diff --git a/3sum/heypaprika.py b/3sum/heypaprika.py
new file mode 100644
index 000000000..dadc7eacf
--- /dev/null
+++ b/3sum/heypaprika.py
@@ -0,0 +1,30 @@
+# 2Sum을 활용한 풀이
+# Note : 정렬 후 푸는 것이 더 직관적이고, 빠름
+
+"""
+복잡도 : 예상 -> 예상한 이유
+
+시간 복잡도 : O(n^2) -> nums 배열 2중 for문
+공간 복잡도 : O(n) -> 최악의 경우 nums 배열 길이만큼의 딕셔너리 생성
+"""
+class Solution:
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+        ans_list = set()
+        added_pair = {}
+        for target_i, target_number in enumerate(nums):
+            cur_ans_list = []
+            num_dict = {}
+            if target_number in added_pair:
+                continue
+
+            for i in range(target_i + 1, len(nums)):
+                num_A = nums[i]
+                num_B = - target_number - num_A
+                if num_B in num_dict:
+                    cur_ans_list.append(sorted([target_number, num_A, num_B]))
+                    added_pair[target_number] = num_A
+                num_dict[num_A] = 1
+            for item in cur_ans_list:
+                ans_list.add(tuple(item))
+        return list(ans_list)
+
diff --git a/construct-binary-tree-from-preorder-and-inorder-traversal/heypaprika.py b/construct-binary-tree-from-preorder-and-inorder-traversal/heypaprika.py
new file mode 100644
index 000000000..675927a0e
--- /dev/null
+++ b/construct-binary-tree-from-preorder-and-inorder-traversal/heypaprika.py
@@ -0,0 +1,59 @@
+# 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
+
+# preorder -> root, left, right
+# inorder -> left, root, right
+
+# Ex1 [3,9,20,null,null,15,7]
+# preorder -> root-1, left, (root-2 - root, left, right)
+# inorder -> left, root-1, (root-2 - left, root, right)
+
+# Ex2 [-1]
+# preorder -> root, left(x), right(x)
+# inorder -> left(x), root, right(x)
+
+# Ex3 [3,9,20,11,8,15,7]
+# preorder -> root-1, (root-2(left) - root, left, right), (root-3(right) - root, left, right)
+# [3,9,11,8,20,15,7]
+# inorder -> (root-1(left) - left, root, right), root-2, (root-3(right) - left, root, right)
+# [11, 9, 8, 3, 15, 20, 7]
+
+# Ex4 [3,9,20,11,8]
+# preorder -> root-1, (root-2(left) - root, left, right), (root-3(right) - root)
+# [3,9,11,8,20]
+# inorder -> (root-1(left) - left, root, right), root-2, (root-3(right) - left(X), root)
+# [11, 9, 8, 3, 20]
+
+# 문제풀이 : divide and conquer
+# preorder의 첫번째 요소를 가지고 inorder split
+# split된 left와 right의 개수에 맞게 preorder을 두 번 째 원소부터 순서대로 할당
+# 예) left 원소 개수 : 3 -> preorder[1:1+3], right 원소 개수 : 2 -> preorder[1+3:]
+# left, right 각각 buildTree로 넣기
+
+"""
+복잡도 : 예상 -> 예상한 이유
+
+시간 복잡도 : O(n) -> 분할해서 처리하지만, 모든 노드를 방문하므로
+공간 복잡도 : O(n) -> 입력 배열만큼의 길이와 거의 동일한 배열이 생성되므로
+"""
+class Solution:
+    def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
+        if len(preorder) == 0:
+            return None
+        if len(preorder) == 1:
+            return TreeNode(preorder[0])
+        root = TreeNode(preorder[0])
+
+        split_idx = inorder.index(preorder[0])
+        left_inorder = inorder[:split_idx]
+        right_inorder = inorder[split_idx+1:]
+        left_preorder = preorder[1:len(left_inorder)+1]
+        right_preorder = preorder[len(left_inorder)+1:]
+        root.left = self.buildTree(left_preorder, left_inorder)
+        root.right = self.buildTree(right_preorder, right_inorder)
+        return root
+
diff --git a/decode-ways/heypaprika.py b/decode-ways/heypaprika.py
new file mode 100644
index 000000000..c850e07c9
--- /dev/null
+++ b/decode-ways/heypaprika.py
@@ -0,0 +1,20 @@
+"""
+복잡도 : 예상 -> 예상한 이유
+
+시간 복잡도 : O(n) -> 배열의 길이 만큼 반복하므로
+공간 복잡도 : O(n) -> n+1 길이의 배열 하나를 생성하므로
+"""
+class Solution:
+    def numDecodings(self, s: str) -> int:
+        if s[0] == '0':
+            return 0
+        a = [0] * (len(s)+1)
+        a[0] = 1
+        a[1] = 1
+        for i in range(2, len(s)+1):
+            if 1 <= int(s[i-1]) <= 9:
+                a[i] += a[i-1]
+            if 10 <= int(s[i-2:i]) <= 26:
+                a[i] += a[i-2]
+        return a[-1]
+