From 9f530a916fe36f6812f83e4386bc936e53849298 Mon Sep 17 00:00:00 2001
From: Won Joon Thomas Choi <113500771+724thomas@users.noreply.github.com>
Date: Tue, 22 Sep 2026 22:27:04 +0900
Subject: [PATCH 1/2] Create 860. Lemonade Change.py

---
 .../860. Lemonade Change.py"                  | 44 +++++++++++++++++++
 1 file changed, 44 insertions(+)
 create mode 100644 "leetcode3/\354\265\234\354\233\220\354\244\200/860. Lemonade Change.py"

diff --git "a/leetcode3/\354\265\234\354\233\220\354\244\200/860. Lemonade Change.py" "b/leetcode3/\354\265\234\354\233\220\354\244\200/860. Lemonade Change.py"
new file mode 100644
index 00000000..5f32513d
--- /dev/null
+++ "b/leetcode3/\354\265\234\354\233\220\354\244\200/860. Lemonade Change.py"	
@@ -0,0 +1,44 @@
+#
+
+'''
+1. 아이디어 :
+-
+
+2. 시간복잡도 :
+    O(n)
+
+3. 자료구조/알고리즘 :
+dict
+
+'''
+class Solution:
+    def lemonadeChange(self, bills: list[int]) -> bool:
+        count = {5:0, 10:0, 20:0} #5, 10, 20
+
+        def is_possible(payment):
+            if payment == 20:
+                if count[10] >= 1 and count[5] >= 1:
+                    count[10] -= 1
+                    count[5] -= 1
+                    count[20] += 1
+                    return True
+                elif count[10] == 0 and count[5] >= 3:
+                    count[5] -= 3
+                    count[20] += 1
+                    return True
+            elif payment == 10:
+                if count[5] >= 1:
+                    count[5] -= 1
+                    count[10] += 1
+                    return True
+            elif payment == 5:
+                count[5] += 1
+                return True
+            
+            return False
+
+        for b in bills:
+            if not is_possible(b):
+                return False
+        return True
+                    

From 7b72c4fb8015729b5a7516f7091276f41e592334 Mon Sep 17 00:00:00 2001
From: Won Joon Thomas Choi <113500771+724thomas@users.noreply.github.com>
Date: Tue, 22 Sep 2026 22:28:02 +0900
Subject: [PATCH 2/2] Create 835. Image Overlap.py

---
 .../835. Image Overlap.py"                    | 42 +++++++++++++++++++
 1 file changed, 42 insertions(+)
 create mode 100644 "leetcode3/\354\265\234\354\233\220\354\244\200/835. Image Overlap.py"

diff --git "a/leetcode3/\354\265\234\354\233\220\354\244\200/835. Image Overlap.py" "b/leetcode3/\354\265\234\354\233\220\354\244\200/835. Image Overlap.py"
new file mode 100644
index 00000000..ee74c4bd
--- /dev/null
+++ "b/leetcode3/\354\265\234\354\233\220\354\244\200/835. Image Overlap.py"	
@@ -0,0 +1,42 @@
+#
+
+'''
+1. 아이디어 :
+한칸씩 옮긴다음 계산한다.
+
+2. 시간복잡도 :
+    O(n**4)
+
+3. 자료구조/알고리즘 :
+-
+
+'''
+class Solution:
+    def largestOverlap(self, img1: list[list[int]], img2: list[list[int]]) -> int:
+        n = len(img1)
+        
+        def get_count(offset_x, offset_y):
+            count = 0
+
+            for x in range(n):
+                for y in range(n):
+                    mx = x + offset_x
+                    my = y + offset_y
+                    if 0<=mx<n and 0<=my<n and img1[x][y] and img2[mx][my]:
+                        count+=1
+                        
+            return count
+        
+        # def get_count(img1, img2):
+        #     count=0
+        #     for x in range(n):
+        #         for y in range(n):
+        #             if img1[x][y] and img2[x][y]:
+        #                 count+=1
+        #     return count
+        
+        ans = 0
+        for i in range(-n, n):
+            for j in range(-n, n):
+                ans = max(ans, get_count(i, j))
+        return ans