Count Sub Islands LeetCode solution in C++

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class Solution {
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 public:
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  bool res;
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  void mark_current_island(vector<vector<int>> &grid1, vector<vector<int>> &grid2, int x, int y,
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                           int r, int c) {
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    if (x < 0 || x >= r || y < 0 || y >= c || grid2[x][y] != 1) // Boundary case for matrix
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      return;
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    // if there is water on  grid1 for the location on B then our current island
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    // can't be counted and so we mark res as false
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    if (grid1[x][y] == 0) {
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      res = false;
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      return;
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    }
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    // Mark current cell as visited
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    grid2[x][y] = 0;
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    // Make recursive call in all 4 adjacent directions
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    mark_current_island(grid1, grid2, x + 1, y, r, c); // DOWN
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    mark_current_island(grid1, grid2, x, y + 1, r, c); // RIGHT
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    mark_current_island(grid1, grid2, x - 1, y, r, c); // TOP
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    mark_current_island(grid1, grid2, x, y - 1, r, c); // LEFT
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  }
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  int countSubIslands(vector<vector<int>> &grid1, vector<vector<int>> &grid2) {
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    // For FAST I/O
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    ios_base::sync_with_stdio(false);
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    cin.tie(NULL);
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    int rows = grid1.size();
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    if (rows == 0) // Empty grid boundary case
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      return 0;
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    int cols = grid1[0].size();
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    // Iterate for all cells of the array
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    int no_of_islands = 0;
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    // dfs on grid2
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    for (int i = 0; i < rows; ++i) {
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      for (int j = 0; j < cols; ++j) {
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        if (grid2[i][j] == 1) {
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          res = true;
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          mark_current_island(grid1, grid2, i, j, rows, cols);
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          // if current island of grid2 meets the requirement update the no of
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          // islands
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          if (res) no_of_islands += 1;
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        }
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      }
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    }
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    return no_of_islands;
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  }
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};