diff --git a/longest-substring-without-repeating-characters/hu6r1s.py b/longest-substring-without-repeating-characters/hu6r1s.py
new file mode 100644
index 000000000..2cf72c135
--- /dev/null
+++ b/longest-substring-without-repeating-characters/hu6r1s.py
@@ -0,0 +1,17 @@
+class Solution:
+    """
+    a       => 세트 {}에 a 없음
+    ab      => 세트 {a}에 b 없음
+    aba     => 세트 {a, b}에 a 있음 => 중복
+    abac    => 더 이상 고려 가치 없음
+    """
+    def lengthOfLongestSubstring(self, s: str) -> int:
+        max_len = 0
+        for start in range(len(s)):
+            chars = set()
+            for end in range(start, len(s)):
+                if s[end] in chars:
+                    break
+                chars.add(s[end])
+                max_len = max(end - start + 1, max_len)
+        return max_len
diff --git a/reverse-linked-list/hu6r1s.py b/reverse-linked-list/hu6r1s.py
new file mode 100644
index 000000000..6fc37bb45
--- /dev/null
+++ b/reverse-linked-list/hu6r1s.py
@@ -0,0 +1,24 @@
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+  """
+  링크드 리스트 학습을 따로 해야할 필요성이 있음
+  """
+  def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
+    stack = []
+    node = head
+    while node:
+        stack.append(node)
+        node = node.next
+    
+    dummy = ListNode()
+    output = dummy
+    while stack:
+        output.next = stack.pop()
+        output = output.next
+
+    output.next = None
+    return dummy.next
diff --git a/set-matrix-zeroes/hu6r1s.py b/set-matrix-zeroes/hu6r1s.py
new file mode 100644
index 000000000..c58e4e84e
--- /dev/null
+++ b/set-matrix-zeroes/hu6r1s.py
@@ -0,0 +1,16 @@
+class Solution:
+    def setZeroes(self, matrix: List[List[int]]) -> None:
+        """
+        Do not return anything, modify matrix in-place instead.
+        """
+        zero = []
+        for i in range(len(matrix)):
+            for j in range(len(matrix[0])):
+                if matrix[i][j] == 0:
+                    zero.append((i, j))
+        
+        for x, y in zero:
+            for i in range(len(matrix[0])):
+                matrix[x][i] = 0
+            for i in range(len(matrix)):
+                matrix[i][y] = 0
diff --git a/unique-paths/hu6r1s.py b/unique-paths/hu6r1s.py
new file mode 100644
index 000000000..77c268ecd
--- /dev/null
+++ b/unique-paths/hu6r1s.py
@@ -0,0 +1,23 @@
+class Solution:
+    """
+    dfs로 풀면 될 것이라고 생각은 했지만 이전 문제와 같은 방식일 줄 알았는데 이러한 방식이 있는 것을 알았음
+    """
+    # def uniquePaths(self, m: int, n: int) -> int:
+    #     def dfs(x, y):
+    #         if x == m - 1 and y == n - 1:
+    #             return 1
+    #         if x >= m or y >= n:
+    #             return 0
+            
+    #         return dfs(x + 1, y) + dfs(x, y + 1)
+        
+    #     return dfs(0, 0)
+
+    def uniquePaths(self, m: int, n: int) -> int:
+        dp = [[1] * n for _ in range(m)]
+
+        for row in range(1, m):
+            for col in range(1, n):
+                dp[row][col] = dp[row - 1][col] + dp[row][col - 1]
+
+        return dp[-1][-1]