diff --git a/longest-substring-without-repeating-characters/hi-rachel.py b/longest-substring-without-repeating-characters/hi-rachel.py
index cdcb6aaf9..baee2e1d8 100644
--- a/longest-substring-without-repeating-characters/hi-rachel.py
+++ b/longest-substring-without-repeating-characters/hi-rachel.py
@@ -58,3 +58,29 @@ def lengthOfLongestSubstring(self, s: str) -> int:
             max_len = max(max_len, right - left + 1)
             
         return max_len
+
+
+# HashMap 풀이
+def lengthOfLongestSubstring(s: str) -> int:
+    if not s:
+        return 0
+ 
+    left = 0  # 윈도우 시작점
+    max_length = 0  # 최대 길이
+    seen = {}  # 문자의 마지막 등장 위치를 저장하는 해시맵
+ 
+    for right in range(len(s)):
+        char = s[right]
+ 
+        # 현재 문자가 윈도우 내에 이미 존재하는 경우
+        if char in seen and seen[char] >= left:
+            # 윈도우 시작점을 중복 문자 다음 위치로 이동
+            left = seen[char] + 1
+ 
+        # 현재 문자의 위치 업데이트
+        seen[char] = right
+ 
+        # 현재 윈도우 길이와 최대 길이 비교 후 업데이트
+        max_length = max(max_length, right - left + 1)
+ 
+    return max_length
diff --git a/number-of-islands/hi-rachel.py b/number-of-islands/hi-rachel.py
new file mode 100644
index 000000000..f3fbf41f4
--- /dev/null
+++ b/number-of-islands/hi-rachel.py
@@ -0,0 +1,72 @@
+"""
+https://leetcode.com/problems/number-of-islands/
+
+Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return the number of islands.
+
+An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
+
+
+Example 1:
+Input: grid = [
+  ["1","1","1","1","0"],
+  ["1","1","0","1","0"],
+  ["1","1","0","0","0"],
+  ["0","0","0","0","0"]
+]
+Output: 1
+
+Example 2:
+Input: grid = [
+  ["1","1","0","0","0"],
+  ["1","1","0","0","0"],
+  ["0","0","1","0","0"],
+  ["0","0","0","1","1"]
+]
+Output: 3
+
+Constraints:
+m == grid.length
+n == grid[i].length
+1 <= m, n <= 300
+grid[i][j] is '0' or '1'.
+
+BFS 풀이
+TC: O(m * n)
+SC: O(m * n)
+"""
+
+from collections import deque
+from typing import List
+
+class Solution:
+    def numIslands(self, grid: List[List[str]]) -> int:
+        ans = 0
+        dx = [-1, 1, 0, 0]
+        dy = [0, 0, -1, 1]
+        rows = len(grid)
+        cols = len(grid[0])
+
+        def bfs(x, y):
+            queue = deque()
+            queue.append((x, y))
+            grid[x][y] = 2
+
+            while queue:
+                x, y = queue.popleft()
+                for i in range(4):
+                    nx = x + dx[i]
+                    ny = y + dy[i]
+                    if 0 <= nx < rows and 0 <= ny < cols:
+                        if grid[nx][ny] == "1":
+                            grid[nx][ny] = "2"
+                            queue.append((nx, ny))
+
+            return True
+
+        for i in range(rows):
+            for j in range(cols):
+                if grid[i][j] == "1":
+                    if bfs(i, j):
+                        ans += 1
+
+        return ans
diff --git a/reverse-linked-list/hi-rachel.py b/reverse-linked-list/hi-rachel.py
index 07d94f9c5..c5eeacec1 100644
--- a/reverse-linked-list/hi-rachel.py
+++ b/reverse-linked-list/hi-rachel.py
@@ -1,10 +1,6 @@
-# Definition for singly-linked list.
-# class ListNode:
-#     def __init__(self, val=0, next=None):
-#         self.val = val
-#         self.next = next
-
 """
+https://leetcode.com/problems/reverse-linked-list/
+
 1. Stack 활용 (LIFO)
 - LinkedList의 모든 원소를 Stack에 넣고 꺼냄
 TC: O(n) time
@@ -16,6 +12,12 @@
 SC: O(1) -> 변수를 포인터 2개만 사용
 """
 
+# Definition for singly-linked list.
