diff --git a/combination-sum/riveroverflows.py b/combination-sum/riveroverflows.py
new file mode 100644
index 000000000..f168757b1
--- /dev/null
+++ b/combination-sum/riveroverflows.py
@@ -0,0 +1,33 @@
+class Solution:
+    """
+    TC: O(N^(T/M))
+        - N = len(candidates), T = target, M = min(candidates)
+    SC: O((T/M) * N^(T/M))
+
+    풀이:
+    백트래킹으로 모든 조합을 탐색하되, 중복 조합 방지를 위해
+    start 인덱스를 활용하여 이전 후보는 건너뛰고, 같은 후보는 재사용 허용.
+    path_sum > target이면 가지치기(pruning)하여 불필요한 탐색을 줄임.
+    """
+    def combinationSum(self, candidates: list[int], target: int) -> list[list[int]]:
+        answer = []
+
+        def backtrack(start, path, path_sum):
+            if path_sum > target:
+                return
+
+            if path_sum == target:
+                answer.append(path[:])
+                return
+
+            for i in range(start, len(candidates)):
+                candidate = candidates[i]
+                path.append(candidate)
+                path_sum += candidate
+                backtrack(i, path, path_sum)
+                path_sum -= candidate
+                path.pop()
+
+        backtrack(0, [], 0)
+
+        return answer
diff --git a/number-of-1-bits/riveroverflows.py b/number-of-1-bits/riveroverflows.py
new file mode 100644
index 000000000..638a6b1de
--- /dev/null
+++ b/number-of-1-bits/riveroverflows.py
@@ -0,0 +1,20 @@
+from typing import *
+
+
+class Solution:
+    """
+    TC: O(log n)
+    SC: O(1)
+
+    풀이:
+    Brian Kernighan의 비트 트릭을 활용.
+    n & (n-1)은 n의 가장 오른쪽 1비트를 하나 제거한다.
+    n이 0이 될 때까지 반복한 횟수가 곧 1비트의 개수.
+    """
+    def hammingWeight(self, n: int) -> int:
+        answer = 0
+        while n > 0:
+            n = n & (n - 1)
+            answer += 1
+
+        return answer
diff --git a/valid-palindrome/riveroverflows.py b/valid-palindrome/riveroverflows.py
new file mode 100644
index 000000000..4cf434e9c
--- /dev/null
+++ b/valid-palindrome/riveroverflows.py
@@ -0,0 +1,36 @@
+class Solution:
+    """
+    풀이 1: 투포인터
+    TC: O(n)
+    SC: O(1)
+
+    양쪽 끝에서 포인터를 좁혀가며 알파벳/숫자만 비교.
+    """
+    def isPalindrome(self, s: str) -> bool:
+        l, r = 0, len(s) - 1
+        while l < r:
+            if not s[l].isalnum():
+                l += 1
+                continue
+            if not s[r].isalnum():
+                r -= 1
+                continue
+
+            if s[l].lower() != s[r].lower():
+                return False
+
+            l += 1
+            r -= 1
+
+        return True
+
+    """
+    풀이 2: 리스트 변환 + 역순 비교
+    TC: O(n)
+    SC: O(n)
+
+    알파벳/숫자만 추출한 리스트를 역순과 비교.
+    """
+    def isPalindrome2(self, s: str) -> bool:
+        s = [c for c in s.lower() if c.isalnum()]
+        return s == s[::-1]