# Pacific Atlantic Water Flow – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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] = [r`_{i}, c_{i}]

*denotes that rain water can flow from cell*

`(r`_{i}, c_{i})

*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]]Explanation:The following cells can flow to the Pacific and Atlantic oceans, as shown below: [0,4]: [0,4] -> Pacific Ocean [0,4] -> Atlantic Ocean [1,3]: [1,3] -> [0,3] -> Pacific Ocean [1,3] -> [1,4] -> Atlantic Ocean [1,4]: [1,4] -> [1,3] -> [0,3] -> Pacific Ocean [1,4] -> Atlantic Ocean [2,2]: [2,2] -> [1,2] -> [0,2] -> Pacific Ocean [2,2] -> [2,3] -> [2,4] -> Atlantic Ocean [3,0]: [3,0] -> Pacific Ocean [3,0] -> [4,0] -> Atlantic Ocean [3,1]: [3,1] -> [3,0] -> Pacific Ocean [3,1] -> [4,1] -> Atlantic Ocean [4,0]: [4,0] -> Pacific Ocean [4,0] -> Atlantic Ocean Note that there are other possible paths for these cells to flow to the Pacific and Atlantic oceans.

**Example 2:**

Input:heights = [[1]]Output:[[0,0]]Explanation:The water can flow from the only cell to the Pacific and Atlantic oceans.

**Constraints:**

`m == heights.length`

`n == heights[r].length`

`1 <= m, n <= 200`

`0 <= heights[r][c] <= 10`

^{5}

# C++ Pacific Atlantic Water Flow LeetCode Solution

```
``````
class Solution {
public:
vector<pair<int, int>> res;
vector<vector<int>> visited;
void dfs(vector<vector<int>>& matrix, int x, int y, int pre, int preval){
if (x < 0 || x >= matrix.size() || y < 0 || y >= matrix[0].size()
|| matrix[x][y] < pre || (visited[x][y] & preval) == preval)
return;
visited[x][y] |= preval;
if (visited[x][y] == 3) res.push_back({x, y});
dfs(matrix, x + 1, y, matrix[x][y], visited[x][y]); dfs(matrix, x - 1, y, matrix[x][y], visited[x][y]);
dfs(matrix, x, y + 1, matrix[x][y], visited[x][y]); dfs(matrix, x, y - 1, matrix[x][y], visited[x][y]);
}
vector<pair<int, int>> pacificAtlantic(vector<vector<int>>& matrix) {
if (matrix.empty()) return res;
int m = matrix.size(), n = matrix[0].size();
visited.resize(m, vector<int>(n, 0));
for (int i = 0; i < m; i++) {
dfs(matrix, i, 0, INT_MIN, 1);
dfs(matrix, i, n - 1, INT_MIN, 2);
}
for (int i = 0; i < n; i++) {
dfs(matrix, 0, i, INT_MIN, 1);
dfs(matrix, m - 1, i, INT_MIN, 2);
}
return res;
}
};
```

# Java Pacific Atlantic Water Flow LeetCode Solution

```
``````
public class Solution {
public List<int[]> pacificAtlantic(int[][] matrix) {
List<int[]> res = new LinkedList<>();
if(matrix == null || matrix.length == 0 || matrix[0].length == 0){
return res;
}
int n = matrix.length, m = matrix[0].length;
boolean[][]pacific = new boolean[n][m];
boolean[][]atlantic = new boolean[n][m];
for(int i=0; i<n; i++){
dfs(matrix, pacific, Integer.MIN_VALUE, i, 0);
dfs(matrix, atlantic, Integer.MIN_VALUE, i, m-1);
}
for(int i=0; i<m; i++){
dfs(matrix, pacific, Integer.MIN_VALUE, 0, i);
dfs(matrix, atlantic, Integer.MIN_VALUE, n-1, i);
}
for (int i = 0; i < n; i++)
for (int j = 0; j < m; j++)
if (pacific[i][j] && atlantic[i][j])
res.add(new int[] {i, j});
return res;
}
int[][]dir = new int[][]{{0,1},{0,-1},{1,0},{-1,0}};
public void dfs(int[][]matrix, boolean[][]visited, int height, int x, int y){
int n = matrix.length, m = matrix[0].length;
if(x<0 || x>=n || y<0 || y>=m || visited[x][y] || matrix[x][y] < height)
return;
visited[x][y] = true;
for(int[]d:dir){
dfs(matrix, visited, matrix[x][y], x+d[0], y+d[1]);
}
}
}
```

# Python 3 Pacific Atlantic Water Flow LeetCode Solution

```
``````
class Solution(object):
def longestIncreasingPath(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: int
"""
if not matrix: return 0
self.directions = [(1,0),(-1,0),(0,1),(0,-1)]
m = len(matrix)
n = len(matrix[0])
cache = [[-1 for _ in range(n)] for _ in range(m)]
res = 0
for i in range(m):
for j in range(n):
cur_len = self.dfs(i, j, matrix, cache, m, n)
res = max(res, cur_len)
return res
def dfs(self, i, j, matrix, cache, m, n):
if cache[i][j] != -1:
return cache[i][j]
res = 1
for direction in self.directions:
x, y = i + direction[0], j + direction[1]
if x < 0 or x >= m or y < 0 or y >= n or matrix[x][y] <= matrix[i][j]:
continue
length = 1 + self.dfs(x, y, matrix, cache, m, n)
res = max(length, res)
cache[i][j] = res
return res
```

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