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
 # Stack 풀이
 class Solution:
     def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
@@ -42,3 +44,27 @@ def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
             curr.next = prev
             prev, curr = curr, temp_next
         return prev
+
+
+"""
+25/9/5 복습
+
+링크드 리스트 뒤집기
+=> 노드를 옮기는 것이 아니라, next 포인터의 방향을 바꾸면 된다!
+
+TC: O(n)
+SC: O(1)
+"""
+class Solution:
+    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        prev = None  # 이전 노드
+        curr = head  # 현재 노드
+
+        while curr:
+            next_node = curr.next  # 다음 노드 기억
+            curr.next = prev       # 현재 노드의 방향을 반대로
+            prev = curr            # 이전 노드를 한 칸 앞으로
+            curr = next_node       # 현재 노드를 다음으로 이동
+
+        # prev가 마지막 노드이자 새 head
+        return prev
diff --git a/set-matrix-zeroes/hi-rachel.py b/set-matrix-zeroes/hi-rachel.py
new file mode 100644
index 000000000..ca93bfa17
--- /dev/null
+++ b/set-matrix-zeroes/hi-rachel.py
@@ -0,0 +1,33 @@
+"""
+https://leetcode.com/problems/set-matrix-zeroes/description/
+
+Given an m x n integer matrix matrix, if an element is 0, set its entire row and column to 0's.
+You must do it in place.
+
+TC: O(m * n)
+SC: O(m + n), set 자료구조 사용
+"""
+
+class Solution:
+    def setZeroes(self, matrix: List[List[int]]) -> None:
+        """
+        Do not return anything, modify matrix in-place instead.
+        """
+        rows = len(matrix)
+        cols = len(matrix[0])
+
+        zero_rows = set() # 최대 m개의 행 인덱스 저장
+        zero_cols = set() # 최대 n개의 열 인덱스 저장
+
+        # 0이 있는 위치 찾기
+        for i in range(rows):
+            for j in range(cols):
+                if matrix[i][j] == 0:
+                    zero_rows.add(i)
+                    zero_cols.add(j)
+
+        # 0이 있는 행과 열 모두 0으로 만들기
+        for i in range(rows):
+            for j in range(cols):
+                if i in zero_rows or j in zero_cols:
+                    matrix[i][j] = 0
diff --git a/unique-paths/hi-rachel.py b/unique-paths/hi-rachel.py
new file mode 100644
index 000000000..d095906f4
--- /dev/null
+++ b/unique-paths/hi-rachel.py
@@ -0,0 +1,51 @@
+"""
+https://leetcode.com/problems/unique-paths/description/
+
+아래로 이동 혹은 (1, 0)
+오른쪽 이동만 가능 (0, 1)
+
+m => rows, n = cols
+로봇이 (0, 0)에서 (m-1, n-1)에 도착 가능한 unique paths 개수를 반환
+
+풀이 시간: 16분
+처음에 어떻게 풀어야 할 줄 몰랐지만, 그림을 그려보며 누적 규칙을 찾음 (위, 왼쪽 값 더해나가기)
+paths[i][j] = paths[i-1][j] + paths[i][j-1]
+
+TC: O(m * n)
+SC: O(m * n)
+"""
+
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        paths = [[0] * n for _ in range(m)]
+        paths[0][0] = 1
+
+        for i in range(m):
+            for j in range(n):
+                if i - 1 >= 0 and j - 1 >= 0:
+                    paths[i][j] = paths[i - 1][j] + paths[i][j - 1]
+                else:
+                    paths[i][j] = 1
+
+        return paths[m - 1][n - 1]
+
+
+"""
+공간 복잡도 최적화 풀이 - 복습 필요
+dp[i][j] = dp[i-1][j] + dp[i][j-1]
+=> dp[j] = dp[j] + dp[j-1]
+
+TC: O(m * n)
+SC: O(n)
+"""
+
+class Solution:
+    def uniquePaths(self, m: int, n: int) -> int:
+        # 첫 행은 모두 1로 초기화
+        dp = [1] * n
+
+        for i in range(1, m):
+            for j in range(1, n):
+                dp[j] = dp[j] + dp[j - 1]
+
+        return dp[-1]