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1444. Number of Ways of Cutting a Pizza 👍

  • Time: $O(MNk \cdot (M + N))$
  • Space: $O(MNk)$
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class Solution {
 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);
  }

 private:
  constexpr static int kMod = 1'000'000'007;
  vector<vector<vector<int>>> dp;
  vector<vector<int>> prefix;

  // 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;
  };

  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];
  }
};
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class Solution {
  public int ways(String[] pizza, int k) {
    final int M = pizza.length;
    final int N = pizza[0].length();
    // dp[m][n][k] := # of ways to cut pizza[m:M][n:N] w/ k cuts
    dp = new int[M][N][k];
    for (int[][] A : dp)
      Arrays.stream(A).forEach(a -> Arrays.fill(a, -1));
    prefix = new int[M + 1][N + 1];

    for (int i = 0; i < M; ++i)
      for (int j = 0; j < N; ++j)
        prefix[i + 1][j + 1] = (pizza[i].charAt(j) == 'A' ? 1 : 0) + prefix[i][j + 1] +
                               prefix[i + 1][j] - prefix[i][j];

    return ways(0, 0, k - 1, M, N);
  }

  private static final int kMod = 1_000_000_007;
  private int[][][] dp;
  private int[][] prefix;

  // hasApple of pizza[row1..row2)[col1..col2)
  private boolean hasApple(int row1, int row2, int col1, int col2) {
    return (prefix[row2][col2] - prefix[row1][col2] - prefix[row2][col1] + prefix[row1][col1]) > 0;
  }

  private int ways(int m, int n, int k, final int M, final 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];
  }
}
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class Solution:
  def ways(self, pizza: List[str], k: int) -> int:
    M = len(pizza)
    N = len(pizza[0])
    prefix = [[0] * (N + 1) for _ in range(M + 1)]

    for i in range(M):
      for j in range(N):
        prefix[i + 1][j + 1] = (pizza[i][j] == 'A') + \
            prefix[i][j + 1] + prefix[i + 1][j] - prefix[i][j]

    # hasApple of pizza[row1..row2)[col1..col2)
    def hasApple(row1: int, row2: int, col1: int, col2: int) -> bool:
      return (prefix[row2][col2] - prefix[row1][col2] -
              prefix[row2][col1] + prefix[row1][col1]) > 0

    # dp(m, n, k) := # of ways to cut pizza[m:M][n:N] w/ k cuts
    @functools.lru_cache(None)
    def dp(m: int, n: int, k: int) -> int:
      if k == 0:
        return 1

      ans = 0

      for i in range(m + 1, M):  # cut horizontally
        if hasApple(m, i, n, N) and hasApple(i, M, n, N):
          ans += dp(i, n, k - 1)

      for j in range(n + 1, N):  # cut vertically
        if hasApple(m, M, n, j) and hasApple(m, M, j, N):
          ans += dp(m, j, k - 1)

      return ans

    return dp(0, 0, k - 1) % (10**9 + 7)