Create 1444. Number of Ways of Cutting a Pizza 31 march

1444. Number of Ways of Cutting a Pizza

Author: Chayandas07

1444. Number of Ways of Cutting a Pizza 31 march

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+class Solution {
+    private:
+    static constexpr int kMod = 1'000'000'007;
+    vector<vector<vector<int>>> dp;
+    vector<vector<int>> prefix;
+
+    public:
+    int ways(vector<string>& pizza, int k) {
+        const int M = pizza.size();
+        const int N = pizza[0].size();
+        // dp[m][n][k] := # of ways to cut pizza[m:M][n:N] w/ k cuts
+        dp.resize(M, vector<vector<int>>(N, vector<int>(k, -1)));
+        prefix.resize(M + 1, vector<int>(N + 1));
+
+        for (int i = 0; i < M; ++i)
+        for (int j = 0; j < N; ++j)
+            prefix[i + 1][j + 1] = (pizza[i][j] == 'A') + prefix[i][j + 1] +
+                                prefix[i + 1][j] - prefix[i][j];
+
+        return ways(0, 0, k - 1, M, N);
+    }
+
+    int ways(int m, int n, int k, const int M, const int N) {
+        if (k == 0)
+        return 1;
+        if (dp[m][n][k] >= 0)
+        return dp[m][n][k];
+
+        dp[m][n][k] = 0;
+
+        for (int i = m + 1; i < M; ++i)  // Cut horizontally
+        if (hasApple(m, i, n, N) && hasApple(i, M, n, N))
+            dp[m][n][k] = (dp[m][n][k] + ways(i, n, k - 1, M, N)) % kMod;
+
+        for (int j = n + 1; j < N; ++j)  // Cut vertically
+        if (hasApple(m, M, n, j) && hasApple(m, M, j, N))
+            dp[m][n][k] = (dp[m][n][k] + ways(m, j, k - 1, M, N)) % kMod;
+
+        return dp[m][n][k];
+    }
+    // HasApple of pizza[row1..row2)[col1..col2)
+    bool hasApple(int row1, int row2, int col1, int col2) {
+        return (prefix[row2][col2] - prefix[row1][col2] - prefix[row2][col1] +
+                prefix[row1][col1]) > 0;
+    };
+};