diff --git a/3sum/devyulbae.py b/3sum/devyulbae.py
new file mode 100644
index 0000000000..a43f85994d
--- /dev/null
+++ b/3sum/devyulbae.py
@@ -0,0 +1,40 @@
+"""
+Blind 75 - LeetCode Problem 15. 3Sum
+https://leetcode.com/problems/3sum/
+시간복잡도 O(n^2)
+"""
+from typing import List
+
+class Solution:
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+        nums.sort()
+        answer = []
+        n = len(nums)
+
+        for i in range(n):
+            if i > 0 and nums[i] == nums[i - 1]:
+                continue  # 중복 원소 스킵
+
+            left, right = i + 1, n - 1
+            while left < right:
+                total = nums[i] + nums[left] + nums[right]
+
+                if total == 0:
+                    answer.append([nums[i], nums[left], nums[right]])
+                    left += 1
+                    right -= 1
+
+                    # 중복값 건너뛰기
+                    while left < right and nums[left] == nums[left - 1]:
+                        left += 1
+                    while left < right and nums[right] == nums[right + 1]:
+                        right -= 1
+
+                elif total < 0:
+                    left += 1  # 합이 작으니 left를 키워야 함
+
+                else:
+                    right -= 1  # 합이 크니 right를 줄여야 함
+
+        return answer
+
diff --git a/climbing-stairs/devyulbae.py b/climbing-stairs/devyulbae.py
new file mode 100644
index 0000000000..a254f3e4fb
--- /dev/null
+++ b/climbing-stairs/devyulbae.py
@@ -0,0 +1,25 @@
+"""
+Blind 75 - LeetCode Problem 70. Climbing Stairs
+https://leetcode.com/problems/climbing-stairs/
+
+계단이 n개, 한 번에 1칸 또는 2칸씩 오를 수 있다.
+combinations를 썼더니 시간초과 발생 (O(2^n))
+
+DP를 이용해 시간복잡도 O(n) 공간복잡도 O(1)로 풀기
+f(n) = f(n-1) + f(n-2)
+"""
+class Solution:
+    def climbStairs(self, n: int) -> int:
+        if n <= 2:
+            return n
+        
+        one_before = 2 # n=2
+        two_before = 1 # n=1
+
+        for _ in range(3, n+1):
+            current = one_before + two_before # f(1) = f(n-1) + f(n-2)
+            two_before = one_before # f 밀기
+            one_before = current # f 밀기
+        
+        return current
+    
diff --git a/product-of-array-except-self/devyulbae.py b/product-of-array-except-self/devyulbae.py
new file mode 100644
index 0000000000..78e1875fde
--- /dev/null
+++ b/product-of-array-except-self/devyulbae.py
@@ -0,0 +1,38 @@
+"""
+Blind75 - 4. Product of Array Except Self
+https://leetcode.com/problems/product-of-array-except-self/
+조건 : 
+- 분할 없이 O(n) 시간 복잡도 이하로 풀어라
+- nums의 모든 원소의 곱은 32비트 정수 범위 내에 들어온다
+- 나눗셈 금지
+
+0이 몇개인가?
+- 2개 이상 : 모두 0인 배열
+- 1개 : 0 본인 제외는 0인 배열
+- 0개 : 좌 우 각각의 메모이제이션을 해주면 O(2n)으로 배열 생성, O(n)으로 정답 배열 생성하므로 O(n)
+"""
+from typing import List
+
+class Solution:
+    def productExceptSelf(self, nums: List[int]) -> List[int]:
+        zero_count = nums.count(0)
+        if zero_count > 1:
+            return [0] * len(nums)
+        elif zero_count == 1:
+            all_product = 1
+            for num in nums:
+                all_product *= num if num != 0 else 1
+            return [0 if num!= 0 else all_product for num in nums]
+            
+        else:
+            left_product = [nums[0]]
+            right_product = [nums[-1]]
+            for i in range(1,len(nums)): 
+                left_product.append(left_product[i-1]*nums[i])
+                right_product.append(right_product[i-1]*nums[-1-i])
+            
+            answer = []
+            for i in range(len(nums)):
+                answer.append((left_product[i-1] if i>0 else 1)* (right_product[-2-i] if i < len(nums)-1 else 1))
+            return answer
+        
diff --git a/valid-anagram/devyulbae.py b/valid-anagram/devyulbae.py
new file mode 100644
index 0000000000..c68d75c85d
--- /dev/null
+++ b/valid-anagram/devyulbae.py
@@ -0,0 +1,15 @@
+"""
+Blind 75 - LeetCode Problem 242: Valid Anagram
+https://leetcode.com/problems/valid-anagram/
+
+t가 s의 애너그램인지 확인하기
+"""
+from collections import Counter
+
+class Solution:
+    def isAnagram(self, s: str, t: str) -> bool:
+        if len(s) != len(t):
+            return False
+
+        return Counter(s) == Counter(t)
+
diff --git a/validate-binary-search-tree/devyulbae.py b/validate-binary-search-tree/devyulbae.py
new file mode 100644
index 0000000000..8943c9eef5
--- /dev/null
+++ b/validate-binary-search-tree/devyulbae.py
@@ -0,0 +1,35 @@
+"""
+Blind 75 - LeetCode Problem 98: Validate Binary Search Tree
+https://leetcode.com/problems/validate-binary-search-tree/
+
+중위 순회 - 왼쪽 서브트리 -> 루트 -> 오른쪽 서브트리
+중위 순회한 값이 오름차순인지 확인하면 된다.
+"""
+from typing import Optional, List
+
+# Definition for a binary tree node.
+class TreeNode:
+    def __init__(self, val=0, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
+
+class Solution:    # root = [2,1,3] answer = True 
+    def traverse(self, node: Optional[TreeNode], inorder: List[int]) -> None:
+        if not node:
+            return
+        self.traverse(node.left, inorder)
+        inorder.append(node.val)
+        self.traverse(node.right, inorder)
+
+
+    def isValidBST(self, root: Optional[TreeNode]) -> bool:
+        inorder = []
+        self.traverse(root, inorder)
+
+        for i in range(1, len(inorder)):
+            if inorder[i] <= inorder[i-1]:
+                return False
+            
+        return True
+