From 7528c20cfc555e6785405eaa1df08f2f1584c826 Mon Sep 17 00:00:00 2001
From: rimogsu <newnyup@gmail.com>
Date: Tue, 13 Oct 2026 22:30:20 +0900
Subject: [PATCH] 1094,4061

---
 .../v3/1094. Car Pooling.py"                  | 32 +++++++++
 ...1. Minimum Queen Moves to Reach Target.py" | 69 +++++++++++++++++++
 2 files changed, 101 insertions(+)
 create mode 100644 "leetcode3/\353\263\200\354\247\200\355\230\221/v3/1094. Car Pooling.py"
 create mode 100644 "leetcode3/\353\263\200\354\247\200\355\230\221/v3/4061. Minimum Queen Moves to Reach Target.py"

diff --git "a/leetcode3/\353\263\200\354\247\200\355\230\221/v3/1094. Car Pooling.py" "b/leetcode3/\353\263\200\354\247\200\355\230\221/v3/1094. Car Pooling.py"
new file mode 100644
index 00000000..ea665b51
--- /dev/null
+++ "b/leetcode3/\353\263\200\354\247\200\355\230\221/v3/1094. Car Pooling.py"	
@@ -0,0 +1,32 @@
+'''
+1. 아이디어 :
+1943. Describe the Painting.py
+4개월전에 코딩테스트 참여했을 때 풀었던 문제인데 구조적으로 100% 동일한 문제임.
+from을 +, to를 -로 둬서 해시의 누적합을 구하고, 여기서 누적합이 capacity를 초과하는지 확인하면 된다.
+
+2. 시간복잡도 :
+o(n)
+
+3. 자료구조/알고리즘 :
+- 배열(누적합), 해시맵(탑승/하차 위치 기록)
+'''
+from collections import defaultdict
+class Solution:
+    def carPooling(self, trips: list[list[int]], capacity: int) -> bool:
+        dic = defaultdict(int)
+        
+        for n, _from, to in trips:
+            dic[_from] += n
+            dic[to] -= n
+        
+        print(dic)
+
+        current = 0
+        for i in [value for key, value in sorted(dic.items(), key=lambda x:x[0])]:
+            current += i
+            if current > capacity:
+                return False
+        
+        return True
+
+                
diff --git "a/leetcode3/\353\263\200\354\247\200\355\230\221/v3/4061. Minimum Queen Moves to Reach Target.py" "b/leetcode3/\353\263\200\354\247\200\355\230\221/v3/4061. Minimum Queen Moves to Reach Target.py"
new file mode 100644
index 00000000..a2ede354
--- /dev/null
+++ "b/leetcode3/\353\263\200\354\247\200\355\230\221/v3/4061. Minimum Queen Moves to Reach Target.py"	
@@ -0,0 +1,69 @@
+'''
+1. 아이디어 :
+따져봤을 때 2번 안에 퀸이 다 갈 수 있음.
+0번: src == target
+1번: 퀸의 이동경로 (src!=target)
+2번: 나머지 전부
+
+2. 시간복잡도 :
+o(1) - 경로만 바로 구하면 됨
+
+3. 자료구조/알고리즘 :
+'''
+from collections import deque
+class Solution:
+    def minQueenMoves(self, source: list[int], target: list[int]) -> int:
+        x1,y1 = tuple(source)
+        x2,y2 = tuple(target)
+        if x1 == x2 and y1 == y2:
+            return 0
+        
+        st = set()
+
+        for i in range(1,9):
+            st.add((x1,i))
+            st.add((i,y1))
+        
+        print(st)
+
+        x = x1
+        y = y1
+        while 1 <= x < 9 and 1<= y < 9:
+            xtmp = x+1
+            ytmp = y+1
+            x = xtmp
+            y = ytmp
+            st.add((x,y))
+
+        x = x1
+        y = y1
+        while 1 <= x < 9 and 1<= y < 9:
+            xtmp = x-1
+            ytmp = y-1
+            x = xtmp
+            y = ytmp
+            st.add((x,y))
+        x = x1
+        y = y1
+        
+        while 1 <= x < 9 and 1<= y < 9:
+            xtmp = x+1
+            ytmp = y-1
+            x = xtmp
+            y = ytmp
+            st.add((x,y))
+        x = x1
+        y = y1
+
+        while 1 <= x < 9 and 1<= y < 9:
+            xtmp = x-1
+            ytmp = y+1
+            x = xtmp
+            y = ytmp
+            st.add((x,y))
+        
+        if (x2,y2) in st:
+            return 1
+        else:
+            return 2        
+