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.
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: These cells can flow to both oceans.
Input: heights = [[1]]
Output: [[0,0]]
Explanation: The water can flow to both oceans from the only cell.
m == heights.lengthn == heights[r].length1 <= m, n <= 2000 <= heights[r][c] <= 10^5Use multi-source DFS from both oceans to find cells reachable from each ocean, then find the intersection.
Algorithm:
Time Complexity: O(m × n) - visit each cell at most twice (once per ocean) Space Complexity: O(m × n) - for the boolean matrices and recursion stack
1import java.util.*;
2
3class Solution {
4 public List<List<Integer>> pacificAtlantic(int[][] heights) {
5 List<List<Integer>> result = new ArrayList<>();
6 if (heights == null || heights.length == 0) return result;
7
8 int m = heights.length;
9 int n = heights[0].length;
10 boolean[][] pacific = new boolean[m][n];
11 boolean[][] atlantic = new boolean[m][n];
12
13 // DFS from Pacific Ocean (top and left edges)
14 for (int i = 0; i < m; i++) {
15 dfs(heights, pacific, i, 0, m, n);
16 }
17 for (int j = 0; j < n; j++) {
18 dfs(heights, pacific, 0, j, m, n);
19 }
20
21 // DFS from Atlantic Ocean (bottom and right edges)
22 for (int i = 0; i < m; i++) {
23 dfs(heights, atlantic, i, n - 1, m, n);
24 }
25 for (int j = 0; j < n; j++) {
26 dfs(heights, atlantic, m - 1, j, m, n);
27 }
28
29 // Find cells that can reach both oceans
30 for (int i = 0; i < m; i++) {
31 for (int j = 0; j < n; j++) {
32 if (pacific[i][j] && atlantic[i][j]) {
33 result.add(Arrays.asList(i, j));
34 }
35 }
36 }
37
38 return result;
39 }
40
41 private void dfs(int[][] heights, boolean[][] ocean, int i, int j, int m, int n) {
42 // Already visited
43 if (ocean[i][j]) return;
44
45 ocean[i][j] = true;
46
47 int[][] directions = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
48 for (int[] dir : directions) {
49 int ni = i + dir[0];
50 int nj = j + dir[1];
51
52 // Check boundaries and water flow (neighbor height >= current height)
53 if (ni >= 0 && ni < m && nj >= 0 && nj < n &&
54 heights[ni][nj] >= heights[i][j]) {
55 dfs(heights, ocean, ni, nj, m, n);
56 }
57 }
58 }
59}
60
61/**
62 * Time Complexity: O(m * n)
63 * - We visit each cell at most twice (once for each ocean)
64 *
65 * Space Complexity: O(m * n)
66 * - Two boolean matrices + recursion stack
67 *
68 * Key Insights:
69 * - Work backwards: find cells reachable TO each ocean FROM the edges
70 * - Water flows from high to low or equal height
71 * - Multi-source DFS from ocean edges
72 * - Intersection of reachable sets gives answer
73 */
74