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2174. Remove All Ones With Row and Column Flips II

Approach 1: Top-down

  • Time: $O(2^{mn} \cdot mn \cdot \max(m, n))$
  • Space: $O(2^{mn})$
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class Solution {
 public:
  int removeOnes(vector<vector<int>>& grid) {
    const int m = grid.size();
    const int n = grid[0].size();
    // dp[i] := min # of operations to remove all 1's from grid
    // Given i representing the state of grid
    dp.resize(1 << m * n, INT_MAX);
    return dfs(encode(grid, m, n), m, n);
  }

 private:
  vector<int> dp;

  int dfs(int mask, int m, int n) {
    if (mask == 0)
      return 0;
    if (dp[mask] < INT_MAX)
      return dp[mask];

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (mask >> i * n + j & 1) {  // grid[i][j] == 1
          int nextMask = mask;
          for (int k = 0; k < n; ++k)  // Set cell in the same row with 0
            nextMask &= ~(1 << i * n + k);
          for (int k = 0; k < m; ++k)  // Set cell in the same col with 0
            nextMask &= ~(1 << k * n + j);
          dp[mask] = min(dp[mask], 1 + dfs(nextMask, m, n));
        }

    return dp[mask];
  }

  int encode(const vector<vector<int>>& grid, int m, int n) {
    int encoded = 0;
    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (grid[i][j])
          encoded |= 1 << i * n + j;
    return encoded;
  }
};

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class Solution {
  public int removeOnes(int[][] grid) {
    final int m = grid.length;
    final int n = grid[0].length;
    // dp[i] := min # of operations to remove all 1's from grid
    // Given i representing the state of grid
    dp = new int[1 << m * n];
    Arrays.fill(dp, Integer.MAX_VALUE);
    return dfs(encode(grid, m, n), m, n);
  }

  private int[] dp;

  private int dfs(int mask, int m, int n) {
    if (mask == 0)
      return 0;
    if (dp[mask] < Integer.MAX_VALUE)
      return dp[mask];

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if ((mask >> i * n + j & 1) == 1) { // grid[i][j] == 1
          int nextMask = mask;
          for (int k = 0; k < n; ++k)
            nextMask &= ~(1 << i * n + k); // Set cell in the same row with 0
          for (int k = 0; k < m; ++k)
            nextMask &= ~(1 << k * n + j);
          dp[mask] = Math.min(dp[mask], 1 + dfs(nextMask, m, n));
        }

    return dp[mask];
  }

  private int encode(int[][] grid, int m, int n) {
    int encoded = 0;
    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (grid[i][j] == 1)
          encoded |= 1 << i * n + j;
    return encoded;
  }
}

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class Solution:
  def removeOnes(self, grid: List[List[int]]) -> int:
    m = len(grid)
    n = len(grid[0])
    # dp[i] := min # Of operations to remove all 1's from grid
    # Given i representing the state of grid

    @functools.lru_cache(None)
    def dp(mask: int) -> int:
      if mask == 0:
        return 0
      ans = math.inf
      for i in range(m):
        for j in range(n):
          if mask >> i * n + j & 1:  # grid[i][j] == 1
            nextMask = mask
            for k in range(n):  # Set cell in the same row with 0
              nextMask &= ~(1 << i * n + k)
            for k in range(m):  # Set cell in the same col with 0
              nextMask &= ~(1 << k * n + j)
            ans = min(ans, 1 + dp(nextMask))
      return ans

    return dp(self.encode(grid, m, n))

  def encode(self, grid: List[List[int]], m: int, n: int) -> int:
    encoded = 0
    for i in range(m):
      for j in range(n):
        if grid[i][j]:
          encoded |= 1 << i * n + j
    return encoded

Approach 2: Bottom-up

  • Time: $O(2^{mn} \cdot mn \cdot \max(m, n))$
  • Space: $O(2^{mn})$
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class Solution {
 public:
  int removeOnes(vector<vector<int>>& grid) {
    const int m = grid.size();
    const int n = grid[0].size();
    const int maxMask = 1 << m * n;
    // dp[i] := min # of operations to remove all 1's from grid
    // Given i representing the state of grid
    vector<int> dp(maxMask, INT_MAX / 2);
    dp[0] = 0;

    for (int mask = 0; mask < maxMask; ++mask)
      for (int i = 0; i < m; ++i)
        for (int j = 0; j < n; ++j)
          if (grid[i][j]) {
            int nextMask = mask;
            for (int k = 0; k < n; ++k)  // Set cell in the same row with 0
              nextMask &= ~(1 << i * n + k);
            for (int k = 0; k < m; ++k)  // Set cell in the same col with 0
              nextMask &= ~(1 << k * n + j);
            dp[mask] = min(dp[mask], 1 + dp[nextMask]);
          }

    return dp[encode(grid, m, n)];
  }

 private:
  int encode(const vector<vector<int>>& grid, int m, int n) {
    int encoded = 0;
    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (grid[i][j])
          encoded |= 1 << i * n + j;
    return encoded;
  }
};

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class Solution {
  public int removeOnes(int[][] grid) {
    final int m = grid.length;
    final int n = grid[0].length;
    final int maxMask = 1 << m * n;
    // dp[i] := min # of operations to remove all 1's from grid
    // Given i representing the state of grid
    int[] dp = new int[1 << m * n];
    Arrays.fill(dp, Integer.MAX_VALUE / 2);
    dp[0] = 0;

    for (int mask = 0; mask < maxMask; ++mask)
      for (int i = 0; i < m; ++i)
        for (int j = 0; j < n; ++j)
          if (grid[i][j] == 1) {
            int nextMask = mask;
            for (int k = 0; k < n; ++k)
              nextMask &= ~(1 << i * n + k); // Set cell in the same row with 0
            for (int k = 0; k < m; ++k)
              nextMask &= ~(1 << k * n + j);
            dp[mask] = Math.min(dp[mask], 1 + dp[nextMask]);
          }

    return dp[encode(grid, m, n)];
  }

  private int encode(int[][] grid, int m, int n) {
    int encoded = 0;
    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (grid[i][j] == 1)
          encoded |= 1 << i * n + j;
    return encoded;
  }
}

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class Solution:
  def removeOnes(self, grid: List[List[int]]) -> int:
    m = len(grid)
    n = len(grid[0])
    maxMask = 1 << m * n
    # dp[i] := min # Of operations to remove all 1's from grid
    # Given i representing the state of grid
    dp = [math.inf] * maxMask
    dp[0] = 0

    for mask in range(maxMask):
      for i in range(m):
        for j in range(n):
          if grid[i][j]:
            nextMask = mask
            for k in range(n):  # Set cell in the same row with 0
              nextMask &= ~(1 << i * n + k)
            for k in range(m):  # Set cell in the same col with 0
              nextMask &= ~(1 << k * n + j)
            dp[mask] = min(dp[mask], 1 + dp[nextMask])

    return dp[self.encode(grid, m, n)]

  def encode(self, grid: List[List[int]], m: int, n: int) -> int:
    encoded = 0
    for i in range(m):
      for j in range(n):
        if grid[i][j]:
          encoded |= 1 << i * n + j
    return encoded