Create number-of-ways-of-cutting-a-pizza.py

Author: kamyu104

Python/number-of-ways-of-cutting-a-pizza.py

@@ -0,0 +1,35 @@
+# Time:  O(m * n * k * (m + n))
+# Space: O(m * n * k)
+
+class Solution(object):
+    def ways(self, pizza, k):
+        """
+        :type pizza: List[str]
+        :type k: int
+        :rtype: int
+        """
+        MOD = 10**9+7
+        prefix = [[0]*len(pizza[0]) for _ in xrange(len(pizza))]
+        for j in reversed(xrange(len(pizza[0]))):
+            accu = 0
+            for i in reversed(xrange(len(pizza))):
+                accu += int(pizza[i][j] == 'A')
+                prefix[i][j] = (prefix[i][j+1] if (j+1 < len(pizza[0])) else 0) + accu
+        dp = [[[0]*k for _ in xrange(len(pizza[0]))] for _ in xrange(len(pizza))]
+        for i in reversed(xrange(len(pizza))):
+            for j in reversed(xrange(len(pizza[0]))):
+                dp[i][j][0] = 1
+                for m in xrange(1, k):
+                    for n in xrange(i+1, len(pizza)):
+                        if prefix[i][j] == prefix[n][j]:
+                            continue
+                        if prefix[n][j] == 0:
+                            break
+                        dp[i][j][m] = (dp[i][j][m] + dp[n][j][m-1]) % MOD
+                    for n in xrange(j+1, len(pizza[0])):
+                        if prefix[i][j] == prefix[i][n]:
+                            continue
+                        if prefix[i][n] == 0:
+                            break
+                        dp[i][j][m] = (dp[i][j][m] + dp[i][n][m-1]) % MOD
+        return dp[0][0][k-1]