[최원준] Day12

leetcode3/최원준/799. Champagne Tower.py

@@ -0,0 +1,32 @@
+#
+
+'''
+1. 아이디어 :
+맨위에 poured 만큼 저장하고 1을 뺀 만큼 아래로 나머지를 /2 보냅니다.
+
+2. 시간복잡도 :
+    O(100 * 100)
+
+3. 자료구조/알고리즘 :
+2차원 배열 / DP
+
+'''
+
+
+class Solution:
+    def champagneTower(self, poured: int, query_row: int, query_glass: int) -> float:
+        glasses = [[0.0] * 101 for a in range(101)]
+        glasses[0][0] = poured
+
+        for row in range(query_row):
+            for col in range(row+1):
+                left_over = glasses[row][col]-1
+
+                if left_over > 0:
+                    glasses[row+1][col] += left_over/2
+                    glasses[row+1][col+1] += left_over/2
+
+                    glasses[row][col] = 1
+
+        target = glasses[query_row][query_glass]
+        return target if target < 1 else 1