35. Pacific Atlantic Water Flow

There is an m x n rectangular island that borders both the Pacific Ocean and Atlantic Ocean. The Pacific Ocean touches the island's left and top edges, and the Atlantic Ocean touches the island's right and bottom edges.

The island is partitioned into a grid of square cells. You are given an m x n integer matrix heights where heights[r][c] represents the height above sea level of the cell at coordinate (r, c).

The island receives a lot of rain, and the rain water can flow to neighboring cells directly north, south, east, and west if the neighboring cell's height is less than or equal to the current cell's height. Water can flow from any cell adjacent to an ocean into the ocean.

Return a 2D list of grid coordinates result where result[i] = [ri, ci] denotes that rain water can flow from cell (ri, ci) to both the Pacific and Atlantic oceans.

Example 1:

Input: heights = [[1,2,2,3,5],[3,2,3,4,4],[2,4,5,3,1],[6,7,1,4,5],[5,1,1,2,4]]
Output: [[0,4],[1,3],[1,4],[2,2],[3,0],[3,1],[4,0]]

Example 2:

Input: heights = [[2,1],[1,2]]
Output: [[0,0],[0,1],[1,0],[1,1]]

Solution:(DFS)

Approach: Now, if we start from the cells connected to atlantic ocean and visit all cells having height greater than current cell (water can only flow from a cell to another one with height equal or lower), we are able to reach some subset of cells (let's call them A).

Next, we start from the cells connected to pacific ocean and repeat the same process, we find another subset (let's call this one B).

The final answer we get will be the intersection of sets A and B (A ∩ B).

class Solution
{
public:
    void dfs(vector<vector<int>> &heights, vector<vector<bool>> &vs, int i, int j, int n, int m)
    {
        if (vs[i][j])
        {
            return;
        }
        
        vs[i][j] = true;

        if (i + 1 < n && heights[i + 1][j] >= heights[i][j])
        {
            dfs(heights, vs, i + 1, j, n, m);
        }

        if (j + 1 < m && heights[i][j + 1] >= heights[i][j])
        {
            dfs(heights, vs, i, j + 1, n, m);
        }

        if (i - 1 >= 0 && heights[i - 1][j] >= heights[i][j])
        {
            dfs(heights, vs, i - 1, j, n, m);
        }

        if (j - 1 >= 0 && heights[i][j - 1] >= heights[i][j])
        {
            dfs(heights, vs, i, j - 1, n, m);
        }

        return;
    }

    vector<vector<int>> pacificAtlantic(vector<vector<int>> &heights)
    {

        int n = heights.size();
        int m = heights[0].size();

        vector<vector<bool>> a(n, vector<bool>(m, false));
        vector<vector<bool>> p(n, vector<bool>(m, false));

        for (int i = 0; i < n; i++)
        {
            dfs(heights, p, i, 0, n, m);
            dfs(heights, a, i, m - 1, n, m);
        }

        for (int i = 0; i < m; i++)
        {
            dfs(heights, p, 0, i, n, m);
            dfs(heights, a, n - 1, i, n, m);
        }

        vector<vector<int>> res;

        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                if (a[i][j] && p[i][j])
                {
                    res.push_back({i, j});
                }
            }
        }

        return res;
    }
};

Time Complexity : O(M * N)