Complete Leetcode 1444

Author: yuchia0221

Backtracking/1444-NumberofWaysofCuttingaPizza/README.md

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+# Number of Ways of Cutting a Pizza
+
+Problem can be found in [here](https://leetcode.com/problems/number-of-ways-of-cutting-a-pizza/)!
+
+### [Solution](/Backtracking/1444-NumberofWaysofCuttingaPizza/solution.py): Backtracking + Prefix Sum + Dyanmic Programming
+
+```python
+def ways(pizza: List[str], k: int) -> int:
+    row_num, col_num = len(pizza), len(pizza[0])
+    remaining_graph = [[0] * (col_num+1) for _ in range(row_num+1)]
+    for row in range(row_num-1, -1, -1):
+        for col in range(col_num-1, -1, -1):
+            remaining_graph[row][col] = remaining_graph[row+1][col] + \
+                remaining_graph[row][col+1] - remaining_graph[row+1][col+1] + (pizza[row][col] == "A")
+
+    @cache
+    def backtracking(row: int, col: int, cut_remained: int) -> int:
+        if remaining_graph[row][col] == 0:
+            return 0
+        if cut_remained == 0:
+            return 1
+
+        answer = 0
+        for new_row in range(row+1, row_num):
+            if remaining_graph[row][col] - remaining_graph[new_row][col] > 0:
+                answer += backtracking(new_row, col, cut_remained-1)
+
+        for new_col in range(col+1, col_num):
+            if remaining_graph[row][col] - remaining_graph[row][new_col] > 0:
+                answer += backtracking(row, new_col, cut_remained-1)
+
+        return answer
+
+    return backtracking(0, 0, k-1) % (pow(10, 9)+7)
+```
+
+Time Complexity: ![O(mnk(m+n))](<https://latex.codecogs.com/svg.image?\inline&space;O(mnk(m+n))>), Space Complexity: ![O(nmk)](<https://latex.codecogs.com/svg.image?\inline&space;O(nmk)>), where m and n is the length of row and column, respecitvely.

Backtracking/1444-NumberofWaysofCuttingaPizza/solution.py

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+from functools import cache
+from typing import List
+
+
+class Solution:
+    def ways(self, pizza: List[str], k: int) -> int:
+        row_num, col_num = len(pizza), len(pizza[0])
+        remaining_graph = [[0] * (col_num+1) for _ in range(row_num+1)]
+        for row in range(row_num-1, -1, -1):
+            for col in range(col_num-1, -1, -1):
+                remaining_graph[row][col] = remaining_graph[row+1][col] + \
+                    remaining_graph[row][col+1] - remaining_graph[row+1][col+1] + (pizza[row][col] == "A")
+
+        @cache
+        def backtracking(row: int, col: int, cut_remained: int) -> int:
+            if remaining_graph[row][col] == 0:
+                return 0
+            if cut_remained == 0:
+                return 1
+
+            answer = 0
+            for new_row in range(row+1, row_num):
+                if remaining_graph[row][col] - remaining_graph[new_row][col] > 0:
+                    answer += backtracking(new_row, col, cut_remained-1)
+
+            for new_col in range(col+1, col_num):
+                if remaining_graph[row][col] - remaining_graph[row][new_col] > 0:
+                    answer += backtracking(row, new_col, cut_remained-1)
+
+            return answer
+
+        return backtracking(0, 0, k-1) % (pow(10, 9)+7)