Introduction

In this blog post, we will explore how to solve the LeetCode problem "1036" 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 a 1 million by 1 million grid on an XY-plane, and the coordinates of each grid square are (x, y). We start at the source = [sx, sy] square and want to reach the target = [tx, ty] square. There is also an array of blocked squares, where each blocked[i] = [xi, yi] represents a blocked square with coordinates (xi, yi). Each move, we can walk one square north, east, south, or west if the square is not in the array of blocked squares. We are also not allowed to walk outside of the grid. Return true if and only if it is possible to reach the target square from the source square through a sequence of valid moves. Example 1: Input: blocked = [[0,1],[1,0]], source = [0,0], target = [0,2] Output: false Explanation: The target square is inaccessible starting from the source square because we cannot move. We cannot move north or east because those squares are blocked. We cannot move south or west because we cannot go outside of the grid. Example 2: Input: blocked = [], source = [0,0], target = [999999,999999] Output: true Explanation: Because there are no blocked cells, it is possible to reach the target square. Constraints: 0 <= blocked.length <= 200 blocked[i].length == 2 0 <= xi, yi < 106 source.length == target.length == 2 0 <= sx, sy, tx, ty < 106 source != target It is guaranteed that source and target are not blocked.

Explanation

Here's a solution approach, complexity analysis, and the code:

• Bounded Exploration: Since the grid is very large (10^6 x 10^6), a standard BFS or DFS would likely time out. The key idea is that if we can escape a certain bounded region around the source/target without encountering any blocked cells, then we can safely assume that we can reach the target from the source (or vice versa).
• Area Check: The maximum number of blocked cells is 200. These blocked cells can form at most a certain area. If the explored area from the source (or target) exceeds a certain limit (related to the number of blocked cells), it means we can escape the blocked region. A crucial insight is recognizing the maximum area that the blocked cells can enclose.

• Bidirectional Search: Perform BFS from both the source and the target simultaneously. This improves the overall efficiency.

• Complexity:

  • Runtime: O(B^2), where B is the number of blocked cells (at most 200).
  • Storage: O(B^2)

Code

    from collections import deque

class Solution:
    def isEscapePossible(self, blocked, source, target):
        """
        :type blocked: List[List[int]]
        :type source: List[int]
        :type target: List[int]
        :rtype: bool
        """
        blocked_set = set(tuple(b) for b in blocked)
        limit = len(blocked) * (len(blocked) - 1) // 2

        def bfs(start, end, blocked_set):
            q = deque([start])
            visited = {tuple(start)}

            directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]

            count = 0
            while q:
                x, y = q.popleft()
                count += 1

                if count > limit:
                    return True # Escaped!

                if (x, y) == tuple(end):
                    return True

                for dx, dy in directions:
                    nx, ny = x + dx, y + dy
                    if 0 <= nx < 10**6 and 0 <= ny < 10**6 and (nx, ny) not in blocked_set and (nx, ny) not in visited:
                        q.append((nx, ny))
                        visited.add((nx, ny))
            return False

        return bfs(source, target, blocked_set) and bfs(target, source, blocked_set)