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1034. Coloring A Border 👎

  • Time: $O(mn)$
  • Space: $O(mn)$
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class Solution {
 public:
  vector<vector<int>> colorBorder(vector<vector<int>>& grid, int r0, int c0,
                                  int color) {
    dfs(grid, r0, c0, grid[r0][c0]);

    for (int i = 0; i < grid.size(); ++i)
      for (int j = 0; j < grid[0].size(); ++j)
        if (grid[i][j] < 0)
          grid[i][j] = color;

    return grid;
  }

 private:
  void dfs(vector<vector<int>>& grid, int i, int j, int startColor) {
    if (i < 0 || i == grid.size() || j < 0 || j == grid[0].size())
      return;
    if (grid[i][j] != startColor)
      return;

    grid[i][j] = -startColor;
    dfs(grid, i + 1, j, startColor);
    dfs(grid, i - 1, j, startColor);
    dfs(grid, i, j + 1, startColor);
    dfs(grid, i, j - 1, startColor);

    // If this cell is already on the boarder, it must be painted later.
    if (i == 0 || i == grid.size() - 1 || j == 0 || j == grid[0].size() - 1)
      return;

    if (abs(grid[i + 1][j]) == startColor &&  //
        abs(grid[i - 1][j]) == startColor &&  //
        abs(grid[i][j + 1]) == startColor &&  //
        abs(grid[i][j - 1]) == startColor)
      grid[i][j] = startColor;
  }
};

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class Solution {
  public int[][] colorBorder(int[][] grid, int r0, int c0, int color) {
    dfs(grid, r0, c0, grid[r0][c0]);

    for (int i = 0; i < grid.length; ++i)
      for (int j = 0; j < grid[0].length; ++j)
        if (grid[i][j] < 0)
          grid[i][j] = color;

    return grid;
  }

  private void dfs(int[][] grid, int i, int j, int startColor) {
    if (i < 0 || i == grid.length || j < 0 || j == grid[0].length)
      return;
    if (grid[i][j] != startColor)
      return;

    grid[i][j] = -startColor;
    dfs(grid, i + 1, j, startColor);
    dfs(grid, i - 1, j, startColor);
    dfs(grid, i, j + 1, startColor);
    dfs(grid, i, j - 1, startColor);

    // If this cell is already on the boarder, it must be painted later.
    if (i == 0 || i == grid.length - 1 || j == 0 || j == grid[0].length - 1)
      return;

    if (Math.abs(grid[i + 1][j]) == startColor && //
        Math.abs(grid[i - 1][j]) == startColor && //
        Math.abs(grid[i][j + 1]) == startColor && //
        Math.abs(grid[i][j - 1]) == startColor)
      grid[i][j] = startColor;
  }
}

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class Solution:
  def colorBorder(
      self,
      grid: list[list[int]],
      r0: int,
      c0: int,
      color: int
  ) -> list[list[int]]:
    def dfs(i: int, j: int, startColor: int) -> None:
      if i < 0 or i == len(grid) or j < 0 or j == len(grid[0]):
        return
      if grid[i][j] != startColor:
        return

      grid[i][j] = -startColor
      dfs(i + 1, j, startColor)
      dfs(i - 1, j, startColor)
      dfs(i, j + 1, startColor)
      dfs(i, j - 1, startColor)

      # If this cell is already on the boarder, it must be painted later.
      if i == 0 or i == len(grid) - 1 or j == 0 or j == len(grid[0]) - 1:
        return

      if (abs(grid[i + 1][j]) == startColor and
          abs(grid[i - 1][j]) == startColor and
          abs(grid[i][j + 1]) == startColor and
              abs(grid[i][j - 1]) == startColor):
        grid[i][j] = startColor

    dfs(r0, c0, grid[r0][c0])

    for i, row in enumerate(grid):
      for j, num in enumerate(row):
        if num < 0:
          grid[i][j] = color

    return grid