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934. Shortest Bridge 👍

Approach 1: Pure DFS

  • Time: $O(n^2)$
  • Space: $O(n^2)$
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class Solution {
 public:
  int shortestBridge(vector<vector<int>>& grid) {
    markGridTwo(grid);

    for (int color = 2;; ++color)
      for (int i = 0; i < grid.size(); ++i)
        for (int j = 0; j < grid[0].size(); ++j)
          if (grid[i][j] == color)
            if (expand(grid, i + 1, j, color) ||  //
                expand(grid, i - 1, j, color) ||  //
                expand(grid, i, j + 1, color) ||  //
                expand(grid, i, j - 1, color))
              return color - 2;
  }

 private:
  // Marks one group to 2s by DFS.
  void markGridTwo(vector<vector<int>>& grid) {
    for (int i = 0; i < grid.size(); ++i)
      for (int j = 0; j < grid[0].size(); ++j)
        if (grid[i][j] == 1) {
          markGridTwo(grid, i, j);
          return;
        }
  }

  void markGridTwo(vector<vector<int>>& grid, int i, int j) {
    if (i < 0 || i == grid.size() || j < 0 || j == grid[0].size())
      return;
    if (grid[i][j] != 1)
      return;
    grid[i][j] = 2;
    markGridTwo(grid, i + 1, j);
    markGridTwo(grid, i - 1, j);
    markGridTwo(grid, i, j + 1);
    markGridTwo(grid, i, j - 1);
  }

  // Returns true if we touch 1s' group through expanding.
  bool expand(vector<vector<int>>& grid, int i, int j, int color) {
    if (i < 0 || i == grid.size() || j < 0 || j == grid[0].size())
      return false;
    if (grid[i][j] == 0)
      grid[i][j] = color + 1;
    return grid[i][j] == 1;
  }
};

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class Solution {
  public int shortestBridge(int[][] grid) {
    markGridTwo(grid);

    for (int color = 2;; ++color)
      for (int i = 0; i < grid.length; ++i)
        for (int j = 0; j < grid[0].length; ++j)
          if (grid[i][j] == color)
            if (expand(grid, i + 1, j, color) || expand(grid, i - 1, j, color) ||
                expand(grid, i, j + 1, color) || expand(grid, i, j - 1, color))
              return color - 2;
  }

  // Marks one group to 2s by DFS.
  private void markGridTwo(int[][] grid) {
    for (int i = 0; i < grid.length; ++i)
      for (int j = 0; j < grid[0].length; ++j)
        if (grid[i][j] == 1) {
          markGridTwo(grid, i, j);
          return;
        }
  }

  private void markGridTwo(int[][] grid, int i, int j) {
    if (i < 0 || i == grid.length || j < 0 || j == grid[0].length)
      return;
    if (grid[i][j] != 1)
      return;
    grid[i][j] = 2;
    markGridTwo(grid, i + 1, j);
    markGridTwo(grid, i - 1, j);
    markGridTwo(grid, i, j + 1);
    markGridTwo(grid, i, j - 1);
  }

  // Returns true if we touch 1s' group through expanding.
  private boolean expand(int[][] grid, int i, int j, int color) {
    if (i < 0 || i == grid.length || j < 0 || j == grid[0].length)
      return false;
    if (grid[i][j] == 0)
      grid[i][j] = color + 1;
    return grid[i][j] == 1;
  }
}

Approach 2: DFS + BFS

  • Time: $O(n^2)$
  • Space: $O(n^2)$
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class Solution {
 public:
  int shortestBridge(vector<vector<int>>& grid) {
    constexpr int dirs[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    const int n = grid.size();
    queue<pair<int, int>> q;

    markGridTwo(grid, q);

    for (int ans = 0; !q.empty(); ++ans)
      for (int sz = q.size(); sz > 0; --sz) {
        const auto [i, j] = q.front();
        q.pop();
        for (const auto& [dx, dy] : dirs) {
          const int x = i + dx;
          const int y = j + dy;
          if (x < 0 || x == n || y < 0 || y == n)
            continue;
          if (grid[x][y] == 2)
            continue;
          if (grid[x][y] == 1)
            return ans;
          grid[x][y] = 2;
          q.emplace(x, y);
        }
      }

    throw;
  }

 private:
  // Marks one group to 2s by DFS.
  void markGridTwo(vector<vector<int>>& grid, queue<pair<int, int>>& q) {
    for (int i = 0; i < grid.size(); ++i)
      for (int j = 0; j < grid[0].size(); ++j)
        if (grid[i][j] == 1) {
          markGridTwo(grid, i, j, q);
          return;
        }
  }

  // Marks one group to 2s by DFS and pushes them to the queue.
  void markGridTwo(vector<vector<int>>& grid, int i, int j,
                   queue<pair<int, int>>& q) {
    if (i < 0 || i == grid.size() || j < 0 || j == grid[0].size())
      return;
    if (grid[i][j] != 1)
      return;
    grid[i][j] = 2;
    q.emplace(i, j);
    markGridTwo(grid, i + 1, j, q);
    markGridTwo(grid, i - 1, j, q);
    markGridTwo(grid, i, j + 1, q);
    markGridTwo(grid, i, j - 1, q);
  }
};

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class Solution {
  public int shortestBridge(int[][] grid) {
    final int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    final int n = grid.length;
    Queue<Pair<Integer, Integer>> q = new ArrayDeque<>();

    markGridTwo(grid, q);

    for (int ans = 0; !q.isEmpty(); ++ans)
      for (int sz = q.size(); sz > 0; --sz) {
        final int i = q.peek().getKey();
        final int j = q.poll().getValue();
        for (int[] dir : dirs) {
          final int x = i + dir[0];
          final int y = j + dir[1];
          if (x < 0 || x == n || y < 0 || y == n)
            continue;
          if (grid[x][y] == 2)
            continue;
          if (grid[x][y] == 1)
            return ans;
          grid[x][y] = 2;
          q.offer(new Pair<>(x, y));
        }
      }

    throw new IllegalArgumentException();
  }

  // Marks one group to 2s by DFS.
  private void markGridTwo(int[][] grid, Queue<Pair<Integer, Integer>> q) {
    for (int i = 0; i < grid.length; ++i)
      for (int j = 0; j < grid[0].length; ++j)
        if (grid[i][j] == 1) {
          markGridTwo(grid, i, j, q);
          return;
        }
  }

  // Marks one group to 2s by DFS and pushes them to the queue.
  private void markGridTwo(int[][] grid, int i, int j, Queue<Pair<Integer, Integer>> q) {
    if (i < 0 || i == grid.length || j < 0 || j == grid[0].length)
      return;
    if (grid[i][j] != 1)
      return;
    grid[i][j] = 2;
    q.offer(new Pair<>(i, j));
    markGridTwo(grid, i + 1, j, q);
    markGridTwo(grid, i - 1, j, q);
    markGridTwo(grid, i, j + 1, q);
    markGridTwo(grid, i, j - 1, q);
  }
}