1]) => MAX(2 + F([1]), F([3,1]))
+    #         F([1]) => MAX(1 + F([]), F([])) => MAX(1 + 0, 0) => 1
+    #             F([]) => 0
+    #             F([]) => 0
+    #         F([3,1]) => MAX(3 + F([]), F([1]))
+    #         F([]) => 0
+    #         F([1]) => MAX(1 + F([]), F([])) => MAX(1 + 0, 0) => 1
+    #             F([]) => 0
+    #             F([]) => 0
+    # 재귀가 불필요하게 반복되고 있다.
+    # 메모이제이션으로 기억해두면 반복은 스킵할 수 있다.
+    # '''
+    # # 달레의 코드 풀이 - 재귀
+    # # def rob(self, nums: List[int]) -> int:
+        
+    # #     def dfs(start):
+    # #         if not start < len(nums):
+    # #             return 0
+    # #         return max(nums[start] + dfs(start + 2), dfs(start + 1))
+    # #     return dfs(0)
+        
+    # # neetcode 풀이 - DP, 이해안됨...
+    # # def rob(self, nums: List[int]) -> int:
+    # #     rob1, rob2 = 0, 0
+    # #     # [rob1, rob2, n, n+1, ...]
+    # #     for n in nums:
+    # #         temp = max(n + rob1, rob2)
+    # #         rob1 = rob2
+    # #         rob2 = temp
+    # #     return rob2
diff --git a/longest-consecutive-sequence/Chaedie.py b/longest-consecutive-sequence/Chaedie.py
new file mode 100644
index 000000000..a41d5183b
--- /dev/null
+++ b/longest-consecutive-sequence/Chaedie.py
@@ -0,0 +1,46 @@
+"""
+Solution: 
+    0. i will use prev, cur pointers in for loop, 
+        so i have to get rid of edge case when len of nums is at most 1
+    1. use hash set for remove duplicate elements 
+    2. sort list
+    3. iterate sorted_nums and use two pointers (prev, cur) 
+        compare prev and cur and count if the difference is 1
+    4. use max method, memorize maxCount
+    5. return maxCount
+
+Time Complexity: 
+    1. remove duplicate by hash set -> O(n)
+    2. sort the list -> O(n log n)
+    3. iterate sorted_nums -> O(n)
+
+    so time complexity of this solution will be O(n log n)
+
+Space Complexity:
+    1. set() -> O(n)
+    2. sorted_nums -> O(n)
+    3. count , maxCount -> O(1)
+
+    space complexity of this solution will be O(n)
+"""
+
+
+class Solution:
+    def longestConsecutive(self, nums: List[int]) -> int:
+        if len(nums) <= 1:
+            return len(nums)
+
+        sorted_nums = sorted(list(set(nums)))
+
+        count = 1
+        maxCount = 1
+        for i in range(1, len(sorted_nums)):
+            prev = sorted_nums[i - 1]
+            cur = sorted_nums[i]
+            if prev + 1 == cur:
+                count += 1
+                maxCount = max(count, maxCount)
+            else:
+                count = 1
+
+        return maxCount
diff --git a/top-k-frequent-elements/Chaedie.py b/top-k-frequent-elements/Chaedie.py
new file mode 100644
index 000000000..8ecf9775e
--- /dev/null
+++ b/top-k-frequent-elements/Chaedie.py
@@ -0,0 +1,30 @@
+"""
+Solution: 
+    1. make a dictionary of (index, Frequency)
+    2. sort the items of dictionary by Frequency
+    3. return list of Top k Frequent Elements
+
+Time Complexity:
+    1. iterate nums for counting -> O(n)
+    2. sort -> O(n log n)
+    3. iterate list for making return value -> O(n)
+
+    So Time complexity of this solution is O(n log n)
+
+Space Complexity: 
+    1. dictionary for counting frequency of nums -> O(n)
+    2. Timsort's space overhead -> O(n) 
+    3. sorted List -> O(n)
+
+    Space complexity of this solution is O(n)
+"""
+
+
+class Solution:
+    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
+        counts = defaultdict(int)
+        for num in nums:
+            counts[num] += 1
+
+        result = sorted(counts.items(), key=lambda x: x[1], reverse=True)
+        return list(map(lambda x: x[0], result[:k]))
diff --git a/valid-palindrome/Chaedie.py b/valid-palindrome/Chaedie.py
new file mode 100644
index 000000000..1a23c5c8b
--- /dev/null
+++ b/valid-palindrome/Chaedie.py
@@ -0,0 +1,31 @@
+"""
+풀이:
+    1) lower case로 변형합니다.
+    2) alpha numeric 인 문자만 string 배열에 담아줍니다.
+    3) string 배열과 string 배열의 역순 배열을 비교해서 return 합니다.
+
+시간 복잡도:
+    1) s.lower() -> O(n) - 각 요소를 순회하며 lower case로 바꿔줍니다.
+    2) for char in s: -> O(n) - 각 요소를 순회하면 isalnum() 여부를 확인합니다.
+    3) string.append(char) -> O(1)
+    4) string[::-1] -> O(n) - 각 요소를 순회합니다.
+
+    결론: O(4n) 이므로 O(n) 입니다.
+
+공간 복잡도: 
+    1) string 배열 - O(n)
+    2) string[::-1] - O(n)
+
+    결론: O(n) 입니다.
+"""
+
+
+class Solution:
+    def isPalindrome(self, s: str) -> bool:
+        string = []
+        s = s.lower()
+        for char in s:
+            if char.isalnum():
+                string.append(char)
+
+        return string == string[::-1]