[변지협] Day22

leetcode3/변지협/v3/835. Image Overlap.py

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+'''
+1. 아이디어:
+문제에 나와있는대로 img1을 전부 하나씩 이동하면서 최대값을 찾음.
+2. 시간복잡도:
+o(m*2*n*2*(m*n*2))
+3. 알고리즘:
+완전탐색
+'''
+class Solution:
+    def move(self, img, dx, dy):
+        m = len(img[0])
+        n = len(img)
+
+        ones = []
+        for y in range(n):
+            for x in range(m):
+                if img[y][x] == 1:
+                    ones.append((x,y))
+
+        tmp = [[0] * m for _ in range(n)]
+        while ones:
+            x,y = ones.pop()
+
+            if 0 <= x+dx < m and 0 <= y+dy < n:
+                tmp[y+dy][x+dx] = 1
+
+        return tmp
+
+    def overlay(self, lst1, lst2):
+        n = len(lst1[0])
+        m = len(lst1)
+        tmp = 0
+        for y in range(m):
+            for x in range(n):
+                if lst1[y][x] == 1 and lst2[y][x] == 1:
+                    tmp += 1
+
+        return tmp 
+
+    def largestOverlap(self, img1: list[list[int]], img2: list[list[int]]) -> int:
+        print(self.overlay(self.move(img1,1,1), img2))
+
+        n = len(img1[0])
+        m = len(img1)
+        tmp = 0
+        for dx in range(-m, m +1):
+            for dy in range(-n, n+ 1):
+                num = self.overlay(self.move(img1,dx,dy), img2)
+                if num > tmp:
+                    tmp = num
+
+        return tmp        

leetcode3/변지협/v3/860. Lemonade Change.py

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+'''
+1. 아이디어:
+문제에 나와있는대로 구현하면 됨.
+2. 시간복잡도:
+o(n)
+3. 알고리즘:
+'''
+class Solution:
+    def lemonadeChange(self, bills: list[int]) -> bool:
+        wallet = {5: 0, 10: 0}
+
+        for bill in bills:
+            if bill == 5:
+                wallet[5] += 1
+            elif bill == 10:
+                wallet[5] -= 1
+                wallet[10] += 1
+
+                if wallet[5] == -1:
+                    return False
+            else:
+                if wallet[10] >= 1 and wallet[5] >= 1:
+                    wallet[10] -= 1
+                    wallet[5] -= 1
+                elif wallet[5] >= 3:
+                    wallet[5] -= 3
+                else:
+                    return False
+
+        return True
+