Description Link to heading

407. Trapping Rain Water II (Hard)

Given an m x n integer matrix heightMap representing the height of each unit cell in a 2D elevation map, return the volume of water it can trap after raining.

Example 1:

Input: heightMap = [[1,4,3,1,3,2],[3,2,1,3,2,4],[2,3,3,2,3,1]]
Output: 4
Explanation: After the rain, water is trapped between the blocks.
We have two small ponds 1 and 3 units trapped.
The total volume of water trapped is 4.

Example 2:

Input: heightMap = [[3,3,3,3,3],[3,2,2,2,3],[3,2,1,2,3],[3,2,2,2,3],[3,3,3,3,3]]
Output: 10

Constraints:

  • m == heightMap.length
  • n == heightMap[i].length
  • 1 <= m, n <= 200
  • 0 <= heightMap[i][j] <= 2 * 10⁴

Solution Link to heading

We can use the Dijkstra algorithm to find the maximum height $h_xy$ in the path from the point $(x, y)$ in the edge to the point $(i, j)$.

The volume of water of each cell is $\min(h_{x, y})^{i, j} - heightMap[i][j]$

Code Link to heading 

class Solution {
  public:
    int trapRainWater(vector<vector<int>> &heightMap) {
        int m = heightMap.size();
        int n = heightMap[0].size();
        auto cmp = [&](vector<int> &v1, vector<int> &v2) {
            return v1[2] > v2[2];
        };
        priority_queue<vector<int>, vector<vector<int>>, decltype(cmp)> pq(cmp);
        for (int i = 0; i < n; ++i) {
            pq.push({0, i, heightMap[0][i]});
            pq.push({m - 1, i, heightMap[m - 1][i]});
        }
        for (int i = 1; i < m - 1; ++i) {
            pq.push({i, 0, heightMap[i][0]});
            pq.push({i, n - 1, heightMap[i][n - 1]});
        }
        vector<vector<int>> dis(m, vector<int>(n));
        vector<vector<int>> vis(m, vector<int>(n));
        vector<vector<int>> neighbor{{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
        while (!pq.empty()) {
            auto vec = pq.top();
            int x = vec[0], y = vec[1], height = vec[2];
            pq.pop();
            if (vis[x][y] != 0) {
                continue;
            }
            vis[x][y] = 1;
            dis[x][y] = height;
            for (int i = 0; i < 4; ++i) {
                int new_x = x + neighbor[i][0];
                int new_y = y + neighbor[i][1];
                if (new_x < m && new_x >= 0 && new_y < n && new_y >= 0) {
                    pq.push({new_x, new_y, std::max(height, heightMap[new_x][new_y])});
                }
            }
        }
        int res = 0;
        for (int i = 1; i < m - 1; ++i) {
            for (int j = 1; j < n - 1; ++j) {
                res += dis[i][j] - heightMap[i][j];
            }
        }
        return res;
    }
};