Pacific Atlantic Water Flow

Problem

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]]

Example 2:

Input: heights = [[2,1],[1,2]]
Output: [[0,0],[0,1],[1,0],[1,1]]

Constraints:

m == heights.length
n == heights[r].length
1 <= m, n <= 200
0 <= heights[r][c] <= 10⁵

Solution

/**
 * @param {number[][]} heights
 * @return {number[][]}
 */
var pacificAtlantic = function(heights) {
    const m = heights.length; // rows
    const n = heights[0].length; // cols
    const pacificVisited = []; // cells that that can be visited from the Pacific
    const atlanticVisited = []; // cells that that can be visited from the Atlantic

    // fill in pacificVisited and atlanticVisited with array of false
    for (let i = 0; i < m; i++) {
        pacificVisited.push(Array(n).fill(false));
        atlanticVisited.push(Array(n).fill(false));
    }

    // dfs starting with left and right of the two oceans
    for (let i = 0; i < m; i++) {
        dfs(heights, pacificVisited, 0, i, 0);
        dfs(heights, atlanticVisited, 0, i, n - 1);
    }

    // dfs starting with top and bottom of the two oceans
    for (let i = 0; i < n; i++) {
        dfs(heights, pacificVisited, 0, 0, i);
        dfs(heights, atlanticVisited, 0, m - 1, i);
    }

    // find cells that can reach both the Pacific and Atlantic
    const res = [];
    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            if (pacificVisited[i][j] && atlanticVisited[i][j]) {
                res.push([i, j]);
            }
        }
    }
    return res;
};

var dfs = function(board, visited, minHeight, row, col) {
    // (row, rol) is not out of range, has greater or equal height, and has not been visited
    if (board[row]?.[col] >= minHeight && !visited[row][col]) {
        visited[row][col] = true;
        const height = board[row][col];
        dfs(board, visited, height, row + 1, col);
        dfs(board, visited, height, row - 1, col);
        dfs(board, visited, height, row, col + 1);
        dfs(board, visited, height, row, col - 1);
    }
};

We will implement a DFS solution. Instead of performing DFS on all cells of the board, we perform it on both the Pacific and Atlantic (ie. starting with all cells directly adjacent to the two oceans). We use pacificVisited[r][c] and atlanticVisited[r][c] to keep track of if (r, c) can reach the Pacific and Atlantic respectively. Finally, we just need to check which cells can reach both the Atlantic and Pacific.

Problem​

Solution​

Problem

Solution