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2123. Minimum Operations to Remove Adjacent Ones in Matrix 👍

  • Time: $O(|V||E|) = O(m^2n^2)$
  • Space: $O(mn)$
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class Solution {
 public:
  int minimumOperations(vector<vector<int>>& grid) {
    const int m = grid.size();
    const int n = grid[0].size();
    int ans = 0;
    vector<vector<int>> seen(m, vector<int>(n));
    vector<vector<int>> match(m, vector<int>(n, -1));

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (grid[i][j] == 1 && match[i][j] == -1) {
          const int sessionId = i * n + j;
          seen[i][j] = sessionId;
          ans += dfs(grid, i, j, sessionId, seen, match);
        }

    return ans;
  }

 private:
  const vector<int> dirs{0, 1, 0, -1, 0};

  int dfs(const vector<vector<int>>& grid, int i, int j, int sessionId,
          vector<vector<int>>& seen, vector<vector<int>>& match) {
    const int m = grid.size();
    const int n = grid[0].size();

    for (int k = 0; k < 4; ++k) {
      const int x = i + dirs[k];
      const int y = j + dirs[k + 1];
      if (x < 0 || x == m || y < 0 || y == n)
        continue;
      if (grid[x][y] == 0 || seen[x][y] == sessionId)
        continue;
      seen[x][y] = sessionId;
      if (match[x][y] == -1 ||
          dfs(grid, match[x][y] / n, match[x][y] % n, sessionId, seen, match)) {
        match[x][y] = i * n + j;
        match[i][j] = x * n + y;
        return 1;
      }
    }

    return 0;
  }
};
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class Solution {
  public int minimumOperations(int[][] grid) {
    final int m = grid.length;
    final int n = grid[0].length;
    int ans = 0;
    int[][] seen = new int[m][n];
    int[][] match = new int[m][n];
    Arrays.stream(match).forEach(A -> Arrays.fill(A, -1));

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (grid[i][j] == 1 && match[i][j] == -1) {
          final int sessionId = i * n + j;
          seen[i][j] = sessionId;
          if (dfs(grid, i, j, sessionId, seen, match))
            ++ans;
        }

    return ans;
  }

  private static final int[] dirs = {0, 1, 0, -1, 0};

  private boolean dfs(final int[][] grid, int i, int j, int sessionId, int[][] seen,
                      int[][] match) {
    final int m = grid.length;
    final int n = grid[0].length;

    for (int k = 0; k < 4; ++k) {
      final int x = i + dirs[k];
      final int y = j + dirs[k + 1];
      if (x < 0 || x == m || y < 0 || y == n)
        continue;
      if (grid[x][y] == 0 || seen[x][y] == sessionId)
        continue;
      seen[x][y] = sessionId;
      if (match[x][y] == -1 ||
          dfs(grid, match[x][y] / n, match[x][y] % n, sessionId, seen, match)) {
        match[x][y] = i * n + j;
        match[i][j] = x * n + y;
        return true;
      }
    }

    return false;
  }
}
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class Solution:
  def minimumOperations(self, grid: List[List[int]]) -> int:
    m = len(grid)
    n = len(grid[0])
    dirs = [0, 1, 0, -1, 0]
    seen = [[0] * n for _ in range(m)]
    match = [[-1] * n for _ in range(m)]

    def dfs(i: int, j: int, sessionId: int) -> int:
      for k in range(4):
        x = i + dirs[k]
        y = j + dirs[k + 1]
        if x < 0 or x == m or y < 0 or y == n:
          continue
        if grid[x][y] == 0 or seen[x][y] == sessionId:
          continue
        seen[x][y] = sessionId
        if match[x][y] == -1 or dfs(*divmod(match[x][y], n), sessionId):
          match[x][y] = i * n + j
          match[i][j] = x * n + y
          return 1
      return 0

    ans = 0

    for i in range(m):
      for j in range(n):
        if grid[i][j] == 1 and match[i][j] == -1:
          sessionId = i * n + j
          seen[i][j] = sessionId
          ans += dfs(i, j, sessionId)

    return ans