LeetCode #417: Pacific Atlantic Water Flow

Problem Statement

(Source) Given an m x n matrix of non-negative integers representing the height of each unit cell in a continent, the “Pacific ocean” touches the left and top edges of the matrix and the “Atlantic ocean” touches the right and bottom edges.

Water can only flow in four directions (up, down, left, or right) from a cell to another one with height equal or lower.

Find the list of grid coordinates where water can flow to both the Pacific and Atlantic ocean.

Note:

• The order of returned grid coordinates does not matter.
• Both m and n are less than 150.

Example:

Given the following 5x5 matrix:

  Pacific ~   ~   ~   ~   ~ 
       ~  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  *
          *   *   *   *   * Atlantic

Return:

[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]] (positions with parentheses in above matrix).

Solution

Tags: Depth-first Search, Breadth-first Search.

A stupid but AC solution is shown as below:

class Solution(object):
    def pacificAtlantic(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[List[int]]
        """
        from collections import deque

        if not matrix or not matrix[0]:
            return []
        res = []
        m, n = len(matrix), len(matrix[0])
        helper = [[0 for j in xrange(n)] for i in xrange(m)]
        # initialisation:
        for j in xrange(n):
            helper[0][j] = 1
            if helper[m - 1][j] != 3:
                helper[m - 1][j] = 3 if helper[m - 1][j] == 1 else 2
        for i in xrange(m):
            if helper[i][0] != 3:
                helper[i][0] = 3 if helper[i][0] == 2 else 1
            if helper[i][n - 1] != 3:
                helper[i][n - 1] = 3 if helper[i][n - 1] == 1 else 2
        # main loop:
        for i in xrange(m):
            for j in xrange(n):
                if helper[i][j] != 3:
                    # bfs:
                    que = deque()
                    explored = set()
                    que.append((i, j))
                    explored.add((i, j))
                    while que:
                        x, y = que.popleft()
                        # left:
                        if y - 1 >= 0 and matrix[x][y] >= matrix[x][y-1] and (x, y-1) not in explored:
                            if helper[x][y-1] == 3 or helper[x][y-1] + helper[i][j] == 3:
                                helper[i][j] = 3
                                break
                            else:
                                helper[i][j] = helper[x][y-1] or helper[i][j]
                                que.append((x, y-1))
                                explored.add((x, y-1))
                        # up:
                        if x - 1 >= 0 and matrix[x][y] >= matrix[x-1][y] and (x-1, y) not in explored:
                            if helper[x-1][y] == 3 or helper[x-1][y] + helper[i][j] == 3:
                                helper[i][j] = 3
                                break
                            else:
                                helper[i][j] = helper[x-1][y] or helper[i][j]
                                que.append((x-1, y))
                                explored.add((x-1, y))
                        # right:
                        if y + 1 < n and matrix[x][y] >= matrix[x][y+1] and (x, y+1) not in explored:
                            if helper[x][y+1] == 3 or helper[x][y+1] + helper[i][j] == 3:
                                helper[i][j] = 3
                                break
                            else:
                                helper[i][j] = helper[x][y+1] or helper[i][j]
                                que.append((x, y+1))
                                explored.add((x, y+1))
                        # down:
                        if x + 1 < m and matrix[x][y] >= matrix[x+1][y] and (x+1, y) not in explored:
                            if helper[x+1][y] == 3 or helper[x+1][y] + helper[i][j] == 3:
                                helper[i][j] = 3
                                break
                            else:
                                helper[i][j] = helper[x+1][y] or helper[i][j]
                                que.append((x+1, y))
                                explored.add((x+1, y))
                if helper[i][j] == 3:
                    res.append([i, j])
        return res

To be optimised :)

A slightly more concise solution, but no idea why TLE (: The idea to simplify above solution is taken from here.

class Solution(object):
    def pacificAtlantic(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[List[int]]
        """
        from collections import deque

        if not matrix or not matrix[0]:
            return []
        res = []
        m, n = len(matrix), len(matrix[0])
        helper = [[0 for j in xrange(n)] for i in xrange(m)]
        # initialisation:
        for j in xrange(n):
            helper[0][j] = 1
            if helper[m - 1][j] != 3:
                helper[m - 1][j] = 3 if helper[m - 1][j] == 1 else 2
        for i in xrange(m):
            if helper[i][0] != 3:
                helper[i][0] = 3 if helper[i][0] == 2 else 1
            if helper[i][n - 1] != 3:
                helper[i][n - 1] = 3 if helper[i][n - 1] == 1 else 2

        idx_arr = [1, 0, -1, 0, 1]
        # main loop:
        for i in xrange(m):
            for j in xrange(n):
                if helper[i][j] != 3:
                    # bfs:
                    que = deque()
                    explored = set()
                    que.append((i, j))
                    explored.add((i, j))
                    while que and helper[i][j] != 3:
                        x, y = que.popleft()
                        for idx in xrange(4):
                            xx, yy = x + idx_arr[idx], y + idx_arr[idx + 1]
                            if 0 <= xx < m and 0 <= yy < n and matrix[x][y] >= matrix[xx][yy] and (xx, yy) not in explored:
                                if helper[xx][yy] == 3 or helper[xx][yy] + helper[i][j] == 3:
                                    helper[i][j] = 3
                                    break
                                else:
                                    helper[i][j] = helper[xx][yy] or helper[i][j]
                                    que.append((xx, yy))
                                    explored.add((xx, yy))
                if helper[i][j] == 3:
                    res.append([i, j])
        return res