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]]
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] <= 105
# https://leetcode.com/problems/pacific-atlantic-water-flow/description/
'''
1. 아이디어 :
Pacific과 Atlantic 둘다 만족하려는 dfs를 만들기보다, pacific을 만족하는 dfs, atlantic을 만족하는 dfs를 따로 만들고,
합집합을 만들면 된다.
2. 시간복잡도 :
O(2mn)
3. 자료구조 :
해시셋
'''
class Solution:
def pacificAtlantic(self, heights: List[List[int]]) -> List[List[int]]:
row = len(heights)
col = len(heights[0])
pac = set()
atl = set()
visited = set()
def dfs(x,y, prev, PorA):
if x<0 or y<0 or x==row or y==col or (x,y) in visited or prev > heights[x][y]:
return
if PorA == "P":
if (x,y) in pac:
return
pac.add((x,y))
elif PorA == "A":
if (x,y) in atl:
return
atl.add((x,y))
visited.add((x,y))
dfs(x+1,y,heights[x][y], PorA)
dfs(x-1,y,heights[x][y], PorA)
dfs(x,y+1,heights[x][y], PorA)
dfs(x,y-1,heights[x][y], PorA)
visited.remove((x,y))
for i in range(row):
dfs(i,0,0,"P")
dfs(i,col-1,0,"A")
for i in range(col):
dfs(0,i,0,"P")
dfs(row-1,i,0,"A")
return pac & atl
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