LeetCode1765. Highest point in the map (BFS)

Power button

Their thinking :

1. Find the location of the water , The team , Bring out the height of the first floor .

2. Any adjacent grid height difference at most by 1, Bring out the height of adjacent grids through the current node , Then team up the adjacent squares ;

3. Continue this process until the queue is empty ; That is, from point to surface .

Code :

class Solution 
{
public:
   int dir[4][2] = {
   {-1, 0}, {1, 0}, {0, -1}, {0, 1}};
   
    vector<vector<int>> highestPeak(vector<vector<int>> &isWater) 
    {
        int row = isWater.size();
        int col = isWater[0].size();
        vector<vector<int>> ans(row, vector<int>(col, -1)); // -1  Indicates that the grid has not been accessed 
        queue<pair<int, int>> q; // Secondary queue 

        for (int i = 0; i < row; ++i) 
        {
            for (int j = 0; j < col; ++j) 
            {
                if (isWater[i][j]) 
                {
                    ans[i][j] = 0;
                    q.push(make_pair(i,j)); //  Team up all waters 
                }
            }
        }

        while (!q.empty()) 
        {
           int sz = q.size();
           while(sz--)
           {
				int t_row = q.front().first;
				int t_col = q.front().second;
				q.pop();

				for (int i = 0; i < 4; ++i) // Four directions 
				{
					int new_row = t_row + dir[i][0];
					int new_col = t_col + dir[i][1];

					if (new_row < 0 || new_row >= row || new_col < 0 || new_col >= col 
                    ||ans[new_row][new_col] != -1)
					{
						continue; // If the position is out of bounds , Visited , Then skip 
					}

                    ans[new_row][new_col] = ans[t_row][t_col] + 1;
                    q.push(make_pair(new_row,new_col));
                }
           }
            
        }

        return ans;
    }
};