Teaching Kids Programming – Nearest Exit from Entrance in Maze via Iterative Deepening Search Algorithm (IDS)


Teaching Kids Programming: Videos on Data Structures and Algorithms

You are given an m x n matrix maze (0-indexed) with empty cells (represented as ‘.’) and walls (represented as ‘+’). You are also given the entrance of the maze, where entrance = [entrancerow, entrancecol] denotes the row and column of the cell you are initially standing at.

In one step, you can move one cell up, down, left, or right. You cannot step into a cell with a wall, and you cannot step outside the maze. Your goal is to find the nearest exit from the entrance. An exit is defined as an empty cell that is at the border of the maze. The entrance does not count as an exit.

find-the-nearest-exit-in-the-maze Teaching Kids Programming - Nearest Exit from Entrance in Maze via Iterative Deepening Search Algorithm (IDS) algorithms Breadth First Search Depth First Search Graph Algorithm Iterative Deepening Search python teaching kids programming youtube video

find-the-nearest-exit-in-the-maze

Return the number of steps in the shortest path from the entrance to the nearest exit, or -1 if no such path exists.

Example 1:
Input: maze = [[“+”,”+”,”.”,”+”],[“.”,”.”,”.”,”+”],[“+”,”+”,”+”,”.”]], entrance = [1,2]
Output: 1
Explanation: There are 3 exits in this maze at [1,0], [0,2], and [2,3].
Initially, you are at the entrance cell [1,2].
– You can reach [1,0] by moving 2 steps left.
– You can reach [0,2] by moving 1 step up.
It is impossible to reach [2,3] from the entrance.
Thus, the nearest exit is [0,2], which is 1 step away.

Example 2:
Input: maze = [[“+”,”+”,”+”],[“.”,”.”,”.”],[“+”,”+”,”+”]], entrance = [1,0]
Output: 2
Explanation: There is 1 exit in this maze at [1,2].
[1,0] does not count as an exit since it is the entrance cell.
Initially, you are at the entrance cell [1,0].
– You can reach [1,2] by moving 2 steps right.
Thus, the nearest exit is [1,2], which is 2 steps away.

Example 3:
Input: maze = [[“.”,”+”]], entrance = [0,0]
Output: -1
Explanation: There are no exits in this maze.

Constraints:
maze.length == m
maze[i].length == n
1 <= m, n <= 100
maze[i][j] is either ‘.’ or ‘+’.
entrance.length == 2
0 <= entrancerow < m
0 <= entrancecol < n
entrance will always be an empty cell.

Hints:
Which type of traversal lets you find the distance from a point?
Try using a Breadth First Search.

Nearest Exit from Entrance in Maze via Iterative Deepening Search Algorithm (IDS)

We talked about using Breadth First Search (BFS) algorithm to find the shortest (optimal) path in a unweighted Graph (could either be directed or undirected). The problem with BFS is the large space complexity as it requires usage of a queue to hold all the elements/nodes in the same level/distance. We could use the Iterative Deepening Search (IDS) which is performing the BFS but in the DFS (Depth First Search) manner.

The core idea of Iterative Deepening Search Algorithm is that we increment the maximum depth limit one by one, and perform a Depth Limit Search (DLS) which is Depth First Search (DFS) with Limit meaning that when the depth exceeds a limit, the DFS has a hard stop.

The advantage of IDS is that we can find the shortest/optimal path, and we just need the similar space complexity as the DFS which is O(D) where D is the maximum depth of the visited Graph/Tree.

Likewise, to avoid visiting same nodes twice, we can mark and unmark the maze, or use a hash set.

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class Solution:
    def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int: 
        rows = len(maze)
        assert rows > 0
        cols = len(maze[0])
        assert cols > 0
        
        @cache
        def isExit(r, c):
            if r == 0 or c == 0 or r == rows - 1 or c == cols - 1:
                return maze[r][c] == '.'
            return False
 
        def dls(r, c, d):
            if (r, c) != tuple(entrance) and isExit(r, c):
                return True
            if d <= 0:
                return False
            maze[r][c] = "+"
            for dr, dc in ((0, 1), (1, 0), (0, -1), (-1, 0)):
                nr, nc = r + dr, c + dc
                if 0 <= nr < rows and 0 <= nc < cols and maze[nr][nc] == '.':                    
                    if dls(nr, nc, d - 1):
                        return True
            maze[r][c] = "."
            return False
 
        d = 0
        D = rows * cols
        while d < D:
            if dls(entrance[0], entrance[1], d):
                return d
            d += 1
        
        return -1
class Solution:
    def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int: 
        rows = len(maze)
        assert rows > 0
        cols = len(maze[0])
        assert cols > 0
        
        @cache
        def isExit(r, c):
            if r == 0 or c == 0 or r == rows - 1 or c == cols - 1:
                return maze[r][c] == '.'
            return False

        def dls(r, c, d):
            if (r, c) != tuple(entrance) and isExit(r, c):
                return True
            if d <= 0:
                return False
            maze[r][c] = "+"
            for dr, dc in ((0, 1), (1, 0), (0, -1), (-1, 0)):
                nr, nc = r + dr, c + dc
                if 0 <= nr < rows and 0 <= nc < cols and maze[nr][nc] == '.':                    
                    if dls(nr, nc, d - 1):
                        return True
            maze[r][c] = "."
            return False

        d = 0
        D = rows * cols
        while d < D:
            if dls(entrance[0], entrance[1], d):
                return d
            d += 1
        
        return -1

Nearest Exit from Entrance in Maze

–EOF (The Ultimate Computing & Technology Blog) —

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