diff --git a/combination-sum/jongwanra.py b/combination-sum/jongwanra.py
new file mode 100644
index 000000000..52bb0845f
--- /dev/null
+++ b/combination-sum/jongwanra.py
@@ -0,0 +1,100 @@
+"""
+[Problem]
+https://leetcode.com/problems/combination-sum/
+
+candidates: unique array of integers
+return a list of all unique combinations of candidates == target
+any order
+
+하나의 숫자는 candidates에서 무제한으로 선택할 수 있다.
+두 조합이 서로 다르다고 간주되는 조건은, 선택된 숫자 중 적어도 하나의 개수가 다를 때이다.
+[Brainstorming]
+DFS를 이용해서 Combination을 만든다.
+종료조건: target == sum || target > sum
+
+[Plan]
+1. candidates를 오름차순 정렬
+2. DFS
+
+[Complexity]
+N -> candidates.length
+M -> approximately target divided by the smallest candidate. -> target / min(candidates)
+Time: O(N^M)
+Space: O(N + M)
+  - 재귀 호출 스택: 깊이는 최대 target / min(candidates)
+  - chosen: 재귀 스택 깊이 만큼 공간 차지
+  - cache: 최악의 경우 answer와 같은 개수의 조합 저장 -> O(number of combinations)
+"""
+
+from typing import List
+
+
+class Solution:
+    def combinationSum1(self, candidates: List[int], target: int) -> List[List[int]]:
+        cache = set()
+        answer = []
+        chosen = []
+
+        def dfs(sum: int) -> None:
+            nonlocal target, candidates, cache, answer, chosen
+            print(chosen)
+            if sum > target:
+                return
+            if sum == target:
+                copied_chosen = chosen[:]
+                copied_chosen.sort()
+
+                cache_key = tuple(copied_chosen)
+                if cache_key in cache:
+                    # print(f"already exists {cache_key} in cache")
+                    return
+                cache.add(cache_key)
+                answer.append(copied_chosen)
+
+            for candidate in candidates:
+                chosen.append(candidate)
+                dfs(sum + candidate)
+                chosen.pop()
+
+        dfs(0)
+        return answer
+
+    """
+    중복 조합 방지 another solution
+    ref: https://www.algodale.com/problems/combination-sum/
+
+    [Complexity]
+    N -> candidates.length
+    M -> target / min(candidates)
+    Time: O(N^M)
+    Space: O(M)
+    """
+
+    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
+        answer = []
+        combi = []
+
+        def dfs(sum: int, start: int) -> None:
+            nonlocal target, answer, combi, candidates
+
+            if sum > target:
+                return
+            if sum == target:
+                answer.append(combi[:])
+                return
+
+            for index in range(start, len(candidates)):
+                candidate = candidates[index]
+                combi.append(candidate)
+                dfs(sum + candidate, index)
+                combi.pop()
+
+        dfs(0, 0)
+        return answer
+
+
+sol = Solution()
+# print(sol.combinationSum([2,3,6,7], 7))
+print(sol.combinationSum([2, 3, 5], 8))
+# print(sol.combinationSum([2], 1))
+
diff --git a/number-of-1-bits/jongwanra.py b/number-of-1-bits/jongwanra.py
new file mode 100644
index 000000000..debfdaa73
--- /dev/null
+++ b/number-of-1-bits/jongwanra.py
@@ -0,0 +1,45 @@
+"""
+[Problem]
+https://leetcode.com/problems/number-of-1-bits/description/
+
+양수 n이 주어졌을 때, 이진법에서 1로 설정된 비트의 개수를 반환하는 함수를 작성해라.
+
+
+[Plan]
+1. 주어진 양수를 이진수로 변환한다.
+2. for-loop을 순회하며 1의 개수를 counting한다.
+
+[Complexity]
+N: bin(n).length - 2
+Time: O(N)
+Space = O(N)
+"""
+class Solution:
+    def hammingWeight(self, n: int) -> int:
+        binary = bin(n)
+        output = 0
+        for index in range(2, len(binary)):
+            if binary[index] == '1':
+                output += 1
+        return output
+"""
+ref: https://www.algodale.com/problems/number-of-1-bits/
+[Complexity]
+Time: O(log n)
+Space: O(1)
+"""
+class AnotherSolution:
+    def hammingWeight(self, n: int) -> int:
+        count = 0
+        while n:
+            quotient, remainder = divmod(n, 2)
+            print(f"n = {n} quotient={quotient}, remainder={remainder}")
+            count += remainder
+            n = quotient
+        return count
+
+sol = AnotherSolution()
+print(sol.hammingWeight(11) == 3)
+print(sol.hammingWeight(128) == 1)
+print(sol.hammingWeight(2147483645) == 30)
+
diff --git a/valid-palindrome/jongwanra.py b/valid-palindrome/jongwanra.py
new file mode 100644
index 000000000..db9d1ac88
--- /dev/null
+++ b/valid-palindrome/jongwanra.py
@@ -0,0 +1,37 @@
+"""
+[Problem]
+https://leetcode.com/problems/valid-palindrome/description/
+
+모든 대문자를 소문자로 변환하고 알파벳과 숫자가 아닌 문자들을 전부 제거하한 이후에 앞에서 부터 일으나 뒤에서 부터 읽나 동일하게 읽힌다면, 그 문장은 회문입니다.
+영숫자 문자들은 알파벳과 숫자들을 포함합니다.
+
+[Brainstorming]
+leftPosition과 rightPosition을 두고, 비교하면서 아닐 경우 false를 return한다. => O(s.length)
+
+[Complexity]
+N: s.length
+Time: O(1/2 * N) => O(N)
+Space: O(1)
+"""
+
+class Solution:
+    def isPalindrome(self, s:str)-> bool:
+        leftPos, rightPos = 0, len(s) - 1
+        while leftPos < rightPos:
+            while not s[leftPos].isalnum() and leftPos < rightPos:
+                leftPos += 1
+            while not s[rightPos].isalnum() and leftPos < rightPos:
+                rightPos -= 1
+
+            if s[leftPos].upper() != s[rightPos].upper():
+                return False
+            leftPos += 1
+            rightPos -= 1
+        return True
+
+sol = Solution()
+print(sol.isPalindrome("A man, a plan, a canal: Panama") == True)
+print(sol.isPalindrome("race a car") == False)
+print(sol.isPalindrome(" ") == True)
+print(sol.isPalindrome("0P") == False)
+print(sol.isPalindrome("a") == True)