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# Trapping Rain Water II – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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:4Explanation: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 * 10`

^{4}

# C++ Trapping Rain Water II LeetCode Solution

```
```October 11, 2016 9:51 AM

5.4K VIEWS

It takes me a long time to figure it out, but actually the idea is quite straightforward. Imagine the pool is surrounded by many bars. The water can only go out from the lowest bar. So we always start from the lowest boundary and keep pushing the bar from boundary towards inside. It works as if we are replacing the old bars with a bar higher than it.

See the following simple example:

4 4 4 4

4 0 1 2

4 4 4 4

it looks like we push the bar of 2 towards left and record the difference. Then you can use the same procedure with the following figure

4 4 4 4

4 0 2 2

4 4 4 4

```
int trapRainWater(vector<vector<int>>& heightMap) {
typedef pair<int,int> cell;
priority_queue<cell, vector<cell>, greater<cell>> q;
int m = heightMap.size();
if (m == 0) return 0;
int n = heightMap[0].size();
vector<int> visited(m*n, false);
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j) {
if (i == 0 || i == m-1 || j == 0 || j == n-1) {
if (!visited[i*n+j])
q.push(cell(heightMap[i][j], i*n+j));
visited[i*n+j] = true;
}
}
int dir[4][2] = {{0,1}, {0, -1}, {1, 0}, {-1, 0}};
int ans = 0;
while(!q.empty()) {
cell c = q.top();
q.pop();
int i = c.second/n, j = c.second%n;
for (int r = 0; r < 4; ++r) {
int ii = i+dir[r][0], jj = j+dir[r][1];
if (ii < 0 || ii >= m || jj < 0 || jj >= n || visited[ii*n+jj])
continue;
ans += max(0, c.first - heightMap[ii][jj]);
q.push(cell(max(c.first, heightMap[ii][jj]), ii*n+jj));
visited[ii*n+jj] = true;
}
}
return ans;
}
```

# Java Trapping Rain Water II LeetCode Solution

```
``````
public class Solution {
public class Cell {
int row;
int col;
int height;
public Cell(int row, int col, int height) {
this.row = row;
this.col = col;
this.height = height;
}
}
public int trapRainWater(int[][] heights) {
if (heights == null || heights.length == 0 || heights[0].length == 0)
return 0;
PriorityQueue<Cell> queue = new PriorityQueue<>(1, new Comparator<Cell>(){
public int compare(Cell a, Cell b) {
return a.height - b.height;
}
});
int m = heights.length;
int n = heights[0].length;
boolean[][] visited = new boolean[m][n];
// Initially, add all the Cells which are on borders to the queue.
for (int i = 0; i < m; i++) {
visited[i][0] = true;
visited[i][n - 1] = true;
queue.offer(new Cell(i, 0, heights[i][0]));
queue.offer(new Cell(i, n - 1, heights[i][n - 1]));
}
for (int i = 0; i < n; i++) {
visited[0][i] = true;
visited[m - 1][i] = true;
queue.offer(new Cell(0, i, heights[0][i]));
queue.offer(new Cell(m - 1, i, heights[m - 1][i]));
}
// from the borders, pick the shortest cell visited and check its neighbors:
// if the neighbor is shorter, collect the water it can trap and update its height as its height plus the water trapped
// add all its neighbors to the queue.
int[][] dirs = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
int res = 0;
while (!queue.isEmpty()) {
Cell cell = queue.poll();
for (int[] dir : dirs) {
int row = cell.row + dir[0];
int col = cell.col + dir[1];
if (row >= 0 && row < m && col >= 0 && col < n && !visited[row][col]) {
visited[row][col] = true;
res += Math.max(0, cell.height - heights[row][col]);
queue.offer(new Cell(row, col, Math.max(heights[row][col], cell.height)));
}
}
}
return res;
}
}
```

# Python 3 Trapping Rain Water II LeetCode Solution

```
``````
class Solution:
def trapRainWater(self, heightMap: List[List[int]]) -> int:
if not heightMap or not heightMap[0]:
return 0
# Initial
# Board cells cannot trap the water
m, n = len(heightMap), len(heightMap[0])
if m < 3 or n < 3:
return 0
# Add Board cells first
heap = []
for i in range(m):
for j in range(n):
if i == 0 or i == m - 1 or j == 0 or j == n - 1:
heapq.heappush(heap, (heightMap[i][j], i, j))
heightMap[i][j] = -1
# Start from level 0
level, res = 0, 0
while heap:
height, x, y = heapq.heappop(heap)
level = max(height, level)
for i, j in [(x - 1, y), (x + 1, y), (x, y - 1), (x, y + 1)]:
if 0 <= i < m and 0 <= j < n and heightMap[i][j] != -1:
heapq.heappush(heap, (heightMap[i][j], i, j))
# If cell's height smaller than the level, then it can trap the rain water
if heightMap[i][j] < level:
res += level - heightMap[i][j]
# Set the height to -1 if the cell is visited
heightMap[i][j] = -1
return res
```

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