Introduction

In this blog post, we will explore how to solve the LeetCode problem "417" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the approach and the Python code:

• Reverse Flow: Instead of trying to find paths from each cell to both oceans, start from the oceans and flow inwards to find which cells can reach them.
• DFS/BFS: Use Depth-First Search (DFS) or Breadth-First Search (BFS) to explore cells reachable from each ocean.

• Intersection: Find the intersection of cells reachable from the Pacific and cells reachable from the Atlantic. These cells can reach both oceans.

• Time Complexity: O(m * n), where m is the number of rows and n is the number of columns.

• Space Complexity: O(m * n) to store the visited sets and the result.

Code

    def pacificAtlantic(heights: list[list[int]]) -> list[list[int]]:
    """
    Finds the cells in a matrix that can flow to both the Pacific and Atlantic oceans.

    Args:
        heights: A 2D list of integers representing the height of each cell.

    Returns:
        A 2D list of coordinates representing the cells that can flow to both oceans.
    """

    if not heights:
        return []

    rows, cols = len(heights), len(heights[0])
    pacific_reachable = set()
    atlantic_reachable = set()

    def dfs(row, col, reachable):
        """
        Performs a Depth-First Search to find cells reachable from a given cell.
        """
        if (row, col) in reachable:
            return

        reachable.add((row, col))

        directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
        for dr, dc in directions:
            new_row, new_col = row + dr, col + dc
            if 0 <= new_row < rows and 0 <= new_col < cols and \
               heights[new_row][new_col] >= heights[row][col]:
                dfs(new_row, new_col, reachable)

    # Start DFS from the Pacific ocean (top and left edges)
    for col in range(cols):
        dfs(0, col, pacific_reachable)
    for row in range(rows):
        dfs(row, 0, pacific_reachable)

    # Start DFS from the Atlantic ocean (bottom and right edges)
    for col in range(cols):
        dfs(rows - 1, col, atlantic_reachable)
    for row in range(rows):
        dfs(row, cols - 1, atlantic_reachable)

    # Find the intersection of cells reachable from both oceans
    result = []
    for row in range(rows):
        for col in range(cols):
            if (row, col) in pacific_reachable and (row, col) in atlantic_reachable:
                result.append([row, col])

    return result