From 3977b20c6b06371c5ac8c99784605ea49c935713 Mon Sep 17 00:00:00 2001
From: Won Joon Thomas Choi <113500771+724thomas@users.noreply.github.com>
Date: Tue, 27 Oct 2026 22:33:14 +0900
Subject: [PATCH 1/2] Create 4070. Minimum Rotations to Dial a Number I.py

---
 .../4070. Minimum Rotations to Dial a Number I.py"   | 12 ++++++++++++
 1 file changed, 12 insertions(+)
 create mode 100644 "leetcode3/\354\265\234\354\233\220\354\244\200/4070. Minimum Rotations to Dial a Number I.py"

diff --git "a/leetcode3/\354\265\234\354\233\220\354\244\200/4070. Minimum Rotations to Dial a Number I.py" "b/leetcode3/\354\265\234\354\233\220\354\244\200/4070. Minimum Rotations to Dial a Number I.py"
new file mode 100644
index 00000000..3456d015
--- /dev/null
+++ "b/leetcode3/\354\265\234\354\233\220\354\244\200/4070. Minimum Rotations to Dial a Number I.py"	
@@ -0,0 +1,12 @@
+class Solution:
+    def minRotations(self, s: str) -> int:
+        ans = 0
+        curr = 0
+
+        for next in s:
+            num = int(next)
+
+            diff = abs(curr-num)
+            ans += min(diff, 10-diff)
+            curr = num
+        return ans

From ff72bbdefdc17b190be87feb4504a539d733b027 Mon Sep 17 00:00:00 2001
From: Won Joon Thomas Choi <113500771+724thomas@users.noreply.github.com>
Date: Tue, 27 Oct 2026 22:35:26 +0900
Subject: [PATCH 2/2] Create 3613. Minimize Maximum Component Cost.py

---
 .../3613. Minimize Maximum Component Cost.py" | 59 +++++++++++++++++++
 1 file changed, 59 insertions(+)
 create mode 100644 "leetcode3/\354\265\234\354\233\220\354\244\200/3613. Minimize Maximum Component Cost.py"

diff --git "a/leetcode3/\354\265\234\354\233\220\354\244\200/3613. Minimize Maximum Component Cost.py" "b/leetcode3/\354\265\234\354\233\220\354\244\200/3613. Minimize Maximum Component Cost.py"
new file mode 100644
index 00000000..2a0a57df
--- /dev/null
+++ "b/leetcode3/\354\265\234\354\233\220\354\244\200/3613. Minimize Maximum Component Cost.py"	
@@ -0,0 +1,59 @@
+#
+
+'''
+1. 아이디어 :
+아무것도 연결하지 않은채로 시작.
+edge를 가장 작은것부터 연결한다.
+k개의 components일때까지 연결한다. (자연스럽게 가장 큰것들은 0이 됨)
+
+2. 시간복잡도 :
+    O(n)
+
+3. 자료구조/알고리즘 :
+union-find
+
+'''
+class Solution:
+    def minCost(self, n: int, edges: List[List[int]], k: int) -> int:
+        #union find
+        # edges.sort(key=lambda x: x[2])
+        # for i in range(k-1):
+        #     edges.pop()
+        # return edges[-1][2] if edges else 0
+
+        def find(x):
+            if par[x] != x:
+                par[x] = find(par[x])
+            return par[x]
+        
+        def union(a, b):
+            a = find(a)
+            b = find(b)
+
+            if a == b:
+                return False
+            
+            if size[a] < size[b]:
+                a, b = b, a
+            
+            par[b] = a
+            size[a] += size[b]
+            return True
+        
+        par = [i for i in range(n)]
+        size = [1] * n
+
+        edges.sort(key=lambda x: x[2])
+
+        components = n
+        ans = 0
+
+        for u, v, w in edges:
+            if components<=k:
+                break
+            
+            if union(u, v):
+                components -= 1
+                ans = w
+        
+        return ans if components <= k else -1