From a05df43c52795751777ba5b086cd05cdd8f44c66 Mon Sep 17 00:00:00 2001
From: Won Joon Thomas Choi <113500771+724thomas@users.noreply.github.com>
Date: Tue, 15 Sep 2026 22:35:40 +0900
Subject: [PATCH 1/2] Create 70. Climbing Stairs.py

---
 .../70. Climbing Stairs.py"                   | 24 +++++++++++++++++++
 1 file changed, 24 insertions(+)
 create mode 100644 "leetcode3/\354\265\234\354\233\220\354\244\200/70. Climbing Stairs.py"

diff --git "a/leetcode3/\354\265\234\354\233\220\354\244\200/70. Climbing Stairs.py" "b/leetcode3/\354\265\234\354\233\220\354\244\200/70. Climbing Stairs.py"
new file mode 100644
index 00000000..52607edc
--- /dev/null
+++ "b/leetcode3/\354\265\234\354\233\220\354\244\200/70. Climbing Stairs.py"	
@@ -0,0 +1,24 @@
+#
+
+'''
+1. 아이디어 :
+x에 도달하기 위해서는 x-1에서 1칸 점프 + x-2에서 1칸 점프이므로, f(x) = f(x-1) + f(x-2)
+길이 2의 배열을 활용하여 공간 절약
+
+2. 시간복잡도 :
+    O(n)
+
+3. 자료구조/알고리즘 :
+dp
+
+'''
+class Solution:
+    def climbStairs(self, n: int) -> int:
+        if n == 1:
+            return n
+
+        ans = [1,2]
+        for i in range(3, n+1):
+            ans = [ans[1],ans[0] + ans[1]]
+        return ans[1]
+

From bff2a7592d38bd52f8ebe1ffba86f03c78f7199b Mon Sep 17 00:00:00 2001
From: Won Joon Thomas Choi <113500771+724thomas@users.noreply.github.com>
Date: Tue, 15 Sep 2026 22:35:54 +0900
Subject: [PATCH 2/2] Create 357. Count Numbers with Unique Digits.py

---
 .../357. Count Numbers with Unique Digits.py" | 42 +++++++++++++++++++
 1 file changed, 42 insertions(+)
 create mode 100644 "leetcode3/\354\265\234\354\233\220\354\244\200/357. Count Numbers with Unique Digits.py"

diff --git "a/leetcode3/\354\265\234\354\233\220\354\244\200/357. Count Numbers with Unique Digits.py" "b/leetcode3/\354\265\234\354\233\220\354\244\200/357. Count Numbers with Unique Digits.py"
new file mode 100644
index 00000000..c40904da
--- /dev/null
+++ "b/leetcode3/\354\265\234\354\233\220\354\244\200/357. Count Numbers with Unique Digits.py"	
@@ -0,0 +1,42 @@
+#
+
+'''
+1. 아이디어 :
+백트래킹으로 모든 경우를 구한다.
+현재 숫자에 방문하지 않은 숫자를 더한다.
+
+2. 시간복잡도 :
+    약 O(10 ** 8)
+
+3. 자료구조/알고리즘 :
+backtracking
+
+'''
+class Solution:
+    def countNumbersWithUniqueDigits(self, n: int) -> int:
+        visited = set()
+    
+        def backtrack(length, total_length):
+            if length == total_length:
+                return 1
+
+            total = 0
+
+            for i in range(10):
+                if length == 0 and i ==0:
+                    continue # 0
+                
+                if i in visited:
+                    continue
+                
+                visited.add(i)
+                total += backtrack(length+1, total_length)
+                visited.remove(i)
+
+            return total
+        
+        ans = 0
+        for length in range(n+1):
+            ans += backtrack(0, length)
+        return ans
+